CHAPTER 15 THERMODYNAMICS
CONCEPTUAL QUESTIONS
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1.
SSM REASONING AND SOLUTION The plunger of a bicycle tire pump is pushed
down rapidly with the end of the pump sealed so that no air escapes. Since the compression
occurs rapidly, there is no time for heat to flow into or out of the system. Therefore, to a
very good approximation, the process may be treated as an adiabatic compression that is
described by Equation 15.4:
W = (3 / 2)nR(Ti − Tf )
The person who pushes the plunger down does work on the system, therefore W is negative.
It follows that the term (Ti − Tf ) must also be negative. Thus, the final temperature Tf must
be greater than the initial temperature Ti. This increase in temperature is evidenced by the
fact that the pump becomes warm to the touch.
Alternate Explanation:
Since the compression occurs rapidly, there is no time for heat to flow into or out of the
system. Therefore, to a very good approximation, the process may be treated as an adiabatic
compression. According to the first law of thermodynamics, the change in the internal
energy is ∆ U = Q − W = −W , since Q = 0 for adiabatic processes. Since work is done on
the system, W is negative; therefore the change in the internal energy, ∆U, is positive. The
work done by the person pushing the plunger is manifested as an increase in the internal
energy of the air in the pump. The internal energy of an ideal gas is proportional to the
Kelvin temperature. Since the internal energy of the gas increases, the temperature of the
air in the pump must also increase. This increase in temperature is evidenced by the fact
that the pump becomes warm to the touch.
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2.
REASONING AND SOLUTION The work done in an isobaric process is given by
Equation 15.2: W = P(Vf − Vi ). According to the first law of thermodynamics, the change
in the internal energy is ∆U = Q − W = Q − P(V f − V i ) .
One hundred joules of heat is added to a gas, and the gas expands at constant pressure
(isobarically). Since the gas expands, the final volume will be greater than the initial
volume. Therefore, the term P(Vf − Vi ) will be positive. Since Q = +100 J, and the term
P(Vf − Vi ) is positive, the change in the internal energy must be less than 100 J. It is not
possible that the internal energy increases by 200 J.
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3.
REASONING AND SOLUTION The internal energy of an ideal gas is proportional to its
Kelvin temperature (see Equation 14.7). In an isothermal process the temperature remains
constant; therefore, the internal energy of an ideal gas remains constant throughout an
isothermal process. Thus, if a gas is compressed isothermally and its internal energy
increases, the gas is not an ideal gas.
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744 THERMODYNAMICS
4.
REASONING AND SOLUTION
According to the first law of thermodynamics
(Equation 15.1), the change in the internal energy is ∆U = Q – W. The process is isochoric,
which means that the volume is constant. Consequently, no work is done, so W = 0. The
process is also adiabatic, which means that no heat enters or leaves the system, so Q = 0.
According to the first law, then, ∆U = Q – W = 0. There is no change in the internal energy,
and the internal energy of the material at the end of the process is the same as it was at the
beginning.
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5.
REASONING AND SOLUTION
a. It is possible for the temperature of a substance to rise without heat flowing into
substance. Consider, for example, the adiabatic compression of an ideal gas. Since
process is an adiabatic process, Q = 0. The work done by the external agent increases
internal energy of the gas. Since the internal energy of an ideal gas is proportional to
Kelvin temperature, the temperature of the gas must increase.
the
the
the
the
b. The temperature of a substance does not necessarily have to change because heat flows
into or out of it. Consider, for example, the isothermal expansion of an ideal gas. Since the
internal energy of an ideal gas is proportional to the Kelvin temperature, the internal energy,
∆U, remains constant during an isothermal process. The first law of thermodynamics gives
∆U = Q − W = 0 , or Q = W. The heat that is added to the gas during the isothermal
expansion is used by the gas to perform the work involved in the expansion. The
temperature of the gas remains unchanged. Similarly, in an isothermal compression, the
work done on the gas as the gas is compressed causes heat to flow out of the gas while the
temperature of the gas remains constant.
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6.
REASONING AND SOLUTION The text drawing shows a pressure-volume graph in
which a gas undergoes a two-step process from A to B and from B to C.
From A to B: The volume V of the gas increases at constant pressure P. According to the
ideal gas law (Equation 14.1), PV = nRT , the temperature T of the gas must increase.
According to Equation 14.7, U = (3 / 2)nRT , if T increases, then ∆U , the change in the
internal energy, must be positive. Since the volume increases at constant pressure (∆V
increases), we know from Equation 15.2, W = P∆V , that the work done is positive. The
first law of thermodynamics (Equation 15.1) states that ∆U = Q − W ; since ∆ U and W are
both positive, Q must also be positive.
From B to C The pressure P of the gas increases at constant volume V. According to the
ideal gas law (Equation 14.1), PV = nRT , the temperature T of the gas must increase.
According to Equation 14.7, U = (3 / 2)nRT , if T increases, then ∆U , the change in the
internal energy, must be positive. Since the process occurs isochorically (∆V = 0), and
according to Equation 15.2, W = P∆V , the work done is zero. The first law of
thermodynamics (Equation 15.1) states that ∆U = Q − W ; since W = 0, Q is also positive
since ∆ U is positive.
Chapter 15 Conceptual Questions
745
These results are summarized in the table below:
∆U
Q
W
A→B
+
+
+
B→ C
+
+
0
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7.
REASONING AND SOLUTION Since the process is an adiabatic process, Q = 0. Since
the gas expands into chamber B under zero external pressure, the work done by the gas is
W = P∆V = 0. According to the first law of thermodynamics, the change in the internal
energy is, therefore, zero: ∆U = Q − W = 0 . The internal energy of an ideal gas is
proportional to the Kelvin temperature of the gas (Equation 14.7). Since the change in the
internal energy of the gas is zero, the temperature change of the gas is zero. The final
temperature of the gas is the same as the initial temperature of the gas.
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8.
SSM REASONING AND SOLUTION A material contracts when it is heated. To
determine the molar specific heat capacities, we first calculate the heat Q needed to raise the
temperature of the material by an amount ∆T. From the first law of thermodynamics,
Q = ∆ U + W . When the heating occurs at constant pressure, the work done is given by
Equation 15.2: W = P∆V = P(V f − V i ) . When the volume is constant, ∆V = 0. Therefore,
we have:
QP = ∆U + P (Vf − Vi )
and
QV = ∆U
Equation 15.6 indicates that the molar heat capacities will be given by C = Q / ( n ∆T ) .
Therefore
∆U + P (Vf − Vi )
∆U
CV =
CP =
and
n ∆T
n ∆T
Since the material contracts when it is heated, Vf is less than Vi. Therefore the term
P (Vf − Vi ) is negative. Hence, the numerator of CP is smaller than the numerator of CV.
Therefore, CV is larger than CP.
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9.
REASONING AND SOLUTION When a solid melts at constant pressure, the volume of
the resulting liquid does not differ much from the volume of the solid. According to the first
law of thermodynamics, ∆U = Q − W = Q − P(V f − V i ) ≈ Q . Hence, the heat that must be
added to melt the solid is used primarily to increase the internal energy of the molecules.
The internal energy of the liquid has increased by an amount Q = mLf compared to that of
the solid, where m is the mass of the material and Lf is the latent heat of fusion.
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746 THERMODYNAMICS
10. REASONING AND SOLUTION According to Equation 14.7, the Kelvin temperature T of
the gas is related to its internal energy U by U = (3 / 2)nRT . The change in the internal
energy is given by the first law of thermodynamics (Equation 15.1), ∆U = Q − W .
It is desired to heat a gas so that its temperature will be as high as possible. If the process
occurs at constant pressure, so that the volume of the gas increases, work is done by the gas.
The available heat is used to do work and to increase the internal energy of the gas. On the
other hand, if the process is carried out at constant volume, the work done is zero, and all of
the heat increases the internal energy of the gas. From Equation 14.7, the internal energy is
directly proportional to the Kelvin temperature of the gas. Since the internal energy
increases by a greater amount when the process occurs at constant volume, the temperature
increase is greatest under conditions of constant volume. Therefore, if it is desired to heat a
gas so that its temperature will be as high as possible, you should heat it under conditions of
constant volume.
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11. REASONING AND SOLUTION A hypothetical device takes 10 000 J of heat from a hot
reservoir and 5000 J of heat from a cold reservoir and produces 15 000 J of work.
a. According to the first law of thermodynamics, ∆U = Q − W . This is a statement of
energy conservation. The hypothetical device does not violate energy conservation. It does
not create or destroy energy. It converts one form of energy (15 000 J of heat) into another
form of energy (15 000 J of work) with no gain or loss.
b. This hypothetical device does violate the second law of thermodynamics. It converts all
of its input heat (15 000 J) into work (15 000 J). Therefore, the efficiency of this device is
1.0 or 100 %. But Equation 15.15 is a consequence of the second law of thermodynamics
and sets the limits of the maximum possible efficiency of any heat device. Since TC is
greater than 0 K, the ratio TC / TH must be positive. Furthermore, since TC < TH, the ratio
TC / TH must be less than one. Therefore, Equation 15.15 implies that the efficiency of any
device must be less than 1 or 100%. Since the efficiency of the hypothetical device is equal
to 100%, it violates the second law of thermodynamics.
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12. REASONING AND SOLUTION According to the second law of thermodynamics, heat
flows spontaneously from a substance at a higher temperature to a substance at a lower
temperature and does not flow spontaneously in the reverse direction. Therefore, according
to the second law of thermodynamics, work must be done to remove heat from a substance
at a lower temperature and deposit it in a substance at a higher temperature. In other words,
the second law requires that energy in the form of work must be supplied to an air
conditioner in order for it to remove heat from a cool space and deposit the heat in a warm
space. An advertisement for an automobile that claimed the same gas mileage with and
without the air conditioner would be suspect. Since the car would use more energy with the
air conditioner on, the car would use more gasoline. Therefore, the mileage should be less
with the air conditioner running.
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Chapter 15 Conceptual Questions
747
13. REASONING AND SOLUTION Carnot's principle states that the most efficient engine
operating between two temperatures is a reversible engine. This means that a reversible
engine operating between the temperatures of 600 and 400 K must be more efficient than an
irreversible engine operating between the same two temperatures. No comparison can be
made with an irreversible engine operating between two temperatures that are different than
600 and 400 K .
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14. REASONING AND SOLUTION
The efficiency of a Carnot engine is given by
Equation 15.15: efficiency = 1 − (TC / TH ) . Three reversible engines A, B, and C, use the
same cold reservoir for their exhaust heats. They use different hot reservoirs with the
following temperatures: (A) 1000 K; (B) 1100 K; and (C) 900 K. We can rank these
engines in order of increasing efficiency according to the following considerations. The
ratio TC / TH is inversely proportional to the value of TH. The ratio TC / TH will be smallest
for engine B; therefore, the quantity 1 – (TC / TH ) will be largest for engine B. Thus, engine
B has the largest efficiency. Similarly, the ratio TC / TH will be largest for engine C;
therefore, the quantity 1 – (TC / TH ) will be smallest for engine C. Thus, engine C has the
smallest efficiency. Hence, the engines are, in order of increasing efficiency: engine C,
engine A, and engine B.
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15. REASONING AND SOLUTION
The efficiency of a Carnot engine is given by
Equation 15.15: efficiency = 1 − (TC / TH ) .
a. Lowering the Kelvin temperature of the cold reservoir by a factor of four makes the ratio
TC / TH one-fourth as great.
b. Raising the Kelvin temperature of the hot reservoir by a factor of four makes the ratio
TC / TH one-fourth as great.
c. Cutting the Kelvin temperature of the cold reservoir in half and doubling the Kelvin
temperature of the hot reservoir makes the ratio TC / TH one-fourth as great.
Therefore, all three possible improvements have the same effect on the efficiency of a
Carnot engine.
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16. SSM REASONING AND SOLUTION A refrigerator is kept in a garage that is not
heated in the cold winter or air-conditioned in the hot summer. In order to make ice cubes,
the refrigerator uses electrical energy to provide the work to remove heat from the interior
of the freezer and deposit the heat outside of the refrigerator. In the summer, the "hot"
reservoir will be at a higher temperature than it is in the winter. Therefore, more work will
be required to remove heat from the interior of the freezer in the summer, and the
refrigerator will use more electrical energy. Hence, it will cost more for the refrigerator to
make a kilogram of ice cubes in the summer.
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748 THERMODYNAMICS
17. REASONING AND SOLUTION The coefficient of performance of a heat pump is given
by Equation 15.17: coefficient of performance = QH / W . From the conservation of
energy, QH = W + QC . Thus, the ratio QH / W can be written 1 + ( QC / W ) . The job of
a heat pump is to remove heat from a cold reservoir, and deliver it to a hot reservoir;
therefore, the ratio QC / W must be nonzero and positive. Hence, the coefficient of
performance, 1 + ( QC / W ) , must always be greater than one.
18. REASONING AND SOLUTION In a refrigerator, the interior of the unit is the cold
reservoir, while the warmer exterior of the room is the hot reservoir. An air conditioner is
like a refrigerator, except that the room being cooled is the cold reservoir, and the outdoor
environment is the hot reservoir. Therefore, an air conditioner cools the inside of the house,
while a refrigerator warms the interior of the house.
19. REASONING AND SOLUTION Heat pumps can deliver more energy into your house
than they consume in operating. A heat pump consumes an amount of energy W , which it
uses to make an amount of heat QC flow from the cold outdoors into the warm house. The
amount of energy the heat pump delivers to the house is QH = W + QC . This is greater
than the amount of energy, W , consumed by the heat pump.
20. REASONING AND SOLUTION A refrigerator is advertised as being easier to "live with"
during the summer, because it puts into your kitchen only the heat that it removes from the
food. The advertisement is describing a refrigerator in which heat is removed from the
interior of the refrigerator and deposited outside the refrigerator without requiring any work.
Since no work is required, the flow must be spontaneous. This violates the second law of
thermodynamics, which states that heat spontaneously flows from a higher-temperature
substance to a lower-temperature substance, and does not flow spontaneously in the reverse
direction. Heat can be made to flow from a cold reservoir to a hot reservoir, but only when
work is done. Both the heat and the work are deposited in the hot reservoir.
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21. REASONING AND SOLUTION On a summer day, a window air conditioner cycles on
and off, according to how the temperature within the room changes. When the unit is on it
will be depositing heat, along with the work required to remove the heat, to the outside.
Therefore, the outside of the unit will be hotter when the unit is on. Hence, you would be
more likely to fry an egg on the outside part of the unit when the unit is on.
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22. REASONING AND SOLUTION The second law of thermodynamics states that the total
entropy of the universe does not change when a reversible process occurs ( ∆S universe = 0)
and increases when an irreversible process occurs ( ∆S universe > 0).
Chapter 15 Conceptual Questions
749
An event happens somewhere in the universe and, as a result, the entropy of an object
changes by –5 J/K. If the event is a reversible process, then the entropy change for the rest
of the universe must be +5 J/K; this results in a total entropy change of zero for the universe.
If the process is irreversible, the only possible choice for the change in the entropy of the
rest of the universe is +10 J/K; this results in a total entropy change of +5 J/K for the
universe. The choices –5 J/K and 0 J/K are not possible choices for the entropy change of
the rest of the universe, because they imply that the total entropy change would be negative.
This would violate the second law of thermodynamics.
23. REASONING AND SOLUTION When water freezes from a less-ordered liquid to a moreordered solid, its entropy decreases. This decrease in entropy does not violate the second
law of thermodynamics, because it is a decrease for only one part of the universe. In terms
of entropy, the second law indicates that the total change in entropy for the entire universe
must be either zero (reversible process) or greater than zero (irreversible process). In the
case of freezing water, heat must be removed from the water and deposited in the
environment. The entropy of the environment increases as a result. If the freezing occurs
reversibly, the increase in entropy of the environment will exactly match the decrease in
entropy of the water, with the result that ∆Suniverse = ∆S water + ∆S environment = 0 . If the
freezing occurs irreversibly, then the increase in entropy of the environment will exceed the
decrease in entropy of the water, with the result that ∆Suniverse = ∆S water + ∆S environment > 0.
24. REASONING AND SOLUTION Since we can interpret the increase of entropy as an
increase in disorder, the more disordered system will have the greater entropy.
a. The popcorn that results from the kernels is more disorderly than a handful of popcorn
kernels; therefore, the popcorn that results from the kernels has the greater entropy.
b. A salad has more disorder after it has been tossed; therefore, the tossed salad has the
greater entropy.
c. A messy apartment is more disorderly than a neat apartment; therefore, a messy
apartment has the greater entropy.
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25. REASONING AND SOLUTION A glass of water contains a teaspoon of dissolved sugar.
After a while, the water evaporates, leaving behind sugar crystals. The entropy of the sugar
crystals is less than the entropy of the dissolved sugar, because the sugar crystals are in a
more ordered state. However, think about the water. The entropy of the gaseous water
vapor is greater than the entropy of the liquid water, because the molecules in the vapor are
in a less ordered state. Since the increase in the entropy of the water is greater than the
decrease in entropy of the sugar, the net change in entropy of the universe is positive. The
process, therefore, does not violate the entropy version of the second law of
thermodynamics.
750 THERMODYNAMICS
26. REASONING AND SOLUTION Since we can interpret the increase of entropy as an
increase in disorder, the more disordered state will have the greater entropy. The finished
building is the most ordered state; therefore it has the smallest entropy. The burned-out
shell of a building is the most disordered state; therefore it has the largest amount of
entropy. The states can be ranked in order of decreasing entropy (largest first) as follows:
(3) the burned-out shell of a building, (1) the unused building material, and (2) the building.
Chapter 15 Problems
751
CHAPTER 15 THERMODYNAMICS
PROBLEMS
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1.
SSM REASONING Energy in the form of work leaves the system, while energy in the
form of heat enters. More energy leaves than enters, so we expect the internal energy of the
system to decrease, that is, we expect the change ∆U in the internal energy to be negative.
The first law of thermodynamics will confirm our expectation. As far as the environment is
concerned, we note that when the system loses energy, the environment gains it, and when
the system gains energy the environment loses it. Therefore, the change in the internal
energy of the environment must be opposite to that of the system.
SOLUTION
a. The system gains heat so Q is positive, according to our convention. The system does
work, so W is also positive, according to our convention. Applying the first law of
thermodynamics from Equation 15.1, we find for the system that
b g b
g
∆U = Q − W = 77 J − 164 J = −87 J
As expected, this value is negative, indicating a decrease.
b. The change in the internal energy of the environment is opposite to that of the system, so
that ∆U environment = +87 J .
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2.
REASONING The change ∆U in the individual’s internal energy is given by the first law of
thermodynamics as ∆U = Q − W (Equation 15.1), where Q is the heat and W is the work.
Note that Q is negative since the individual gives off heat and that W is positive since the
individual does work. We will apply the first law twice, once to the individual walking and
again to the individual jogging. By taking advantage of the fact that ∆U is the same in each
case, we will be able to obtain the heat given off during the walking.
SOLUTION Applying the first law of thermodynamics to walking and jogging gives
∆U walking = Qwalking − Wwalking
and
∆U jogging = Q jogging − Wjogging
Since ∆U walking = ∆U jogging , we have
Qwalking − Wwalking = Q jogging − Wjogging
(
) (
) (
)
Qwalking = Wwalking + Qjogging − W jogging = 8.2 × 105 J + −4.9 × 105 J − 6.4 × 105 J = −3.1× 105 J
The value for Qwalking is negative, indicating that the individual gives off heat. The
magnitude of this heat is 3.1×105 J .
752 THERMODYNAMICS
3.
REASONING Since the student does work, W is positive, according to our convention.
Since his internal energy decreases, the change ∆U in the internal energy is negative. The
first law of thermodynamics will allow us to determine the heat Q.
SOLUTION
a. The work is W = +1.6 × 104 J .
b. The change in internal energy is ∆U = –4.2 × 104 J .
c. Applying the first law of thermodynamics from Equation 15.1, we find that
c
h c
h
Q = ∆U + W = −4.2 × 10 4 J + 1.6 × 10 4 J = −2 .6 × 10 4 J
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4.
REASONING According to the discussion in Section 14.3, the internal energy U of a
monatomic ideal gas is given by U = 32 nRT (Equation 14.7), where n is the number of
moles, R is the universal gas constant, and T is the Kelvin temperature. When the
temperature changes to a final value of Tf from an initial value of Ti, the internal energy
changes by an amount
U f − U i = 32 nR (Tf − Ti )
∆U
⎛ 2 ⎞
Solving this equation for the final temperature yields Tf = ⎜
⎟ ∆U + Ti . We are given n
⎝ 3nR ⎠
and Ti , but must determine ∆U. The change ∆U in the internal energy of the gas is related to
the heat Q and the work W by the first law of thermodynamics, ∆U = Q − W
(Equation 15.1). Using these two relations will allow us to find the final temperature of the
gas.
SOLUTION Substituting ∆U = Q − W into the expression for the final temperature gives
⎛ 2 ⎞
Tf = ⎜
⎟ ( Q − W ) + Ti
⎝ 3nR ⎠
⎧⎪
⎫⎪
2
=⎨
⎬ ⎡⎣ +2438 J − ( −962 J ) ⎤⎦ + 345 K = 436 K
⎪⎩ 3 ( 3.00 mol ) ⎡⎣8.31 J/ ( mol ⋅ K ) ⎤⎦ ⎪⎭
Note that the heat is positive (Q = +2438 J) since the system (the gas) gains heat, and the
work is negative (W = −962 J), since it is done on the system.
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Chapter 15 Problems
5.
753
SSM REASONING Since the change in the internal energy and the heat released in the
process are given, the first law of thermodynamics (Equation 15.1) can be used to find the
work done. Since we are told how much work is required to make the car go one mile, we
can determine how far the car can travel. When the gasoline burns, its internal energy
decreases and heat flows into the surroundings; therefore, both ∆U and Q are negative.
SOLUTION According to the first law of thermodynamics, the work that is done when one
gallon of gasoline is burned in the engine is
W = Q − ∆U = −1.00 × 10 8 J – (–1.19 × 10 8 J) = 0.19 × 10 8 J
Since 6.0 × 10 5 J of work is required to make the car go one mile, the car can travel
0.19 × 10 8 J
FG 1 mile IJ =
H 6.0 × 10 J K
5
32 miles
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6.
REASONING AND SOLUTION
a. For the weight lifter
∆U = Q − W
= −mLv − W = −(0.150 kg)(2.42 × 106 J/kg) − 1.40 × 105 J = −5.03 × 105 J
b. Since 1 nutritional calorie = 4186 J, the number of nutritional calories is
Calorie ⎞
(5.03 × 105 J ) ⎛⎜⎝ 14186
⎟=
J ⎠
1.20 × 102 nutritional calories
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7.
REASONING We will apply the first law of thermodynamics as given in Equation 15.1
(∆U = Q – W) to the overall process. First, however, we add the changes in the internal
energy to obtain the overall change ∆U and add the work values to get the overall work W.
SOLUTION
In both steps the internal energy increases, so overall we have
∆U = 228 J + 115 J = +343 J. In both steps the work is negative according to our
convention, since it is done on the system.
Overall, then, we have
W = –166 J – 177 J = –343 J. Using the first law of thermodynamics from Equation 15.1,
we find
∆U = Q − W or Q = ∆U + W = +343 J + −343 J = 0 J
b
g b
g
Since the heat is zero, the overall process is adiabatic .
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754 THERMODYNAMICS
8.
REASONING When a gas expands under isobaric conditions, its pressure remains
constant. The work W done by the expanding gas is W = P (Vf − Vi), Equation 15.2, where P
is the pressure and Vf and Vi are the final and initial volumes. Since all the variables in this
relation are known, we can solve for the final volume.
SOLUTION Solving W = P (Vf − Vi) for the final volume gives
W
480 J
+ Vi =
+ 1.5 × 10−3 m3 = 4.5 × 10−3 m3
5
P
1.6 × 10 Pa
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Vf =
9.
SSM REASONING According to Equation 15.2, W = P∆V, the average pressure P of
the expanding gas is equal to P = W / ∆V , where the work W done by the gas on the bullet
can be found from the work-energy theorem (Equation 6.3). Assuming that the barrel of the
gun is cylindrical with radius r, the volume of the barrel is equal to its length L multiplied
by the area (π r2) of its cross section. Thus, the change in volume of the expanding gas is
∆V = Lπ r 2 .
SOLUTION The work done by the gas on the bullet is given by Equation 6.3 as
2
2
W = 12 m(vfinal
− vinitial
) = 12 (2.6 × 10−3 kg)[(370 m/s)2 − 0] = 180 J
The average pressure of the expanding gas is, therefore,
W
180 J
=
= 1.2 × 107 Pa
∆V (0.61 m)π (2.8 × 10−3 m) 2
______________________________________________________________________________
P=
10. REASONING For segment AB, there is no work, since the volume is constant. For
segment BC the process is isobaric and Equation 15.2 applies. For segment CA, the work
can be obtained as the area under the line CA in the graph.
SOLUTION
a. For segment AB, the process is isochoric, that is, the volume is constant. For a process in
which the volume is constant, no work is done, so W = 0 J .
b. For segment BC, the process is isobaric, that is, the pressure is constant. Here, the
volume is increasing, so the gas is expanding against the outside environment. As a result,
the gas does work, which is positive according to our convention. Using Equation 15.2 and
the data in the drawing, we obtain
Chapter 15 Problems
c
755
h
W = P V f − Vi
c
= 7 .0 × 10 5 Pa
h c5.0 × 10
−3
h c
m 3 − 2 .0 × 10 −3 m 3
h=
+2 .1 × 10 3 J
c. For segment CA, the volume of the gas is decreasing, so the gas is being compressed and
work is being done on it. Therefore, the work is negative, according to our convention. The
magnitude of the work is the area under the segment CA. We estimate that this area is 15 of
the squares in the graphical grid. The area of each square is
(1.0 × 105 Pa)(1.0 × 10–3 m3) = 1.0 × 102 J
The work, then, is
W = – 15 (1.0 × 102 J) = −1.5 × 10 3 J
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11. REASONING AND SOLUTION
a. Starting at point A, the work done during the first (vertical) straight-line segment is
W1 = P1∆V1 = P1(0 m3) = 0 J
For the second (horizontal) straight-line segment, the work is
W2 = P2∆V2 = 10(1.0 × 104 Pa)6(2.0 × 10–3 m3) = 1200 J
For the third (vertical) straight-line segment the work is
W3 = P3∆V3 = P3(0 m3) = 0 J
For the fourth (horizontal) straight-line segment the work is
W4 = P4∆V4 = 15(1.0 × 104 Pa)6(2.0 × 10–3 m3) = 1800 J
The total work done is
W = W1 + W2 + W3 + W4 = +3.0 × 103 J
b. Since the total work is positive, work is done by the system .
______________________________________________________________________________
12. REASONING The pressure P of the gas remains constant while its volume increases by an
amount ∆V. Therefore, the work W done by the expanding gas is given by W = P∆V
(Equation 15.2). ∆V is known, so if we can obtain a value for W, we can use this expression
to calculate the pressure. To determine W, we turn to the first law of thermodynamics
∆U = Q − W (Equation 15.1), where Q is the heat and ∆U is the change in the internal
energy. ∆U is given, so to use the first law to determine W we need information about Q.
According to Equation 12.4, the heat needed to raise the temperature of a mass m of material
by an amount ∆T is Q = c m ∆T where c is the material’s specific heat capacity.
756 THERMODYNAMICS
SOLUTION According to Equation 15.2, the pressure P of the expanding gas can be
determined from the work W and the change ∆V in volume of the gas according to
P=
W
∆V
Using the first law of thermodynamics, we can write the work as W = Q − ∆U
(Equation 15.1). With this substitution, the expression for the pressure becomes
P=
W
Q − ∆U
=
∆V
∆V
(1)
Using Equation 12.4, we can write the heat as Q = c m ∆T , which can then be substituted
into Equation (1). Thus,
P=
Q − ∆U c m ∆T − ∆U
=
∆V
∆V
(
)
⎡⎣1080 J/ ( kg ⋅ C° ) ⎤⎦ 24.0 ×10−3 kg ( 53.0 C° ) − 939 J
=
= 3.1× 105 Pa
−3
3
1.40 × 10 m
13. SSM REASONING The work done in an isobaric process is given by Equation 15.2,
W = P∆V ; therefore, the pressure is equal to P = W / ∆V . In order to use this expression,
we must first determine a numerical value for the work done; this can be calculated using
the first law of thermodynamics (Equation 15.1), ∆U = Q − W .
SOLUTION Solving Equation 15.1 for the work W, we find
W = Q − ∆U = 1500 J–(+4500 J) = − 3.0 × 103 J
Therefore, the pressure is
P=
W
−3.0 × 103 J
=
= 3.0 × 105 Pa
3
∆V −0.010 m
The change in volume ∆V, which is the final volume minus the initial volume, is negative
because the final volume is 0.010 m3 less than the initial volume.
______________________________________________________________________________
14. REASONING AND SOLUTION According to the first law of thermodynamics, the change
in internal energy is ∆U = Q – W. The work can be obtained from the area under the graph.
There are sixty squares of area under the graph, so the positive work of expansion is
Chapter 15 Problems
c
hc
757
h
W = 60 1.0 × 10 4 Pa 2.0 × 10 –3 m 3 = 1200 J
Since Q = 2700 J, the change in internal energy is
∆U = Q – W = 2700 J – 1200 J = 1500 J
______________________________________________________________________________
15. SSM WWW REASONING AND SOLUTION The first law of thermodynamics states
that ∆U = Q − W . The work W involved in an isobaric process is, according to Equation
15.2, W = P∆V . Combining these two expressions leads to ∆U = Q − P∆V . Solving for Q
gives
Q = ∆U + P∆V
(1)
Since this is an expansion, ∆V > 0 , so P∆V > 0 . From the ideal gas law, PV = nRT , we
have P ∆V = nR ∆T . Since P∆V > 0 , it follows that nR ∆T > 0 . The internal energy of an
ideal gas is directly proportional to its Kelvin temperature T. Therefore, since nR ∆T > 0 , it
follows that ∆U > 0 . Since both terms on the right hand side of Equation (1) are positive,
the left hand side of Equation (1) must also be positive. Thus, Q is positive. By the
convention described in the text, this means that
heat can only flow into an ideal gas during an isobaric expansion
______________________________________________________________________________
16. REASONING AND SOLUTION The rod's volume increases by an amount ∆V = βV0∆T,
according to Equation 12.3. The work done by the expanding aluminum is, from
Equation 15.2,
W = P∆V = PβV0∆T = (1.01 × 105 Pa)(69 × 10–6/C°)(1.4 × 10–3 m3)(3.0 × 102 C°) = 2.9 J
______________________________________________________________________________
17. REASONING AND SOLUTION Since the pan is open, the process takes place at constant
(atmospheric) pressure P0. The work involved in an isobaric process is given by Equation
15.2: W = P0∆V. The change in volume of the liquid as it is heated is given according to
Equation 12.3 as ∆V = βV0∆T, where β is the coefficient of volume expansion. Table 12.1
b g
gives β = 207 × 10 –6 C °
–1
for water.
The heat absorbed by the water is given by
Equation 12.4 as Q = cm∆T, where c = 4186 J/ ( kg ⋅ C° ) is the specific heat capacity of
liquid water according to Table 12.2. Therefore,
P0 β
Pβ
W P0 ∆V P0 β V0 ∆T
=
=
=
= 0
Q cm∆T
cm∆T
c(m / V0 ) c ρ
758 THERMODYNAMICS
where ρ = 1.00 × 10 3 kg / m 3 is the density of the water (see Table 11.1). Thus, we find
(
)(
)
1.01 × 105 Pa 207 × 10 –6 C° –1
W P0 β
=
=
= 4.99 × 10 –6
3
3
Q c ρ ⎡ 4186 J/ ( kg ⋅ C° ) ⎤ 1.00 × 10 kg/m
⎣
⎦
______________________________________________________________________________
(
)
18. REASONING We can use the first law of thermodynamics, ∆U = Q − W (Equation 15.1) to
find the work W. The heat is Q = −4700 J, where the minus sign denotes that the system (the
gas) loses heat. The internal energy U of a monatomic ideal gas is given by
U = 32 nRT (Equation 14.7), where n is the number of moles, R is the universal gas constant,
and T is the Kelvin temperature. If the temperature remains constant during the process, the
internal energy does not change, so ∆U = 0 J.
SOLUTION The work done during the isothermal process is
W = Q − ∆U = − 4700 J + 0 J = −4700 J
The negative sign indicates that work is done on the system.
______________________________________________________________________________
19. SSM REASONING According to the first law of thermodynamics (Equation 15.1),
∆U = Q − W . For a monatomic ideal gas (Equation 14.7), U = 23 nRT . Therefore, for the
process in question, the change in the internal energy is ∆U = 23 nR∆T . Combining the last
expression for ∆U with Equation 15.1 yields
3
2
nR∆T = Q − W
This expression can be solved for ∆T .
SOLUTION
a. The heat is Q = +1200 J , since it is absorbed by the system. The work is W = +2500 J ,
since it is done by the system. Solving the above expression for ∆T and substituting the
values for the data given in the problem statement, we have
∆T =
Q −W
1200 J − 2500 J
= 3
= –2.1 × 10 2 K
3
nR
⋅
(0.50
mol)[8.31
J
/
(mol
K)]
2
2
b. Since ∆T = Tfinal − Tinitial is negative, Tinitial must be greater than Tfinal ; this change
represents a decrease
in temperature.
Chapter 15 Problems
759
Alternatively, one could deduce that the temperature decreases from the following physical
argument. Since the system loses more energy in doing work than it gains in the form of
heat, the internal energy of the system decreases. Since the internal energy of an ideal gas
depends only on the temperature, a decrease in the internal energy must correspond to a
decrease in the temperature.
______________________________________________________________________________
20. REASONING Since the gas is expanding adiabatically, the work done is given by
Equation 15.4 as W = 32 nR (Ti − Tf ) . Once the work is known, we can use the first law of
thermodynamics to find the change in the internal energy of the gas.
SOLUTION
a. The work done by the expanding gas is
W = 32 nR (Ti − Tf ) =
3
2
( 5.0 mol ) ⎡⎣8.31 J/ ( mol ⋅ K ) ⎤⎦ ( 370 K − 290 K ) =
+5.0 × 103 J
b. Since the process is adiabatic, Q = 0, and the change in the internal energy is
∆U = Q − W = 0 − 5.0 × 103 J = − 5.0 × 103 J
______________________________________________________________________________
21. REASONING During an adiabatic process, no heat flows into or out of the gas (Q = 0 J).
For an ideas gas, the final pressure and volume (Pf and Vf) are related to the initial pressure
γ
γ
and volume (Pi and Vi) by PV
i i = Pf Vf
( Equation 15.5) ,
where γ is the ratio of the specific
heat capacities at constant pressure and constant volume (γ =
5
3
in this problem). We will
use this relation to find Vf /Vi.
γ
γ
SOLUTION Solving PV
for Vf /Vi and noting that the pressure doubles
i i = Pf Vf
(Pf /Pi = 2.0) during the compression, we have
1
1
Vf ⎛ Pi ⎞ γ ⎛ 1 ⎞ ( 5 / 3)
=⎜ ⎟ =⎜
= 0.66
⎟
Vi ⎜⎝ Pf ⎟⎠
⎝ 2.0 ⎠
______________________________________________________________________________
22. REASONING According to the first law of thermodynamics ∆U = Q − W (Equation 15.1),
where ∆U is the change in the internal energy, Q is the heat, and W is the work. This
expression may be solved for the heat. ∆U can be evaluated by remembering that the
3
2
internal energy of a monatomic ideal gas is U = nRT (Equation 14.7), where n is the
number of moles, R = 8.31 J/(mol⋅K) is the universal gas constant, and T is the Kelvin
temperature. Since heat is being added isothermally, the temperature remains constant and
760 THERMODYNAMICS
so does the internal energy of the gas. Therefore, ∆U = 0 J. To evaluate W we use
⎛V ⎞
W = nRT ln ⎜⎜ f ⎟⎟ (Equation 15.3), where Vf and Vi are the final and initial volumes of the
⎝ Vi ⎠
gas, respectively.
SOLUTION According to the first law of thermodynamics, as given in Equation 15.1, the
heat added to the gas is
Q = ∆U + W
Using the fact that ∆U = 0 J for an ideal gas undergoing an isothermal process and the fact
⎛V ⎞
that W = nRT ln ⎜⎜ f ⎟⎟ (Equation 15.3), we can rewrite the expression for the heat as
⎝ Vi ⎠
follows:
⎛V ⎞
Q = ∆U + W = W = nRT ln ⎜⎜ f ⎟⎟
⎝ Vi ⎠
Since the volume of the gas doubles, we know that Vf = 2 Vi. Thus, it follows that
⎛V
Q = nRT ln ⎜ f
⎜V
⎝ i
⎛ 2 Vi
⎞
⎟⎟ = ( 2.5 mol ) ⎣⎡8.31 J/ ( mol ⋅ K ) ⎦⎤ ( 430 K ) ln ⎜
⎜ Vi
⎠
⎝
⎞
⎟ = 6200 J
⎟
⎠
23. SSM REASONING When the expansion is isothermal, the work done can be calculated
from Equation (15.3): W = nRT ln (Vf / Vi ) . When the expansion is adiabatic, the work done
can be calculated from Equation 15.4: W = 32 nR(Ti − Tf ) .
Since the gas does the same amount of work whether it expands adiabatically or
isothermally, we can equate the right hand sides of these two equations. We also note that
since the initial temperature is the same for both cases, the temperature T in the isothermal
expansion is the same as the initial temperature Ti for the adiabatic expansion. We then
have
⎛ Vf ⎞ 3
⎛ Vf ⎞ 32 (Ti − Tf )
nRTi ln ⎜⎜ ⎟⎟ = nR(Ti − Tf ) or ln ⎜⎜ ⎟⎟ =
2
Ti
⎝ Vi ⎠
⎝ Vi ⎠
SOLUTION Solving for the ratio of the volumes gives
3 (T − T ) / T
Vf
3
= e 2 i f i = e 2 (405 K − 245 K)/(405 K) = 1.81
Vi
______________________________________________________________________________
Chapter 15 Problems
761
Pressure (×105 Pa)
24. REASONING
a. The work done by the gas is equal to the area under the pressure-versus-volume curve.
We will measure this area by using the graph given with the problem.
6.00
4.00
0
B
A
2.00
2.00
0
4.00
6.00
10.0
8.00
12.0
Volume, m3
b. Since the gas is an ideal gas, it obeys the ideal gas law, PV = nRT (Equation 14.1). This
implies that PAVA / TA = PBVB / TB . All the variables except for TB in this relation are
known. Therefore, we can use this expression to find the temperature at point B.
c. The heat Q that has been added to or removed from the gas can be obtained from the first
law of thermodynamics, Q = ∆U + W (Equation 15.1), where ∆U is the change in the
internal energy of the gas and W is the work done by the gas. The work W is known from
part (a) of the problem. The change ∆U in the internal energy of the gas can be obtained
from Equation 14.7, ∆U = U B − U A = 32 nR (TB − TA ) , where n is the number of moles, R is
the universal gas constant, and TB and TA are the final and initial Kelvin temperatures. We
do not know n, but we can use the ideal gas law (PV = nRT ) to replace nRTB by PBVB and
to replace nRTA by PAVA.
SOLUTION
a. From the drawing we see that the area under the curve is 5.00 “squares,” where each
square has an area of 2.00 × 105 Pa 2.00 m3 = 4.00 ×105 J. Therefore, the work W done
(
by the gas is
)(
(
)
)
W = ( 5.00 squares ) 4.00 × 105 J/square = 2.00 × 106 J
b. In the Reasoning section, we have seen that PAVA / TA = PBVB / TB . Solving this relation
for the temperature TB at point B, using the fact that PA = PB (see the graph), and taking the
values for VB and VA from the graph, we have that
⎛PV
TB = ⎜⎜ B B
⎝ PAVA
⎞
⎛ VB ⎞
⎛ 10.0 m3 ⎞
⎟⎟ TA = ⎜⎜
⎟⎟ TA = ⎜⎜
3⎟
⎟ (185 K ) = 925 K
⎝ 2.00 m ⎠
⎠
⎝ VA ⎠
762 THERMODYNAMICS
c. From the Reasoning section we know that the heat Q that has been added to or removed
from the gas is given by Q = ∆U + W. The change ∆U in the internal energy of the gas is
∆U = U B − U A = 32 nR (TB − TA ) . Thus, the heat can be expressed as
Q = ∆U + W = 32 nR (TB − TA ) + W
We now use the ideal gas law (PV = nRT ) to replace nRTB by PBVB and nRTA by PAVA.
The result is
Q = 32 ( PBVB − PAVA ) + W
Taking the values for PB, VB, PA, and VA from the graph and using the result from part a
that W = 2.00 × 106 J, we find that the heat is
Q=
3
2
( PBVB − PAVA ) + W
(
)(
) (
)(
)
= 32 ⎡ 2.00 ×105 Pa 10.0 m3 − 2.00 × 105 Pa 2.00 m3 ⎤ + 2.00 × 106 J = 4.40 × 106 J
⎣
⎦
______________________________________________________________________________
25. REASONING AND SOLUTION
a. Since the curved line between A and C is an isotherm, the initial and final temperatures
are the same. Since the internal energy of an ideal monatomic gas is U = (3/2)nRT, the
initial and final energies are also the same, and the change in the internal energy is ∆U = 0.
The first law of thermodynamics, then, indicates that for the process A→B→C, we have
∆U = 0 = Q − W
or
Q =W
The heat is equal to the work. Determining the work from the area beneath the straight line
segments AB and BC, we find that
(
)
Q = W = – ( 4.00 × 105 Pa ) 0.400 m3 – 0.200 m3 = – 8.00 × 104 J
b. The minus sign is included because the gas is compressed, so that work is done on the
gas. Since the answer for Q is negative, we conclude that heat flows out of the gas .
______________________________________________________________________________
26. REASONING AND SOLUTION
Step A → B
The internal energy of a monatomic ideal gas is U = (3/2)nRT. Thus, the change is
∆U = 32 nR ∆T =
3
2
(1.00 mol ) ⎡⎣8.31 J/ ( mol ⋅ K )⎤⎦ (800.0 K – 400.0 K ) =
4990 J
The work for this constant pressure step is W = P∆V. But the ideal gas law applies, so
W = P∆V = nR ∆T = (1.00 mol ) ⎡⎣8.31 J/ ( mol ⋅ K ) ⎤⎦ ( 800.0 K – 400.0 K ) = 3320 J
Chapter 15 Problems
763
The first law of thermodynamics indicates that the heat is
Q = ∆U + W = 32 nR ∆T + nR ∆T
=
5
2
(1.00 mol ) ⎡⎣8.31 J/ ( mol ⋅ K )⎤⎦ (800.0 K – 400.0 K ) =
8310 J
Step B → C
The internal energy of a monatomic ideal gas is U = (3/2)nRT. Thus, the change is
∆U =
3 nR ∆T
2
=
3
2
(1.00 mol ) ⎡⎣8.31 J/ ( mol ⋅ K )⎤⎦ ( 400.0 K – 800.0 K ) =
– 4990 J
The volume is constant in this step, so the work done by the gas is W = 0 J .
The first law of thermodynamics indicates that the heat is
Q = ∆U + W = ∆U = – 4990 J
Step C → D
The internal energy of a monatomic ideal gas is U = (3/2)nRT. Thus, the change is
∆U =
3 nR ∆T
2
=
3
2
(1.00 mol ) ⎡⎣8.31 J/ ( mol ⋅ K )⎤⎦ ( 200.0 K – 400.0 K ) =
– 2490 J
The work for this constant pressure step is W = P∆V. But the ideal gas law applies, so
W = P∆V = nR ∆T = (1.00 mol ) ⎡⎣8.31 J/ ( mol ⋅ K ) ⎤⎦ ( 200.0 K – 400.0 K ) = – 1660 J
The first law of thermodynamics indicates that the heat is
Q = ∆U + W = 32 nR ∆T + nR ∆T
=
5
2
(1.00 mol ) ⎡⎣8.31 J/ ( mol ⋅ K )⎤⎦ ( 200.0 K – 400.0 K ) =
– 4150 J
Step D → A
The internal energy of a monatomic ideal gas is U = (3/2)nRT. Thus, the change is
∆U =
3 nR ∆T
2
=
3
2
(1.00 mol ) ⎡⎣8.31 J/ ( mol ⋅ K )⎤⎦ ( 400.0 K – 200.0 K ) =
2490 J
The volume is constant in this step, so the work done by the gas is W = 0 J
The first law of thermodynamics indicates that the heat is
Q = ∆U + W = ∆U = 2490 J
______________________________________________________________________________
764 THERMODYNAMICS
γ
γ
27. SSM REASONING
According to Equation 15.5, PV
i i = Pf Vf . The ideal gas law
states that V = nRT / P for both the initial and final conditions. Thus, we have
γ
⎛ nRTi ⎞
⎛ nRTf
Pi ⎜⎜
⎟⎟ = Pf ⎜⎜
⎝ Pi ⎠
⎝ Pf
γ
⎞
⎟⎟
⎠
⎛T
= ⎜⎜ i
Pi ⎝ Tf
Pf
or
γ / (1–γ )
⎞
⎟⎟
⎠
Since the ratio of the temperatures is known, the last expression can be solved for the final
pressure Pf .
SOLUTION Since Ti / Tf = 1/ 2 , and γ = 5 / 3 , we find that
γ / (1–γ )
( 5/3)/ ⎣1–( 5/3)⎦
⎛ Ti ⎞
⎛1⎞
5
Pf = Pi ⎜⎜ ⎟⎟
= 1.50 × 10 Pa ⎜ ⎟
= 8.49 × 105 Pa
T
2
⎝ ⎠
⎝ f⎠
______________________________________________________________________________
(
⎡
)
28. REASONING AND SOLUTION The
three-step process is shown on the P-V
diagram at the right. From the first law
of thermodynamics,
Q = ∆U + W
⎤
P
1
(1)
3
However, the ideal gas is back in its
initial state at the end of the three-step
process, so that ∆U = 0 overall. With this
value for ∆U, Equation (1) becomes
Q = W, and we conclude that
2
3V0
V0
V
Q = W1→2 + W2→3 + W3→1
N N N
isothermal
isobaric
isochoric
Using Equations 15.3 for the isothermal work and Equation 15.2 for the isobaric work, and
remembering that there is no work done in an isochoric process, we find that
⎛ 3V ⎞
Q = nRT ln ⎜ 0 ⎟ + P (V0 – 3V0 ) = nRT ln 3 – 2 PV0
⎜V ⎟
⎝ 0 ⎠
In this result, T = 438 K, P is the pressure for step 2 → 3 , and 2PV0 = 2(P3V0)/3 = 2nRT/3.
In addition, we know that n = 1 mol. Therefore,
(
Q = nRT ln 3 – 2 PV0 = nRT ln 3 − 23 nRT = nRT ln 3 – 23
(
)
)
= (1 mol ) ⎡⎣8.31 J ( mol ⋅ K ) ⎤⎦ ( 438 K ) ln 3 – 23 = 1.57 × 103 J
Since this answer is positive, heat is absorbed by the gas.
______________________________________________________________________________
Chapter 15 Problems
765
29. SSM REASONING AND SOLUTION
a. The final temperature of the adiabatic process is given by solving Equation 15.4 for Tf.
W
Tf = Ti −
3 nR
2
= 393 K –
3 (1.00
2
825 J
= 327 K
mol) [8.31 J/(mol ⋅ K) ]
γ
γ
b. According to Equation 15.5 for the adiabatic expansion of an ideal gas, PV
i i = Pf Vf .
Therefore,
⎛P⎞
Vfγ = Viγ ⎜⎜ i ⎟⎟
⎝ Pf ⎠
From the ideal gas law, PV = nRT ; therefore, the ratio of the pressures is given by
Pi ⎛ Ti ⎞ ⎛ Vf
= ⎜ ⎟⎜
Pf ⎜⎝ Tf ⎟⎠ ⎜⎝ Vi
⎞
⎟⎟
⎠
Combining the previous two equations gives
⎛T
Vfγ = Viγ ⎜⎜ i
⎝ Tf
⎞ ⎛ Vf
⎟⎟ ⎜⎜
⎠ ⎝ Vi
⎞
⎟⎟
⎠
Solving for Vf we obtain
Vfγ ⎛ Ti
=⎜
Vf ⎜⎝ Tf
⎞ ⎛ Viγ
⎟⎟ ⎜⎜
⎠ ⎝ Vi
⎡⎛ T
Vf = ⎢⎜⎜ i
⎢⎣⎝ Tf
⎞
⎟
⎟
⎠
or
1/ ( γ −1)
⎞ (γ −1) ⎤
⎥
⎟⎟ Vi
⎥⎦
⎠
Vf(
γ −1)
⎛T
= Vi ⎜⎜ i
⎝ Tf
⎛T
= ⎜⎜ i
⎝ Tf
⎞ (γ −1)
⎟⎟ Vi
⎠
1/ ( γ −1)
⎞
⎟⎟
⎠
Therefore,
1/(γ −1)
1/(2 / 3)
3/ 2
⎛ Ti ⎞
3 ⎛ 393 K ⎞
3 ⎛ 393 K ⎞
3
= (0.100 m ) ⎜
= (0.100 m ) ⎜
Vf = Vi ⎜⎜ ⎟⎟
⎟
⎟ = 0.132 m
T
327
K
327
K
⎝
⎠
⎝
⎠
⎝ f ⎠
______________________________________________________________________________
30. REASONING When the temperature of a gas changes as a result of heat Q being added, the
change ∆T in temperature is related to the amount of heat according to
Q = Cn ∆T ( Equation 15.6 ) , where C is the molar specific heat capacity, and n is the
number of moles. The heat QV added under conditions of constant volume is
QV = CV n ∆TV , where CV is the specific heat capacity at constant volume and is given by
CV = 32 R ( Equation 15.8 ) and R is the universal gas constant. The heat QP added under
766 THERMODYNAMICS
conditions of constant pressure is QP = CP n ∆TP , where CP is the specific heat capacity at
constant pressure and is given by CP = 52 R ( Equation 15.7 ) . It is given that QV = QP, and
this fact will allow us to find the change in temperature of the gas whose pressure remains
constant.
SOLUTION Setting QV = QP, gives
CV n ∆TV = CP n ∆TP
QP
QV
Algebraically eliminating n and solving for ∆TP, we obtain
⎛ 3R⎞
⎛C ⎞
∆TP = ⎜ V ⎟ ∆TV = ⎜ 52 ⎟ ( 75 K ) = 45 K
⎜C ⎟
⎜ R⎟
⎝ P⎠
⎝2 ⎠
______________________________________________________________________________
31. SSM REASONING AND SOLUTION According to the first law of thermodynamics
(Equation 15.1), ∆U = U f − U i = Q − W . Since the internal energy of this gas is doubled by
the addition of heat, the initial and final internal energies are U and 2U, respectively.
Therefore,
∆U = U f − U i = 2U − U = U
Equation 15.1 for this situation then becomes U = Q − W . Solving for Q gives
Q = U +W
(1)
The initial internal energy of the gas can be calculated from Equation 14.7:
3
2
U = nRT =
3
2
( 2.5 mol ) ⎡⎣8.31 J/ ( mol ⋅ K )⎤⎦ ( 350 K ) = 1.1×104 J
a. If the process is carried out isochorically (i.e., at constant volume), then W = 0, and the
heat required to double the internal energy is
Q = U + W = U + 0 = 1.1× 104 J
b. If the process is carried out isobarically (i.e., at constant pressure), then W = P ∆V , and
Equation (1) above becomes
(2)
Q = U + W = U + P ∆V
From the ideal gas law, PV = nRT , we have that P ∆V = nR ∆T , and Equation (2) becomes
Q = U + nR ∆T
(3)
Chapter 15 Problems
767
The internal energy of an ideal gas is directly proportional to its Kelvin temperature. Since
the internal energy of the gas is doubled, the final Kelvin temperature will be twice the
initial Kelvin temperature, or ∆T = 350 K. Substituting values into Equation (3) gives
Q = 1.1× 104 J + (2.5 mol)[8.31 J/(mol ⋅ K)](350 K) = 1.8 ×104 J
______________________________________________________________________________
32. REASONING AND SOLUTION
a. The amount of heat needed to raise the temperature of the gas at constant volume is given
by Equations 15.6 and 15.8, Q = n CV ∆T. Solving for ∆T yields
Q
5.24 × 103 J
∆T =
=
= 1.40 × 102 K
3
nCv ( 3.00 mol ) 2 R
( )
b. The change in the internal energy of the gas is given by the first law of thermodynamics
with W = 0, since the gas is heated at constant volume:
∆U = Q − W = 5.24 × 103 J − 0 = 5.24 × 103 J
c. The change in pressure can be obtained from the ideal gas law,
(
)
nR ∆T ( 3.00 mol ) R 1.40 × 10 K
∆P =
=
= 2.33 × 103 Pa
3
V
1.50 m
______________________________________________________________________________
2
33. REASONING AND SOLUTION The heat added at constant pressure is
Q = Cpn∆T = (5R/2) n∆T
The work done during the process is W = P∆V. The ideal gas law requires that
∆V = nR∆T/P, so W = nR∆T. The required ratio is then
Q/W = 5/2
______________________________________________________________________________
34. REASONING The heat QP that needs to be added under constant-pressure conditions is
given by QP = CP n∆T , according to Equation 15.6, where CP is the molar specific heat
capacity at constant pressure, n is the number of moles of the gas, and ∆T is the change in
the temperature. We have values for n and ∆T, but not for CP. For an ideal gas of any type,
however, Equation 15.10 applies, so that CP = CV + R , where CV is the molar specific heat
capacity at constant volume and R = 8.31 J/(mol⋅K) is the universal gas constant. This result
is useful, because we can evaluate CV from the heat added under constant-volume
conditions.
768 THERMODYNAMICS
SOLUTION
According to Equation 15.6, we know that QP = CP n∆T .
Substituting
CP = CV + R (Equation 15.10) into this expression, we obtain
QP = CP n∆T = ( CV + R ) n∆T
(1)
Using Equation 15.6 again, this time for constant-volume conditions, we have CV =
QV
n∆T
,
where QV is the heat added. Substituting this result into Equation (1) gives
⎛ Q
⎞
QP = ( CV + R ) n∆T = ⎜ V + R ⎟ n∆T
⎝ n∆T
⎠
= QV + nR∆T = 3500 J + (1.6 mol ) ⎡⎣8.31 J/ ( mol ⋅ K ) ⎤⎦ ( 75 K ) = 4500 J
35. SSM REASONING AND SOLUTION The amount of heat required to change the
temperature of the gas is given by Equation 15.6, where CP is given by Equation 15.7.
Q = C P n ∆T = 25 Rn ∆T = 25 [8.31 J / (mol ⋅ K)] (1.5 mol) (77 K) = 2400 J
____________________________________________________________________________________________
36. REASONING AND SOLUTION The total heat generated by the students is
Q = (200)(130 W)(3000 s) = 7.8 × 107 J
For the isochoric process.
Q = Cvn∆T = (5R/2)n ∆T
The number of moles of air in the room is found from the ideal gas law to be
(
)(
)
1.01 × 105 Pa 1200 m3
PV
n=
=
= 5.0 × 104 mol
RT ⎡⎣8.31 J/ ( mol ⋅ K ) ⎤⎦ ( 294 K )
Now
∆T =
Q
=
5 Rn
2
7.8 × 107 J
= 75 K
⎡⎣8.31 J/ ( mol ⋅ K ) ⎤⎦ 5.0 × 104 mol
______________________________________________________________________________
5
2
(
)
37. REASONING The power rating P of the heater is equal to the heat Q supplied to the gas
divided by the time t the heater is on, P = Q / t (Equation 6.10b). Therefore, t = Q / P . The
heat required to change the temperature of a gas under conditions of constant pressure is
given by Q = CP n ∆T ( Equation 15.6 ) , where CP is the molar specific heat capacity at
constant pressure, n is the number of moles, and ∆T = Tf − Ti is the change in temperature.
Chapter 15 Problems
769
For a monatomic ideal gas, the specific heat capacity at constant pressure is CP = 52 R,
Equation (15.7), where R is the universal gas constant. We do not know n, Tf and Ti, but we
can use the ideal gas law, PV = nRT , (Equation 14.1) to replace nRTf by PfVf and to replace
nRTi by PiVi, quantities that we do know.
SOLUTION Substituting Q = CP n ∆T = CP n (Tf − Ti ) into t = Q / P and using the fact that
CP = 52 R give
t=
Q CP n (Tf − Ti )
=
=
P
P
5
2
R n (Tf − Ti )
P
Replacing RnTf by PfVf and RnTi by PiVi and remembering that Pi = Pf , we find
t=
5P
2 i
(Vf − Vi )
P
Since the volume of the gas increases by 25.0%, Vf = 1.250Vi. The time that the heater is on
is
5 P V −V
5 P 1.250V − V
5 P 0.250 V
(
) i
(
(
i)
i
i)
t= 2 i f
= 2 i
= 2 i
P
P
P
=
5
2
( 7.60 ×105 Pa ) ( 0.250 ) (1.40 ×10−3 m3 ) = 44.3 s
15.0 W
______________________________________________________________________________
38. REASONING AND SOLUTION The change in volume is ∆V = – sA, where s is the
distance through which the piston drops and A is the piston area. The minus sign is included
because the volume decreases. Thus,
– ∆V
s=
A
The ideal gas law states that ∆V = nR∆T/P. But
∆T = Q /
Q = C p n ∆T = 52 R n ∆T . Thus,
c Rn h . Using these expressions for ∆V and ∆T, we find that
– nR ∆T / P – nR Q / b Rn g
–Q
s=
=
=
5
2
5
2
A
=
b
– –2093 J
5
2
5
2
PA
g
c1.01 × 10 Pa hc 3.14 × 10
5
–2
m2
PA
h=
0 . 264 m
______________________________________________________________________________
770 THERMODYNAMICS
39. REASONING AND SOLUTION Let P, V, and T represent the initial values of pressure,
volume, and temperature. The first process is isochoric, so
Q1 = CVn ∆T1 = (3R/2)n ∆T1
The ideal gas law for this process gives ∆T1 = 2PV/(nR), so Q1 = 3PV.
The second process is isobaric, so
Q2 = CPn ∆T2 = (5R/2)n ∆T2
The ideal gas law for this process gives ∆T2 = 3PV/nR, so Q2 = (15/2) PV. The total heat is
Q = Q1 + Q2 = (21/2) PV.
But at conditions of standard temperature and pressure (see Section 14.2), P = 1.01 × 105 Pa
and V = 22.4 liters = 22.4 × 10–3 m3, so
Q=
21 PV
2
=
21
2
(1.01 × 105 Pa )( 22.4 × 10−3 m3 ) =
2.38 × 104 J
______________________________________________________________________________
40. REASONING According to Equation 15.11, the efficiency of a heat engine is e = W / QH ,
where W is the magnitude of the work and QH is the magnitude of the input heat. Thus,
the magnitude of the work is W = e QH . We can apply this result before and after the tuneup to compute the extra work produced.
SOLUTION Using Equation 15.11, we find the work before and after the tune-up as
follows:
WBefore = eBefore QH
and WAfter = eAfter QH
Subtracting the “before” equation from the “after” equation gives
WAfter − WBefore = eAfter QH − eBefore QH = ( eAfter − eBefore ) QH = 0.050 (1300 J ) = 65 J
______________________________________________________________________________
41. SSM REASONING AND SOLUTION The efficiency of a heat engine is defined by
Equation 15.11 as e = W / QH , where W is the magnitude of the work done and QH is
the magnitude of the heat input. The principle of energy conservation requires that
QH = W + QC , where QC is the magnitude of the heat rejected to the cold reservoir
(Equation 15.12). Combining Equations 15.11 and 15.12 gives
e=
W
W + QC
=
16 600 J
= 0.631
16 600 J + 9700 J
Chapter 15 Problems
771
______________________________________________________________________________
42. REASONING According to the definition of efficiency given in Equation 15.11, an engine
with an efficiency e does work of magnitude W = e QH , where QH is the magnitude of
the input heat. We will apply this expression to each engine and utilize the fact that in each
case the same work is done. We expect to find that engine 2 requires less input heat to do
the same amount of work, because it has the greater efficiency.
SOLUTION Applying Equation 15.11 to each engine gives
W = e1 QH1
and
W = e2 QH2
Engine 1
Engine 2
Since the work is the same for each engine, we have
e1 QH1 = e2 QH2
or
⎛e ⎞
QH2 = ⎜⎜ 1 ⎟⎟ QH1
⎝ e2 ⎠
It follows, then, that
⎛e ⎞
⎛ 0.18 ⎞
QH2 = ⎜⎜ 1 ⎟⎟ QH1 = ⎜
⎟ ( 5500 J ) = 3800 J
⎝ 0.26 ⎠
⎝ e2 ⎠
which is less than the input heat for engine 1, as expected.
43. REASONING AND SOLUTION
a. The efficiency is e = W / QH , where W is the magnitude of the work and QH is the
magnitude of the input heat. It follows that
QH =
W 5500 J
=
= 8600 J
e
0.64
b. The magnitude of the rejected heat is
QC = QH − W = 8600 J − 5500 J = 3100 J
______________________________________________________________________________
44. REASONING The efficiency e of an engine can be expressed as (see Equation 15.13)
e = 1 − QC / QH , where QC is the magnitude of the heat delivered to the cold reservoir
and QH is the magnitude of the heat supplied to the engine from the hot reservoir. Solving
this equation for QC gives QC = (1 − e ) QH . We will use this expression twice, once for
772 THERMODYNAMICS
the improved engine and once for the original engine. Taking the ratio of these expressions
will give us the answer that we seek.
SOLUTION Taking the ratio of the heat rejected to the cold reservoir by the improved
engine to that for the original engine gives
QC, improved
=
QC, original
(1 − eimproved ) QH, improved
(1 − eoriginal ) QH, original
But the input heat to both engines is the same, so QH, improved = QH, original . Thus, the ratio
becomes
QC, improved
QC, original
=
1 − eimproved
1 − eoriginal
=
1 − 0.42
= 0.75
1 − 0.23
______________________________________________________________________________
45. SSM
WWW
(
REASONING The efficiency of either engine is given by Equation
)
15.13, e = 1 − QC / QH . Since engine A receives three times more input heat, produces
five times more work, and rejects two times more heat than engine B, it follows that
QHA = 3 QHB , WA = 5 WB , and QCA = 2 QCB . As required by the principle of energy
conservation for engine A (Equation 15.12),
QHA = QCA + W
A
N
3 QHB
5W
2 QCB
B
Thus,
3 QHB = 2 QCB + 5 WB
(1)
Since engine B also obeys the principle of energy conservation (Equation 15.12),
QHB = QCB + WB
Substituting QHB from Equation (2) into Equation (1) yields
3( QCB + WB ) = 2 QCB + 5 WB
Solving for WB gives
WB =
1
2
QCB
Therefore, Equation (2) predicts for engine B that
(2)
Chapter 15 Problems
QHB = QCB + WB =
3
2
773
QCB
SOLUTION
a. Substituting QCA = 2 QCB and QHA = 3 QHB into Equation 15.13 for engine A, we
have
eA = 1 −
b. Substituting QHB =
QCA
QHA
3
2
= 1−
2 QCB
3 QHB
= 1−
2 QCB
3
( 32 QCB )
= 1−
4
5
=
9
9
QCB into Equation 15.13 for engine B, we have
eB = 1 −
QCB
QHB
= 1−
QCB
3
2
QCB
= 1−
2
1
=
3
3
______________________________________________________________________________
46. REASONING AND SOLUTION The efficiency is given by
e = 1 − (TC/TH) = 1 − [(200 K)/(500 K)] = 0.6 = W / QH
The work is
W = e QH = (0.6)(5000 J) = 3000 J
______________________________________________________________________________
47. REASONING We will use the subscript “27” to denote the engine whose efficiency is
27.0% (e27 = 0.270) and the subscript “32” to denote the engine whose efficiency is 32.0%
(e32 = 0.320). In general, the efficiency eCarnot of a Carnot engine depends on the Kelvin
temperatures, TC and TH, of its cold and hot reservoirs through the relation (see
Equation 15.15) eCarnot = 1 − (TC/TH). Solving this equation for the temperature TC, 32 of the
engine whose efficiency is e32 gives TC, 32 = (1 − e32)TH, 32. We are given e32, but do not
know the temperature TH, 32. However, we are told that this temperature is the same as that
of the hot reservoir of the engine whose efficiency is e27, so TH, 32 = TH, 27. The temperature
TH, 27 can be determined since we know the efficiency and cold reservoir temperature of this
engine.
SOLUTION The temperature of the cold reservoir for engine whose efficiency is e32 is
TC, 32 = (1 − e32)TH, 32. Since TH, 32 = TH, 27, we have that
TC, 32 = (1 − e32)TH, 27
(1
774 THERMODYNAMICS
The efficiency e27 is given by Equation 15.15 as e27 = 1 − (TC, 27/TH, 27). Solving this
equation for the temperature TH, 27 of the hot reservoir and substituting the result into
Equation 1 yields
⎛ 1 − e32 ⎞
⎛ 1 − 0.320 ⎞
TC, 32 = ⎜
TC, 27 = ⎜
⎟
⎟ ( 275 K ) = 256 K
⎜ 1− e ⎟
−
1
0.270
⎝
⎠
27 ⎠
⎝
______________________________________________________________________________
48. REASONING
The efficiency eCarnot of a Carnot engine is
eCarnot = 1 −
TC
TH
(Equation 15.15), where TC and TH are, respectively, the Kelvin temperatures of the cold
and hot reservoirs. After the changes are made to the temperatures, this same equation still
applies, except that the variables must be labeled to denote the new values. We will use a
“prime” for this purpose. From the original efficiency and the information given about the
changes made to the temperatures, we will be able to obtain the new temperature ratio and,
hence, the new efficiency.
SOLUTION After the reservoir temperatures are changed, the engine has an efficiency that,
according to Equation 15.15, is
T′
′
eCarnot
= 1− C
TH′
where the “prime” denotes the new engine. Using unprimed symbols to denote the original
engine, we know that TC′ = 2TC and TH′ = 4TH . With these substitutions, the efficiency of
the new engine becomes
T′
2T
1⎛T ⎞
′
eCarnot
= 1 − C = 1 − C = 1 − ⎜⎜ C ⎟⎟
(1)
2 T
TH′
4TH
⎝ H⎠
To obtain the original ratio TC / TH , we use Equation 15.15:
eCarnot = 1 −
TC
TH
TC
or
TH
= 1 − eCarnot
Substituting this original temperature ratio into Equation (1) gives
1⎛T
′
= 1 − ⎜⎜ C
eCarnot
2 T
⎝ H
⎞
1
⎟⎟ = 1 − 2 (1 − eCarnot ) =
⎠
1
2
(1 + eCarnot ) = 12 (1 + 0.40 ) = 0.70
Chapter 15 Problems
775
49. SSM REASONING The efficiency e of a Carnot engine is given by Equation 15.15,
e = 1 − (TC / TH ) , where, according to Equation 15.14, QC / QH = TC / TH . Since the
efficiency is given along with TC and QC , Equation 15.15 can be used to calculate TH .
Once TH is known, the ratio TC / TH is thus known, and Equation 15.14 can be used to
calculate QH .
SOLUTION
a. Solving Equation 15.15 for TH gives
TH =
TC
1–e
=
378 K
= 1260 K
1–0.700
b. Solving Equation 15.14 for QH gives
⎛T ⎞
⎛ 1260 K ⎞
4
QH = QC ⎜ H ⎟ = (5230 J) ⎜
⎟ = 1.74 × 10 J
⎜T ⎟
⎝ 378 K ⎠
⎝ C⎠
______________________________________________________________________________
50. REASONING AND SOLUTION In order to find out how many kilograms of ice in the tub
are melted, we must determine QC , the amount of exhaust heat delivered to the cold
reservoir. Since the hot reservoir consists of boiling water (TH = 373.0 K) and the cold
reservoir consists of ice and water (TC = 273.0 K), the efficiency of this engine is
e = 1−
TC
TH
= 1−
273.0 K
= 0.2681
373.0 K
The work done by the engine is
W = e QH = (0.2681)(6800 J) = 1823 J
Therefore, the amount of heat delivered to the cold reservoir is
QC = QH − W = 6800 J – 1823 J = 4977 J
Using the definition of the latent heat of fusion (melting) Lf , we find that the amount of ice
that melts is
Q
4977 J
m = C =
= 0.015 kg
Lf
33.5 ×104 J/kg
______________________________________________________________________________
776 THERMODYNAMICS
51. REASONING AND SOLUTION The efficiency of the engine is e = 1 − (TC/TH) so
(i) Increase TH by 40 K; e = 1 − [(350 K)/(690 K)] = 0.493
(ii) Decrease TC by 40 K; e = 1 − [(310 K)/(650 K)] = 0.523
The greatest improvement is made by lowering the temperature of the cold reservoir.
______________________________________________________________________________
52. REASONING AND SOLUTION The amount of work delivered by the engines can be
determined from Equation 15.12, QH = W + QC . Solving for W for each engine gives:
W1 = QH1 − QC1
and
W2 = QH2 − QC2
The total work delivered by the two engines is
(
) (
W = W1 + W2 = QH1 − QC1 + QH2 − QC2
)
But we know that QH2 = QC1 , so that
(
) (
)
W = QH1 − QC1 + QC1 − QC2 = QH1 − QC2
(1)
Since these are Carnot engines,
QC1
QH1
=
TC1
TH1
⇒
QC1 = QH1
TC1
TH1
⎛ 670 K ⎞
3
= ( 4800 J ) ⎜
⎟ = 3.61×10 J
⎝ 890 K ⎠
Similarly, noting that QH2 = QC1 and that TH2 = TC1, we have
QC2 = QH2
TC2
TH2
= QC1
TC2
TC1
(
)
⎛ 420 K ⎞
3
= 3.61× 103 J ⎜
⎟ = 2.26 × 10 J
⎝ 670 K ⎠
Substituting into Equation (1) gives
W = 4800 J − 2.26 × 103 J = 2.5 × 103 J
______________________________________________________________________________
53. SSM REASONING The maximum efficiency e at which the power plant can operate is
given by Equation 15.15, e = 1 − ( TC / TH ) . The power output is given; it can be used to
find the magnitude W of the work output for a 24 hour period. With the efficiency and
W known, Equation 15.11, e = W / QH , can be used to find QH , the magnitude of the
input heat. The magnitude QC of the exhaust heat can then be found from Equation 15.12,
QH = W + QC .
Chapter 15 Problems
777
SOLUTION
a. The maximum efficiency is
e = 1−
TC
TH
= 1−
323 K
= 0.360
505 K
b. Since the power output of the power plant is P = 84 000 kW, the required heat input
QH for a 24 hour period is
QH =
W
Pt (8.4 ×107 J/s)(24 h) ⎛ 3600 s ⎞
13
=
=
⎜
⎟ = 2.02 × 10 J
e
e
0.360
1
h
⎝
⎠
Therefore, solving Equation 15.12 for QC , we have
QC = QH − W = 2.02 × 1013 J − 7.3 × 1012 J = 1.3 × 1013 J
______________________________________________________________________________
54. REASONING AND SOLUTION The temperature of the gasoline engine input is
T1 = 904 K, the exhaust temperature is T2 = 412 K, and the air temperature is T3 = 300 K.
The efficiency of the engine/exhaust is
e1 = 1 − (T2/T1) = 0.544
The efficiency of the second engine is
e2 = 1 − (T3/T2) = 0.272
The
magnitude
of
the
work
done
by
each
segment
is
W1 = e1 QH1
and
W2 = e2 QH2 = e2 QC1 since
QH2 = QC1
Now examine
( W1 + W2 ) / W1
to find the ratio of the total work produced by both engines
to that produced by the first engine alone.
W1 + W2
W1
(
=
e1 QH1 + e2 QC1
e1 QH1
⎛ e ⎞⎛ Q
= 1 + ⎜ 2 ⎟ ⎜ C1
⎜ e ⎟⎜ Q
⎝ 1 ⎠ ⎝ H1
⎞
⎟
⎟
⎠
)
But, e1 = 1 − QC1 / QH1 , so that QC1 / QH1 = 1 − e1 . Therefore,
W1 + W2
W1
= 1 +
e2
e
1 − e1 ) = 1 + 2 − e2 = 1 + 0.500 − 0.272 = 1.23
(
e1
e1
______________________________________________________________________________
778 THERMODYNAMICS
55. SSM REASONING AND SOLUTION The efficiency e of the power plant is threefourths its Carnot efficiency so, according to Equation 15.15,
⎛ T
e = 0.75 ⎜⎜ 1 − C
⎝ TH
⎞
40 K + 273 K ⎞
⎛
⎟⎟ = 0.75 ⎜ 1 −
⎟ = 0.33
⎝ 285 K + 273 K ⎠
⎠
The power output of the plant is 1.2 × 109 watts. According to Equation 15.11,
e = W / QH = ( Power ⋅ t ) / QH . Therefore, at 33% efficiency, the magnitude of the heat
input per unit time is
QH
t
=
Power 1.2 ×109 W
=
= 3.6 × 109 J/s
0.33
e
From the principle of conservation of energy, the heat output per unit time must be
QC
t
=
QH
t
− Power = 2.4 × 109 J/s
The rejected heat is carried away by the flowing water and, according to Equation 12.4,
QC = cm ∆T . Therefore,
QC
t
=
cm ∆T
t
⎛m
= c⎜
t
⎝ t
QC
or
⎞
⎟ ∆T
⎠
Solving the last equation for ∆t, we have
QC
QC / t
c(m / t )
=
2.4 × 109 J/s
= 5.7 C°
⎡⎣ 4186 J/ ( kg ⋅ C° ) ⎤⎦ 1.0 × 105 kg/s
______________________________________________________________________________
tc(m / t )
=
56. REASONING The expansion from point a to
point b and the compression from point c to point
d occur isothermally, and we will apply the first
law of thermodynamics to these parts of the cycle
in order to obtain expressions for the input and
rejected heats, magnitudes QH and QC ,
respectively. In order to simplify the resulting
expression for QC / QH , we will then use the
fact that the expansion from point b to point c and
the compression from point d to point a are
adiabatic.
(
)
a
Pressure →
∆T =
b
TC
TH
d
c
Volume →
Chapter 15 Problems
779
SOLUTION According to the first law of thermodynamics, the change in internal energy
∆U is given by ∆U = Q − W (Equation 15.1), where Q is the heat and W is the work. Since
the internal energy of an ideal gas is proportional to the Kelvin temperature and the
temperature is constant for an isothermal process, it follows that ∆U = 0 J for such a case.
The work of isothermal expansion or compression for an ideal gas is
W = nRT ln (Vf / Vi ) (Equation 15.3), where n is the number of moles, R is the universal gas
constant, T is the Kelvin temperature, Vf is the final volume of the gas, and Vi is the initial
volume. We have, then, that
⎛V ⎞
⎛V ⎞
∆U = Q − W or 0 = Q − nRT ln ⎜⎜ f ⎟⎟ or Q = nRT ln ⎜⎜ f ⎟⎟
⎝ Vi ⎠
⎝ Vi ⎠
Applying this result for Q to the isothermal expansion (temperature = TH) from point a to
point b and the isothermal compression (temperature = TC) from point c to point d, we have
⎛V ⎞
QH = nRTH ln ⎜ b ⎟
⎜V ⎟
⎝ a⎠
and
⎛V
QC = nRTC ln ⎜ d
⎜V
⎝ c
⎞
⎟⎟
⎠
where Vf = Vb and Vi = Va for the isotherm at TH and Vf = Vd and Vi = Vc for the isotherm
at TC. In this problem, we are interested in the magnitude of the heats. For QH, this poses
no problem, since Vb > Va , ln (Vb / Va ) is positive, and we have
⎛V ⎞
QH = nRTH ln ⎜ b ⎟
(1)
⎜V ⎟
⎝ a⎠
However, for QC, we need to be careful, because Vc > Vd and ln (Vd / Vc ) is negative. Thus,
we write for the magnitude of QC that
⎛V
QC = −nRTC ln ⎜ d
⎜V
⎝ c
⎞
⎛ Vc ⎞
⎟⎟ = nRTC ln ⎜⎜ ⎟⎟
⎠
⎝ Vd ⎠
(2)
According to Equations (1) and (2), the ratio of the magnitudes of the rejected and input
heats is
⎛V ⎞
⎛V ⎞
nRTC ln ⎜⎜ c ⎟⎟ TC ln ⎜⎜ c ⎟⎟
QC
⎝ Vd ⎠ =
⎝ Vd ⎠
=
(3)
⎛ Vb ⎞
⎛ Vb ⎞
QH
nRTH ln ⎜⎜ ⎟⎟ TH ln ⎜⎜ ⎟⎟
⎝ Va ⎠
⎝ Va ⎠
We now consider the adiabatic parts of the Carnot cycle. For the adiabatic expansion or
compression of an ideal gas the initial pressure and volume (Pi and Vi) are related to the
final pressure and volume (Pf and Vf) according to
780 THERMODYNAMICS
Pi Viγ = Pf Vfγ
(15.5)
where γ is the ratio of the specific heats at constant pressure and constant volume. It is also
true that P = nRT / V (Equation 14.1), according to the ideal gas law. Substituting this
expression for the pressure into Equation 15.5 gives
⎛ nRTi
⎜⎜
⎝ Vi
⎞ γ ⎛ nRTf
⎟⎟ Vi = ⎜⎜
⎠
⎝ Vf
⎞ γ
⎟⎟ Vf
⎠
γ −1
TV
= Tf Vfγ −1
i i
or
Applying this result to the adiabatic expansion from point b to point c and to the adiabatic
compression from point d to point a, we obtain
THVbγ −1 = TCVcγ −1
TCVdγ −1 = THVaγ −1
and
Dividing the first of these equations by the second shows that
THVbγ −1
THVaγ −1
=
TCVcγ −1
TCVdγ −1
or
Vbγ −1
Vcγ −1
=
Vaγ −1
Vdγ −1
or
Vb
Va
=
Vc
Vd
With this result, Equation (3) becomes
QC
QH
57. REASONING
=
⎛V
TC ln ⎜ c
⎜V
⎝ d
⎞
⎟⎟
⎠
⎛V ⎞
TH ln ⎜⎜ b ⎟⎟
⎝ Va ⎠
=
TC
TH
The coefficient of performance COP is defined as COP = QC / W
(Equation 15.16), where QC is the magnitude of the heat removed from the cold reservoir
and W is the magnitude of the work done on the refrigerator. The work is related to the
magnitude QH of the heat deposited into the hot reservoir and QC by the conservation of
energy, W = QH − QC . Thus, the coefficient of performance can be written as (after some
algebraic manipulations)
COP =
1
QH
QC
−1
781
Chapter 15 Problems
The maximum coefficient of performance occurs when the refrigerator is a Carnot
refrigerator. For a Carnot refrigerator, the ratio QH / QC is equal to the ratio TH / TC of the
Kelvin temperatures of the hot and cold reservoirs, QH / QC = TH / TC (Equation 15.14).
SOLUTION Substituting QH / QC = TH / TC into the expression above for the COP gives
1
COP =
=
1
= 13
296 K
−1
275 K
TH
−1
TC
______________________________________________________________________________
58. REASONING AND SOLUTION We know that
QC = QH − W = 14 200 J − 800 J = 13 400 J
Therefore,
⎛Q ⎞
⎛ 13 400 J ⎞
TC = TH ⎜ C ⎟ = ( 301 K ) ⎜
⎟ = 284 K
⎜Q ⎟
14
200
J
⎝
⎠
⎝ H ⎠
______________________________________________________________________________
59. SSM WWW REASONING AND SOLUTION Equation 15.14 holds for a Carnot air
conditioner as well as a Carnot engine. Therefore, solving Equation 15.14 for QC , we
have
⎛T ⎞
⎛ 299 K ⎞
5
QC = QH ⎜⎜ C ⎟⎟ = 6.12 × 105 J ⎜
⎟ = 5.86 × 10 J
⎝ 312 K ⎠
⎝ TH ⎠
______________________________________________________________________________
(
60. REASONING
)
Since the refrigerator is a Carnot device, we know that
QC
QH
=
TC
TH
(Equation 15.14). We have values for TH (the temperature of the kitchen) and QC (the
magnitude of the heat removed from the food). Thus, we can use this expression to
determine TC (the temperature inside the refrigerator), provided that a value can be obtained
for QH (the magnitude of the heat that the refrigerator deposits into the kitchen). Energy
conservation dictates that QH = W + QC (Equation 15.12), where W is the magnitude of
the work that the appliance uses and is known.
SOLUTION From Equation 15.14, it follows that
782 THERMODYNAMICS
QC
QH
=
TC
⎛Q ⎞
TC = TH ⎜ C ⎟
⎜Q ⎟
⎝ H ⎠
or
TH
Substituting QH = W + QC (Equation 15.12) into this result for TC gives
⎛Q ⎞
⎛ QC
TC = TH ⎜ C ⎟ = TH ⎜
⎜Q ⎟
⎜W +Q
C
⎝ H ⎠
⎝
⎞
2561 J
⎛
⎞
⎟ = ( 301 K ) ⎜
⎟ = 275 K
⎟
⎝ 241 J + 2561 J ⎠
⎠
61. REASONING AND SOLUTION
a. The work is
W = QH − QC = 3140 J − 2090 J = 1050 J
b. The coefficient of performance (COP) is
3140 J
= 2.99
W
1050 J
______________________________________________________________________________
COP =
QH
=
62. REASONING The coefficient of performance of an air conditioner is QC / W , according
to Equation 15.16, where QC is the magnitude of the heat removed from the house and W
is the magnitude of the work required for the removal. In addition, we know that the first
law of thermodynamics (energy conservation) applies, so that W = QH − QC , according to
Equation 15.12. In this equation QH is the magnitude of the heat discarded outside. While
we have no direct information about QC and QH , we do know that the air conditioner is a
Carnot device. This means that Equation 15.14 applies: QC / QH = TC / TH . Thus, the
given temperatures will allow us to calculate the coefficient of performance.
SOLUTION Using Equation 15.16 for the definition of the coefficient of performance and
Equation 15.12 for the fact that W = QH − QC , we have
Coefficient of performance =
QC
W
=
QC
QH − QC
=
QC / QH
1 − QC / QH
Equation 15.14 applies, so that QC / QH = TC / TH . With this substitution, we find
Chapter 15 Problems
Coefficient of performance =
QC / QH
1 − QC / QH
=
783
TC / TH
1 − TC / TH
297 K
= 21
TH − TC ( 311 K ) − ( 297 K )
______________________________________________________________________________
=
TC
=
63. REASONING The conservation of energy applies to the air conditioner, so that
QH = W + QC , where QH is the amount of heat put into the room by the unit, QC is the
amount of heat removed from the room by the unit, and W is the amount of work needed to
operate the unit. Therefore, a net heat of QH − QC = W is added to and heats up the room.
To find the temperature rise of the room, we will use the COP to determine W and then use
the given molar specific heat capacity.
SOLUTION Let COP denote the coefficient of performance. By definition (Equation
15.16), COP = QC / W , so that
W =
QC
COP
=
7.6 × 104 J
= 3.8 ×104 J
2.0
The temperature rise in the room can be found as follows: QH − QC = W = CV n ∆T .
Solving for ∆T gives
∆T =
W
CV n
=
W
( 52 R ) n
=
3.8 × 104 J
= 0.48 K
5 ⎡8.31 J/ mol ⋅ K ⎤ 3800 mol
(
)
(
)
⎦
2⎣
______________________________________________________________________________
64. REASONING AND SOLUTION The amount of heat removed when the ice freezes is
QC = cm ∆T + mLf
= [4186 J/(kg⋅C°)](1.50 kg)(20.0 C°) + (1.50 kg)(33.5 × 104 J/kg) = 6.28 × 105 J
Since the coefficient of performance is COP = QC / W , the magnitude of the work done by
the refrigerator is
W =
QC
COP
=
6.28 ×105 J
= 2.09 ×105 J
3.00
The magnitude of the heat delivered to the kitchen is
QH = QC + W = 6.28 ×105 J + 2.09 × 105 J = 8.37 × 105 J
784 THERMODYNAMICS
The space heater has a power output P of
Q
P = H = 3.00 × 103 J/s
t
Therefore,
Q
8.37 × 105 J
t= H =
= 279 s
P
3.00 ×103 J/s
______________________________________________________________________________
WWW REASONING Let the coefficient of performance be represented by the
symbol COP. Then according to Equation 15.16, COP = QC / W . From the statement of
65. SSM
energy conservation for a Carnot refrigerator (Equation 15.12), W = QH − QC . Combining
Equations 15.16 and 15.12 leads to
COP =
QC
QH − QC
=
QC / QC
( QH − QC ) / QC
=
1
( QH / QC ) − 1
Replacing the ratio of the heats with the ratio of the Kelvin temperatures, according to
Equation 15.14, leads to
1
COP =
(1)
TH / TC − 1
The heat QC that must be removed from the refrigerator when the water is cooled can be
calculated using Equation 12.4, QC = cm∆T ; therefore,
W =
QC
COP
=
cm ∆T
COP
(2)
SOLUTION
a. Substituting values into Equation (1) gives
COP =
1
TH
−1
TC
=
1
=
( 20.0 + 273.15) K − 1
( 6.0 + 273.15 ) K
2.0 ×101
b. Substituting values into Equation (2) gives
cm ∆T ⎣⎡ 4186 J/ ( kg ⋅ C° ) ⎦⎤ ( 5.00 kg )( 20.0 °C − 6.0 °C )
=
= 1.5 × 104 J
1
COP
2.0 × 10
______________________________________________________________________________
W =
Chapter 15 Problems
785
66. REASONING According to the conservation of energy, the work W done by the electrical
energy is W = QH − QC , where QH is the magnitude of the heat delivered to the outside
(the hot reservoir) and QC is the magnitude of the heat removed from the house (the cold
reservoir). Dividing both sides of this relation by the time t, we have
W
=
t
QH
−
t
QC
t
The term W / t is the magnitude of the work per second that must be done by the electrical
energy, and the terms QH / t and QC / t are, respectively, the magnitude of the heat per
second delivered to the outside and removed from the house. Since the air conditioner is a
Carnot air conditioner, we know that QH / QC is equal to the ratio TH / TC of the Kelvin
temperatures of the hot and cold reservoirs, QH / QC = TH / TC (Equation 15.14). This
expression, along with the one above for W / t , will allow us to determine the magnitude of
the work per second done by the electrical energy.
SOLUTION Solving the expression QH / QC = TH / TC for QH , substituting the result
into the relation
W
=
t
QH
t
−
QC
t
, and recognizing that QC / t = 10 500 J/s, give
QC TH
W
t
=
QH
t
−
QC
t
=
TC
t
−
QC
t
⎛ Q ⎞⎛ T
⎞
⎛ 306.15 K ⎞
= ⎜ C ⎟ ⎜ H − 1⎟ = (10 500 J/s ) ⎜
− 1⎟ = 5.0 × 102 J/s
⎟
⎜ t ⎟ ⎜ TC
⎝ 292.15 K ⎠
⎠
⎝
⎠⎝
In this result we have used the fact that TH = 273.15 + 33.0 °C = 306.15 K and
TC = 273.15 + 19.0 °C = 292.15 K.
______________________________________________________________________________
67. SSM REASONING The efficiency of the Carnot engine is, according to Equation 15.15,
e = 1−
TC
TH
= 1−
842 K 1
=
1684 K 2
Therefore, the magnitude of the work delivered by the engine is, according to Equation
15.11,
786 THERMODYNAMICS
W = e QH =
1
2
QH
The heat pump removes an amount of heat QH from the cold reservoir. Thus, the amount
of heat Q ′ delivered to the hot reservoir of the heat pump is
1
2
Q′ = QH + W = QH + QH =
3
2
QH
Therefore, Q′ / QH = 3 / 2 . According to Equation 15.14, Q′ / QH = T ′ / TC , so T ′ / TC = 3 / 2 .
SOLUTION Solving for T ′ gives
3
2
3
2
T ′ = TC = (842 K) = 1.26 × 103 K
______________________________________________________________________________
68. REASONING According to the discussion on Section 15.11, the change ∆Suniverse in the
entropy of the universe is the sum of the change in entropy ∆SC of the cold reservoir and the
change in entropy ∆SH of the hot reservoir, or ∆Suniverse = ∆SC + ∆SH. The change in entropy
of each reservoir is given by Equation 15.18 as ∆S = (Q/T)R, where Q is the heat removed
from or delivered to the reservoir and T is the Kelvin temperature of the reservoir. In
applying this equation we imagine a process in which the heat is lost by the house and
gained by the outside in a reversible fashion.
SOLUTION Since heat is lost from the hot reservoir (inside the house), the change in
entropy is negative: ∆SH = −QH/TH. Since heat is gained by the cold reservoir (the
outdoors), the change in entropy is positive: ∆SC = +QC/TC. Here we are using the symbols
QH and QC to denote the magnitudes of the heats. The change in the entropy of the universe
is
∆Suniverse = ∆SC + ∆SH =
QC
TC
−
QH 24 500 J 24 500 J
=
−
= 11.6 J/K
258 K
294 K
TH
In this calculation we have used the fact that TC = 273 − 15 °C = 258 K and
TH = 273 + 21 °C = 294 K.
______________________________________________________________________________
69. REASONING AND SOLUTION Equation 15.19 gives the unavailable work as
Wunavailable = T0 ∆S
(1)
Chapter 15 Problems
787
where T0 = 248 K . We also know that Wunavailable = 0.300 Q . Furthermore, we can apply
Equation 15.18 to the heat lost from the 394-K reservoir and the heat gained by the reservoir
at temperature T, with the result that
–Q
Q
∆S =
+
394 K
T
With these substitutions for T0, Wunavailable, and ∆S, Equation (1) becomes
⎛ –Q
Q⎞
0.300 Q = ( 248 K ) ⎜
or
T = 267 K
+ ⎟
394
K
T
⎝
⎠
______________________________________________________________________________
70. REASONING The total entropy change ∆Suniverse of the universe is the sum of the entropy
changes of the hot and cold reservoir. For each reservoir, the entropy change is given by
⎛Q⎞
∆S = ⎜ ⎟ (Equation 15.18). As indicated by the label “R,” this equation applies only to
⎝ T ⎠R
reversible processes. For the two irreversible engines, therefore, we apply this equation to
an imaginary process that removes the given heat from the hot reservoir reversibly and
rejects the given heat to the cold reservoir reversibly. According to the second law of
thermodynamics stated in terms of entropy (see Section 15.11), the reversible engine is the
one for which ∆Suniverse = 0 J/K , and the irreversible engine that could exist is the one for
which ∆Suniverse > 0 J/K . The irreversible engine that could not exist is the one for which
∆Suniverse < 0 J/K .
SOLUTION Using Equation 15.18, we write the total entropy change of the universe as the
sum of the entropy changes of the hot (H) and cold (C) reservoirs.
∆Suniverse =
− QH
TH
+
QC
TC
In this expression, we have used − QH for the heat from the hot reservoir because that
reservoir loses heat. We have used + QC for the heat rejected to the cold reservoir because
that reservoir gains heat. Applying this expression to the three engines gives the following
results:
788 THERMODYNAMICS
Engine I
∆S universe =
Engine II
∆S universe =
Engine III
∆S universe =
− QH
TH
− QH
TH
− QH
TH
+
QC
+
QC
+
QC
TC
TC
TC
=
−1650 J 1120 J
+
= +0.4 J/K
550 K
330 K
=
−1650 J 990 J
+
= 0 J/K
550 K 330 K
=
−1650 J 660 J
+
= −1.0 J/K
550 K 330 K
Since ∆Suniverse = 0 J/K for Engine II, it is reversible. Since ∆Suniverse > 0 J/K for Engine I,
it is irreversible and could exist. Since ∆Suniverse < 0 J/K for Engine III, it is irreversible
and could not exist.
71. SSM REASONING AND SOLUTION
a. Since the energy that becomes unavailable for doing work is zero for the process, we
have from Equation 15.19, Wunavailable = T0 ∆S universe = 0 . Therefore, ∆Suniverse = 0 and
according to the discussion in Section 15.11, the process is reversible .
b. Since the process is reversible, we have (see Section 15.11)
∆Suniverse = ∆Ssystem + ∆Ssurroundings = 0
Therefore,
∆Ssurroundings = −∆Ssystem = –125 J/K
______________________________________________________________________________
72. REASONING AND SOLUTION
a. We know that the hot and cold waters exchange equal amounts of heat, i.e.,
∆Qhw = ∆Qcw, so that (mc∆T)hw = (mc∆T)cw, or
(1.00 kg)[4186 J/(kg⋅C°)](373 K − Tf) = (2.00 kg)[4186 J/(kg⋅C°)](Tf − 283 K)
Solving for Tf, we obtain Tf = 313K .
b. Since ∆S = mc ln(Tf/Ti):
∆Shw = mhwc ln[(313 K)/(373 K)] = −734 J/K
∆Scw = mcwc ln[(313 K)/(283 K)] = +844 J/K
Chapter 15 Problems
789
Therefore,
∆Suniverse = ∆Shw + ∆Scw = 1.10 × 102 J/K
c. The energy unavailable for doing work is, therefore,
Wunavailable = T0 ∆Suniverse = (273 K)(1.10 × 102 J/K) = 3.00 × 104 J
______________________________________________________________________________
73. REASONING The change ∆Suniverse in entropy of the universe for this process is the sum
of the entropy changes for (1) the warm water (∆Swater) as it cools down from its initial
temperature of 85.0 °C to its final temperature Tf , (2) the ice (∆Sice) as it melts at 0 °C, and
(3) the ice water (∆Sice
water
) as it warms up from 0 °C to the final temperature Tf:
∆Suniverse = ∆Swater + ∆Sice + ∆Sice water.
To find the final temperature Tf , we will follow the procedure outlined in Sections 12.7 and
12.8, where we set the heat lost by the warm water as it cools down equal to the heat gained
by the melting ice and the resulting ice water as it warms up. The heat Q that must be
supplied or removed to change the temperature of a substance of mass m by an amount ∆T is
Q = cm∆T (Equation 12.4), where c is the specific heat capacity. The heat that must be
supplied to melt a mass m of a substance is Q = mLf (Equation 12.5), where Lf is the latent
heat of fusion.
SOLUTION
a. We begin by finding the final temperature Tf of the water. Setting the heat lost equal to
the heat gained gives
cmwater ( 85.0 °C − Tf ) =
mice Lf + micec (Tf − 0.0 °C )
Heat lost by water
Heat gained
by melting ice
Heat gained by ice water
Solving this relation for the final temperature Tf yields
Tf =
cmwater ( 85.0 °C ) − mice Lf
c ( mice + mwater )
(
)
⎡⎣ 4186 J/ ( kg ⋅ C° ) ⎤⎦ ( 6.00 kg )( 85.0 °C ) − ( 3.00 kg ) 33.5 × 104 J/kg
=
= 30.0 °C
⎡⎣ 4186 J/ ( kg ⋅ C° ) ⎤⎦ ( 3.00 kg + 6.00 kg )
790 THERMODYNAMICS
We have taken the specific heat capacity of 4186 J/ ( kg ⋅ C° ) for water from Table 12.2 and
the latent heat of 33.5 ×104 J/kg from Table 12.3. This temperature is equivalent to
Tf = (273 + 30.0 °C) = 303 K.
The change ∆Suniverse in the entropy of the universe is the sum of three contributions:
[Contribution 1]
⎛T
∆S water = mwater c ln ⎜ f
⎜T
⎝ i
⎞
⎛ 303 K ⎞
⎟⎟ = (6.00 kg)[4186 J/ ( kg ⋅ C° ) ] ln ⎜
⎟ = − 4190 J/K
358
K
⎝
⎠
⎠
where Ti = 273 + 85.0 °C = 358 K.
[Contribution 2]
∆Sice
mLf
Q
=
=
=
T
T
( 3.00 kg ) ( 33.5 ×104 J/kg )
273 K
= +3680 J/K
[Contribution 3]
⎛T
∆Sice water = mice c ln ⎜⎜ f
⎝ Ti
⎞
⎛ 303 K ⎞
⎟⎟ = (3.00 kg)[4186 J/ ( kg ⋅ C° ) ] ln ⎜
⎟ = + 1310 J/K
273
K
⎝
⎠
⎠
The change in the entropy of the universe is
∆Suniverse = ∆Swater + ∆Sice + ∆Sice water = +8.0 × 102 J/K
b. The entropy of the universe increases , because the mixing process is irreversible.
______________________________________________________________________________
74. REASONING An adiabatic process is one for which no heat enters or leaves the system, so
Q = 0 J. The work is given as W = +610 J, where the plus sign denotes that the gas does
work, according to our convention. Knowing the heat and the work, we can use the first law
of thermodynamics to find the change ∆U in internal energy as ∆U = Q – W (Equation
15.1). Knowing the change in the internal energy, we can find the change in the
temperature by recalling that the internal energy of a monatomic ideal gas is U = 23 nRT,
according to Equation 14.7. As a result, it follows that ∆U =
3
2
nR∆T.
SOLUTION Using the first law from Equation 15.5 and the change in internal energy from
Equation 14.7, we have
∆U = Q − W
Therefore, we find
or
3
2
nR∆T = Q − W
Chapter 15 Problems
∆T =
b
2 Q −W
3nR
g=
b g b g
3b 0.50 mol g 8.31 J / b mol ⋅ K g
2 0 J − 610 J
791
= −98 K
The change in temperature is a decrease.
______________________________________________________________________________
75. SSM REASONING AND SOLUTION
a. Since the temperature of the gas is kept constant at all times, the process is isothermal;
therefore, the internal energy of an ideal gas does not change and ∆U = 0 .
b. From the first law of thermodynamics (Equation 15.1), ∆U = Q − W . But ∆U = 0 , so
that Q = W . Since work is done on the gas, the work is negative, and Q = −6.1× 103 J .
c. The work done in an isothermal compression is given by Equation 15.3:
⎛V
W = nRT ln ⎜⎜ f
⎝ Vi
Therefore, the temperature of the gas is
T=
W
(
nR ln Vf / Vi
)
=
⎞
⎟⎟
⎠
–6.1×103 J
= 310 K
(3.0 mol)[8.31 J/(mol ⋅ K)] ln ⎡⎣ (2.5 × 10 –2 m3 )/(5.5 × 10 –2 m3 ) ⎤⎦
76. REASONING According to Equation 15.11, the efficiency e of a heat engine is
e = W / QH , where W is the magnitude of the work done by the engine and QH is the
magnitude of the input heat that the engine uses.
QH is given, but W is unknown.
However, energy conservation requires that QH = W + QC (Equation 15.12), where QC
is the magnitude of the heat rejected by the engine and is given. From this equation,
therefore, a value for W can be obtained.
SOLUTION According to Equation 15.11, the efficiency is
e=
Since
QH = W + QC
W
QH
(Equation 15.12), we can solve for
W
W = QH − QC . Substituting this result into the efficiency expression gives
e=
W
QH
=
QH − QC
QH
=
5.6 ×104 J − 1.8 ×104 J
= 0.68
5.6 × 104 J
to show that
792 THERMODYNAMICS
77. REASONING AND SOLUTION The work done by the expanding gas is
W = Q − ∆U = 2050 J − 1730 J = 320 J
The work, according to Equation 6.1, is also the magnitude F of the force exerted on the
piston times the magnitude s of its displacement. But the force is equal to the weight mg of
the block and piston, so that the work is W = Fs = mgs. Thus, we have
s=
320 J
W
=
= 0.24 m
mg
135 kg 9.80 m / s 2
b
gc
h
______________________________________________________________________________
78. REASONING For any refrigerator, the first law of thermodynamics (Equation 15.12)
indicates that W = QH − QC . In this expression, we know that W = 2500 J and wish to
find QC . To do so, we need information about QC . But the refrigerator is a Carnot
device, so we know in addition that QC / QH = TC / TH (Equation 15.14).
With this
additional equation, we can solve for QH and substitute into the first law, obtaining in the
process an equation that contains only QC as an unknown.
SOLUTION From Equation 15.14 we have
QC
QH
=
TC
or
TH
⎛T
QH = QC ⎜ H
⎜T
⎝ C
⎞
⎟⎟
⎠
Substituting this expression for QH into the first law of thermodynamics gives
⎛T
W = QH − QC = QC ⎜ H
⎜T
⎝ C
⎞
⎟⎟ − QC
⎠
Solving for QC , we find
QC =
W
=
2500 J
= 3.1× 104 J
299 K
−1
277 K
TH
−1
TC
______________________________________________________________________________
79. SSM REASONING AND SOLUTION The change in entropy ∆S of a system for a
process in which heat Q enters or leaves the system reversibly at a constant temperature T is
given by Equation 15.18, ∆S = (Q / T ) R . For a phase change, Q = mL , where L is the
latent heat (see Section 12.8).
Chapter 15 Problems
793
a. If we imagine a reversible process in which 3.00 kg of ice melts into water at 273 K, the
change in entropy of the water molecules is
(
)
(3.00 kg) 3.35 ×105 J/kg
⎛ mLf ⎞
⎛Q⎞
∆S = ⎜ ⎟ = ⎜
= 3.68 × 103 J/K
⎟ =
273 K
⎝ T ⎠R ⎝ T ⎠R
b. Similarly, if we imagine a reversible process in which 3.00 kg of water changes into
steam at 373 K, the change in entropy of the water molecules is
(
)
(3.00 kg) 2.26 ×106 J/kg
⎛ mLv ⎞
⎛Q⎞
∆S = ⎜ ⎟ = ⎜
= 1.82 × 104 J/K
⎟ =
373 K
⎝ T ⎠R ⎝ T ⎠R
c. Since the change in entropy is greater for the vaporization process than for the fusion
process, the vaporization process creates more disorder in the collection of water
molecules.
______________________________________________________________________________
80. REASONING AND SOLUTION The amount of heat removed at constant volume is
Q = nCV ∆T = ( 2.5 mol )
( 32 R ) ( 35 K ) =
1100 J
______________________________________________________________________________
81. REASONING AND SOLUTION Suppose this device were a Carnot engine instead of a
heat pump. We know that its efficiency e would be
e = 1 − (TC/TH) = 1 − [(265 K)/(298 K)] = 0.111
The efficiency, however, is also given by
e=
W
QH
Since the heat pump’s coefficient of performance COP is COP = QH / W , we have that
1
1
=
= 9.03
W
e
0.111
______________________________________________________________________________
COP =
QH
=
82. REASONING The smallest possible temperature of the hot reservoir would occur when the
engine is a Carnot engine, since it has the greatest efficiency of any engine operating
between the same hot and cold reservoirs. The efficiency eCarnot of a Carnot engine is (see
794 THERMODYNAMICS
Equation 15.15) eCarnot = 1 − (TC/TH), where TC and TH are the Kelvin temperatures of its
cold and hot reservoirs. Solving this equation for TH gives TH = TC/(1 − eCarnot). We are
given TC, but do not know eCarnot. However, the efficiency is defined as the magnitude W
of the work done by the engine divided by the magnitude QH of the input heat from the hot
reservoir, so eCarnot = W / QH (Equation 15.11). Furthermore, the conservation of energy
requires that the magnitude QH of the input heat equals the sum of the magnitude W of
the work done by the engine and the magnitude QC of the heat it rejects to the cold
reservoir, QH = W + QC . By combining these relations, we will be able to find the
temperature of the hot reservoir of the Carnot engine.
SOLUTION From the Reasoning section, the temperature of the hot reservoir is
TH = TC/(1 − eCarnot). Writing the efficiency of the engine as eCarnot = W / QH , the
expression for the temperature becomes
TH =
TC
1 − eCarnot
=
TC
W
1−
QH
From the conservation of energy, we have that QH = W + QC . Substituting this expression
for QH into the one above for TH gives
TH =
TC
W
1−
W
=
1−
TC
W
=
W + QC
285 K
= 1090 K
18 500 J
1−
18 500 J + 6550 J
______________________________________________________________________________
83. SSM REASONING The work done in the process is equal to the "area" under the curved
line between A and B in the drawing. From the graph, we find that there are about 78
"squares" under the curve. Each square has an "area" of
( 2.0 ×104 Pa )( 2.0 ×10−3 m3 ) = 4.0 ×101 J
SOLUTION
a. The work done in the process has a magnitude of
(
)
W = ( 78 ) 4.0 ×101 J = 3100 J
b. The final volume is smaller than the initial volume, so the gas is compressed. Therefore,
work is done on the gas so the work is negative .
Chapter 15 Problems
795
______________________________________________________________________________
84. REASONING AND SOLUTION
a. The work is the area under the path ACB. There are 48 "squares" under the path, so that
(
)(
)
W = − 48 2.0 × 104 Pa 2.0 × 10 –3 m3 = −1900 J
The minus sign is included because the gas is compressed, so that work is done on it. Since
there is no temperature change between A and B (the line AB is an isotherm) and the gas is
ideal, ∆U = 0, so
Q = ∆U + W = W = −1900 J
b. The negative answer for W means that heat flows out of the gas.
______________________________________________________________________________
85. SSM REASONING According to Equations 15.6 and 15.7, the heat supplied to a
monatomic ideal gas at constant pressure is Q = CP n ∆T , with CP = 52 R . Thus,
Q = 52 nR ∆T . The percentage of this heat used to increase the internal energy by an amount
∆U is
⎛ ∆U
⎛ ∆U ⎞
Percentage = ⎜
⎟ × 100 % = ⎜⎜ 5
⎝ Q ⎠
⎝ 2 nR ∆T
⎞
⎟⎟ × 100 %
⎠
(1)
But according to the first law of thermodynamics, ∆U = Q − W . The work W is W = P ∆V ,
and for an ideal gas P ∆V = nR ∆T . Therefore, the work W becomes W = P ∆V = nR ∆T
and the change in the internal energy is ∆U = Q − W = 52 nR ∆T − nR∆T = 32 nR∆T .
Combining this expression for ∆U with Equation (1) above yields a numerical value for the
percentage of heat being supplied to the gas that is used to increase its internal energy.
SOLUTION
a. The percentage is
⎛ ∆U
Percentage = ⎜ 5
⎜ nR ∆T
⎝2
⎞
⎟⎟ × 100 % =
⎠
⎛ 32 nR ∆T
⎜⎜ 5
⎝ 2 nR ∆T
⎞
⎛ 3⎞
⎟⎟ × 100 % = ⎜ ⎟ × 100 % = 60.0 %
⎝5⎠
⎠
b. The remainder of the heat, or 40.0 % , is used for the work of expansion.
______________________________________________________________________________
796 THERMODYNAMICS
86. REASONING AND SOLUTION The magnitude of the heat removed from the ice QC is
QC = mLf = (2.0 kg)(33.5 × 104 J/kg) = 6.7 × 105 J
The magnitude of the heat leaving the refrigerator QH is therefore,
QH = QC (TH/TC) = (6.7 × 105 J)(300 K)/(258 K) = 7.8 × 105 J
The magnitude of the work done by the refrigerator is therefore,
W = QH − QC = 1.1 × 105 J
At $0.10 per kWh (or $0.10 per 3.6 × 106 J), the cost is
(
)
$0.10
1.1 × 105 J = $3.0 × 10−3 = 0.30 cents
6
3.6 × 10 J
______________________________________________________________________________
87. REASONING During an adiabatic process, no heat flows into or out of the gas (Q = 0 J).
For an ideas gas, the final pressure and volume (Pf and Vf) are related to the initial pressure
γ
γ
and volume (Pi and Vi) by PV
i i = Pf Vf
( Equation 15.5) ,
where γ is the ratio of the specific
heat capacities at constant pressure and constant volume (γ =
7
5
in this problem). The initial
and final pressures are not given. However, the initial and final temperatures are known, so
we can use the ideal gas law, PV = nRT (Equation 14.1) to relate the temperatures to the
pressures. We will then be able to find Vi/Vf in terms of the initial and final temperatures.
γ
γ
SOLUTION Substituting the ideal gas law, PV = nRT, into PV
i i = Pf Vf gives
⎛ nRTi
⎜⎜
⎝ Vi
⎞ γ ⎛ nRTf
⎟⎟ Vi = ⎜⎜
⎠
⎝ Vf
⎞ γ
⎟⎟ Vf
⎠
or
Ti Viγ −1 = Tf Vfγ −1
Solving this expression for the ratio of the initial volume to the final volume yields
Vi ⎛ Tf
=⎜ ⎟
Vf ⎜⎝ Ti ⎟⎠
1
⎞ γ −1
The initial and final Kelvin temperatures are Ti = (21 °C + 273) = 294 K and
Tf = (688 °C + 273) = 961 K. The ratio of the volumes is
Chapter 15 Problems
1
797
1
Vi ⎛ Tf ⎞ γ −1 ⎛ 961 K ⎞ ( 75 − 1)
= 19.3
= ⎜ ⎟ =⎜
⎟
Vf ⎜⎝ Ti ⎟⎠
⎝ 294 K ⎠
______________________________________________________________________________
88. REASONING The change in the internal energy of the gas can be found using the first law
of thermodynamics, since the heat added to the gas is known and the work can be calculated
by using Equation 15.2, W = P ∆V. The molar specific heat capacity at constant pressure can
be evaluated by using Equation 15.6 and the ideal gas law.
SOLUTION
a. The change in the internal energy is
∆U = Q − W = Q − P ∆V
(
)(
)
= 31.4 J − 1.40 × 104 Pa 8.00 × 10−4 m3 − 3.00 × 10−4 m3 = 24.4 J
b. According to Equation 15.6, the molar specific heat capacity at constant pressure is
CP = Q/(n ∆T). The term n ∆T can be expressed in terms of the pressure and change in
volume by using the ideal gas law:
P ∆V = n R ∆T
or
n ∆T = P ∆V/R
Substituting this relation for n ∆T into CP = Q/(n ∆T), we obtain
Q
31.4 J
=
= 37.3 J/(mol ⋅ K)
P∆V
1.40 × 104 Pa 5.00 × 10−4 m3
R
R
______________________________________________________________________________
CP =
(
)(
)
89. REASONING The heat added is given by Equation 15.6 as Q = CV n ∆T, where CV is the
molar specific heat capacity at constant volume, n is the number of moles, and ∆T is the
change in temperature. But the heat is supplied by the heater at a rate of ten watts, or ten
joules per second, so Q = (10.0 W)t, where t is the on-time for the heater. In addition, we
know that the ideal gas law applies: PV = nRT (Equation 14.1). Since the volume is
constant while the temperature changes by an amount ∆T, the amount by which the pressure
changes is ∆P. This change in pressure is given by the ideal gas law in the form
(∆P)V = nR(∆T).
SOLUTION Using Equation 15.6 and the expression Q = (10.0 W)t for the heat delivered
by the heater, we have
C n∆T
Q = CV n∆T or (10.0 W ) t = CV n∆T or t = V
10.0 W
798 THERMODYNAMICS
Using the ideal gas law in the form (∆P)V = nR(∆T), we can express the change in
temperature as ∆T = (∆P)V/nR. With this substitution for ∆T, the expression for the time
becomes
C n ∆P V
t= V
10.0 W nR
b g
b
g
According to Equation 15.8, CV =
3
2
R for a monatomic ideal gas, so we find
(
)(
)
3 5.0 × 104 Pa 1.00 ×10−3 m3
3 ( ∆P ) V
3Rn ( ∆P ) V
t=
=
=
= 7.5 s
2 (10.0 W ) nR 2 (10.0 W )
2 (10.0 W )
____________________________________________________________________________________________
90. REASONING AND SOLUTION From Equation 15.14 we know that the magnitude of the
input heat QH and the magnitude of the exhaust heat QC of a Carnot engine are related to
the Kelvin temperatures of the hot and cold reservoirs according to QH / QC = TH / TC . We
also know that QC = QH − W , according to Equation 15.12.
a. Therefore, we find T1 as follows:
QH
QC
=
T
5550 J
= 1
5550 J − 1750 J 503 K
T1 = 735 K
or
b. Similarly, we find T2 as follows:
QH
QC
=
5550 J − 1750 J
503 K
=
5550 J − 1750 J − 1750 J
T2
or
T2 = 271 K
______________________________________________________________________________
91. SSM REASONING AND SOLUTION We wish to find an expression for the overall
efficiency e in terms of the efficiencies e1 and e2 . From the problem statement, the overall
efficiency of the two-engine device is
W + W2
e= 1
(1)
QH
where QH is the input heat to engine 1. The efficiency of a heat engine is defined by
Equation 15.11, e = W / QH , so we can write
W1 = e1 QH
and
W2 = e2 QH2
(2)
Chapter 15 Problems
799
Since the heat rejected by engine 1 is used as input heat for the second engine, QH2 = QC1 ,
and the expression above for W2 can be written as
W2 = e2 QC1
(3)
According to Equation 15.12, we have QC1 = QH – W1 , so that Equation (3) becomes
(
W2 = e2 Q − W
H
1
)
(4)
Substituting Equations (2) and (4) into Equation (1) gives
e=
(
e1 QH + e2 QH − W1
QH
) = e1 QH + e2 ( QH − e1 QH )
QH
Algebraically canceling the QH 's in the right hand side of the last expression gives the
desired result:
e = e1 + e2 − e1e2
______________________________________________________________________________
92. REASONING AND SOLUTION Let the left be side 1 and the right be side 2. Since the
partition moves to the right, side 1 does work on side 2, so that the work values involved
satisfy the relation W1 = −W2 . Using Equation 15.4 for each work value, we find that
3
nR
2
(T1i –T1f ) = − 32 nR (T2i –T2f )
or
T1f + T2f = T1i + T2i = 525 K + 275 K = 8.00 × 102 K
We now seek a second equation for the two unknowns T1f and T2f . Equation 15.5 for an
adiabatic process indicates that P1iV1iγ = P1f V1fγ and P2iV2iγ = P2f V2fγ . Dividing these two
equations and using the facts that V1i = V2i and P1f = P2f , gives
P1iV1iγ
P2iV2iγ
=
P1f V1fγ
P2f V2fγ
or
P1i ⎛ V1f
=⎜
P2i ⎜⎝ V2f
γ
⎞
⎟
⎟
⎠
Using the ideal gas law, we find that
P1i ⎛ V1f ⎞
=⎜
⎟
P2i ⎜⎝ V2f ⎟⎠
γ
becomes
nRT1i /V1i ⎛ nRT1f /P1f
=⎜
nRT2i /V2i ⎜⎝ nRT2f /P2f
Since V1i = V2i and P1f = P2f , the result above reduces to
⎞
⎟⎟
⎠
γ
800 THERMODYNAMICS
⎛T
= ⎜⎜ 1f
T2i ⎝ T2f
T1i
γ
⎞
⎟⎟
⎠
or
T1f
T2f
1/ γ
⎛T ⎞
= ⎜⎜ 1i ⎟⎟
⎝ T2i ⎠
1/ γ
⎛ 525 K ⎞
=⎜
⎟
⎝ 275 K ⎠
= 1.474
Using this expression for the ratio of the final temperatures in T1f + T2f = 8.00 × 102 K , we
find that
a. T1f = 477 K
and
b. T2f = 323 K
______________________________________________________________________________
93. CONCEPT QUESTIONS
a. By itself, the work would decrease the internal energy of the system. This is because the
system does work and would use some of its internal energy in the process.
b. By itself, the heat would increase the internal energy of the system, because it flows into
the system. Thus, it would add to the supply of internal energy that the system already has.
c. The conservation-of-energy principle indicates that energy can neither be created nor
destroyed, but can only be converted from one form to another. Therefore, the internal
energy of the system increases, because more energy enters the system as heat than leaves
the system as work.
SOLUTION Using the first law of thermodynamics (conservation of energy) from Equation
15.1, we obtain
(
) (
)
∆U = Q − W = 7.6 × 104 J − 4.8 × 104 J = +2.8 × 104 J
The plus sign indicates that the internal energy increases, as expected.
______________________________________________________________________________
94. CONCEPT QUESTIONS
a. The energy gain in the form of heat means that the internal energy of the system would
increase by an equal amount in the absence of work. This follows from the first law of
thermodynamics:∆U = Q – W (Equation 15.1). But the internal energy increases by an even
greater amount, which means that energy also enters the system because work is being done
on it.
b. According to our convention, work done on the system is negative.
c. The volume of the system decreases. This is because work is done on the system. In
other words, the environment is pushing inward on the system, compressing it.
Alternatively, Equation 15.2 indicates that work W done at constant pressure P is W = P∆V,
where ∆V is the change in volume. Since we know that the work is negative, the change in
volume must also be negative. But ∆V = Vf – Vi, so the final volume Vf is less than the
initial volume Vi.
Chapter 15 Problems
801
SOLUTION Using Equations 15.1 (∆U = Q – W) and 15.2 (W = P∆V), we get
∆U = Q − W = Q − P∆V
Solving for ∆V gives
∆V =
Q − ∆U ( 2780 J ) − ( 3990 J )
=
= −9.60 × 10−3 m3
5
P
1.26 × 10 Pa
As expected, ∆V is negative, reflecting a decrease in volume.
______________________________________________________________________________
95. CONCEPT QUESTIONS
a. The internal energy of an ideal gas remains the same during an isothermal process. The
temperature is constant in an isothermal process, and the internal energy of an ideal gas is
proportional to the kelvin temperature, as Section 14.3 discusses. Since the temperature is
constant, the internal energy is constant.
b. The work done is equal to the heat that flows into the gas. According to the first law of
thermodynamics, the change ∆U in the internal energy is given by Equations 15.1 as
∆U = Q – W. Since the internal energy U is constant, we have ∆U = 0 = Q – W, or W = Q.
SOLUTION According to Equation 15.3, the work done in an isothermal process involving
an ideal gas is
⎛V ⎞
W
W = nRT ln ⎜⎜ f ⎟⎟ or T =
⎛V ⎞
⎝ Vi ⎠
nR ln ⎜ f ⎟
⎝ Vi ⎠
Since W = Q for an isothermal process utilizing an ideal gas, we find
4.75 × 103 J
Q
=
= 208 K
⎛ Vf ⎞
⎛ 0.250 m3 ⎞
nR ln ⎜ ⎟ ( 3.00 mol ) ⎡⎣8.31 J/ ( mol ⋅ K ) ⎤⎦ ln ⎜
3⎟
⎝ 0.100 m ⎠
⎝ Vi ⎠
______________________________________________________________________________
T=
96. CONCEPT QUESTIONS
a. The 4.1 × 106 J of energy equals the magnitude QH of the input heat. It makes no sense
for this energy to be the magnitude W of the work, because then the work would be
independent of the climb. Common sense indicates that more work is done when the climb
is through a greater versus a smaller height.
b. The work done in climbing upward is related to the vertical height of the climb via the
work-energy theorem (Equation 6.8), which is
802 THERMODYNAMICS
c
h
Wnc = KE f + PE f − KE 0 + PE 0
Final total
mechanical energy
Initial total
mechanical energy
Here, Wnc is the net work done by nonconservative forces, in this case the work done by the
climber in going upward. Since the climber starts at rest and finishes at rest, the final
kinetic energy KEf and the initial kinetic energy KE0 are zero. As a result, we have
Wnc = W = PEf – PE0, where PEf and PE0 are the final and initial gravitational potential
energies, respectively. Equation 6.5 gives the gravitational potential energy as PE = mgh,
where h is the vertical height. Taking the height at her starting point to be zero, we then
have Wnc = W = mgh.
SOLUTION Using Equation 15.11 for the efficiency and relating the work to the height via
the work-energy theorem, we find
(
)
2
mgh ( 52 kg ) 9.80 m/s ( 730 m )
e=
=
=
= 0.091
QH
QH
4.1×106 J
______________________________________________________________________________
W
97. CONCEPT QUESTIONS
a. According to Equation 15.11, the efficiency of a heat engine is e = W / QH , where W
is the magnitude of the work and QH is the magnitude of the input heat. Thus, the
magnitude of the work is W = e QH . The engine that delivers more work for a given heat
input is the engine with the higher efficiency. In this case, that is engine B.
b. The efficiency of a Carnot engine is given by Equation 15.15 as eCarnot = 1 – TC/TH.
Smaller values of the cold-reservoir temperature TC mean greater efficiencies for a given
value of the hot-reservoir temperature TH. Thus, the engine with the greater efficiency has
the lower cold-reservoir temperature. In this case, that is engine B.
SOLUTION Using Equation 15.11, we find the work delivered by each engine as follows:
Engine A
W = e QH = ( 0.60 )(1200 J ) = 720 J
Engine B
W = e QH = ( 0.80 )(1200 J ) = 960 J
Equation 15.15 for the efficiency eCarnot of a Carnot engine can be solved for the
temperature of the cold reservoir:
T
eCarnot = 1 − C
or TC = (1 − eCarnot ) TH
TH
Chapter 15 Problems
803
Applying this result to each engine gives
Engine A
TC = (1 − 0.60 )( 650 K ) = 260 K
Engine B
TC = (1 − 0.80 )( 650 K ) = 130 K
As expected, engine B delivers more work and has the lower cold-reservoir temperature.
______________________________________________________________________________
98. CONCEPT QUESTIONS
a. An air conditioner removes heat from a room by doing work to make the heat flow up the
temperature “hill” from cold to hot. More work is required to remove a given amount of
heat when the temperature difference against which the air conditioner is working is greater.
Here the hot temperature outside is the same for each unit, but the room serviced by unit A
is kept colder. Thus, unit A must work against the greater temperature difference and uses
more work than unit B.
b. The heat deposited outside is equal to the heat removed plus the work done. Since both
units remove the same amount of heat, the unit that deposits more heat outside is the unit
that uses the greater amount of work. That is unit A.
SOLUTION According to the first law of thermodynamics (Equation 15.12), we know that
W = QH − QC . In addition, since the air conditioners are Carnot devices, we know that
the ratio of the magnitude of the heat QC
removed from the cold reservoir to the
magnitude of the heat QH deposited in the hot reservoir is equal to the ratio of the reservoir
temperatures or QC / QH = TC / TH (Equation 15.14). Using these two equations, we have
⎛Q
⎞
⎛T
⎞
W = QH − QC = ⎜ H − 1⎟ QC = ⎜ H − 1⎟ QC
⎜T
⎟
⎜Q
⎟
⎝ C
⎠
⎝ C
⎠
Applying this result to each air conditioner gives
Unit A
⎛ 309.0 K ⎞
− 1⎟ ( 4330 J ) = 220 J
W =⎜
⎝ 294.0 K ⎠
Unit B
⎛ 309.0 K ⎞
− 1⎟ ( 4330 J ) = 120 J
W =⎜
⎝ 301.0 K ⎠
We can find the heat deposited outside directly from Equation 15.14 by solving it for QH .
804 THERMODYNAMICS
QC
QH
=
TC
TH
or
⎛T
QH = ⎜ H
⎜T
⎝ C
⎞
⎟⎟ QC
⎠
Applying this result to each air conditioner gives
Unit A
⎛ 309.0 K ⎞
QH = ⎜
⎟ ( 4330 J ) = 4550 J
⎝ 294.0 K ⎠
Unit B
⎛ 309.0 K ⎞
QH = ⎜
⎟ ( 4330 J ) = 4450 J
⎝ 301.0 K ⎠
As expected, unit A uses more work and deposits more heat outside.
______________________________________________________________________________
Chapter 15 Problems
B
C
Pressure
99. CONCEPT QUESTIONS
a. There is no work done for the process
A→B. The reason is that the volume is
constant (see the drawing), which means
that the change ∆V in the volume is
zero. In other words, the area under the
plot of pressure versus volume is zero,
and we see that WA→B = 0 J.
805
A
b. If the change ∆UB→C in the internal
energy of the gas and the work WB→C
Volume
are known for the process B→C, the heat QB→C can be determined by using the first law of
thermodynamics as QB→C = ∆UB→C + WB→C (Equation 15.1).
c. Yes. For the process C→A it is possible to find the change in the internal energy of the
gas if the change in the internal energies for the processes A→B and B→C are known. The
total change ∆Utotal in the internal energy for the three processes is
∆Utotal = ∆UA→B + ∆UB→C + UC→A. We can use this equation to find ∆UC→A. The
quantities ∆UA→B and ∆UB→C are known. We also know ∆Utotal, which is the change in the
internal energy for the total process A→B→C→A. This process begins and ends at the same
place on the pressure-versus-volume plot. Therefore, the value of the internal energy U is
the same at the start and the end, with the result that ∆Utotal = 0 J.
SOLUTION
a. Since the change in volume is zero (∆V = 0 m3), the area under the plot of pressure versus
volume is zero, with the result that the work is WA→B = 0 J .
b. The change in the internal energy of the gas for the process A→B is ∆UA→B =
QA→B − WA→B = +561 J − 0 J = +561 J .
c. According to the first law of thermodynamics, the change in the internal energy of the gas
for the process B→C is ∆UB→C = QB→C − WB→C. Thus, the heat added to the gas is QB→C
= ∆UB→C + WB→C = +4303 J + 2867 J = +7170 J .
d. According to the first law of thermodynamics, the change ∆Utotal in the total internal
energy for the three processes is ∆Utotal = ∆UA→B + ∆UB→C + ∆UC→A. Solving this
equation for ∆UC→A gives ∆UC→A = ∆Utotal − ∆UA→B − ∆UB→C. As discussed in part c of
the Concept Questions, ∆Utotal = 0 J, since the third process ends at point A, which is the
start of the first process. Therefore, ∆UC→A = 0 J − 561 J − 4303 J = −4864 J .
806 THERMODYNAMICS
e. According to the first law of thermodynamics, the change in the internal energy of the gas
for the process C→A is ∆UC→A = QC→A − WC→A. The heat removed during this process is
QC→A = ∆UC→A + WC→A = −4864 J −3740 J = −8604 J .
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100. CONCEPT QUESTIONS
a. According to the discussion in Section 15.11, the change ∆Suniverse in the entropy of the
universe is the sum of the change in entropy ∆SH of the hot reservoir and the change in
entropy ∆SC of the cold reservoir, so ∆Suniverse = ∆SH + ∆SC. The change in entropy of each
reservoir is given by Equation 15.18 as ∆S = (Q/T)R . The engine is irreversible, so we must
imagine a process in which the heat Q is added to or removed from the reservoirs reversibly.
T is the Kelvin temperature of a reservoir. Since heat is lost from the hot reservoir, the
change in entropy is negative: ∆SH = − QH /TH. Since heat is gained by the cold reservoir,
the change in entropy is positive: ∆SC = + QC /TC. The change in entropy of the universe is
∆Suniverse = ∆SH + ∆SC = −
QH
TH
+
QC
TC
.
b. The change in the entropy of the universe is greater than zero, ∆Suniverse > 0 J/K , as it
must be for any irreversible process. (This is the second law of thermodynamics stated in
terms of entropy.)
c. When a reversible engine (a Carnot engine) operates between the same hot and cold
temperatures as the irreversible engine, the reversible engine produces more work, assuming
that the magnitude QH of the input heat to both engines is the same. This is because the
reversible engine is more efficient, according to the second law of thermodynamics.
d. The difference in the work produced by the two engines is labeled Wunavailable in
Section15.11, where Wunavailable = Wreversible − Wirreversible. The difference in the work is
related to the change in the entropy of the universe by Wunavailable = T0 ∆Suniverse (Equation
15.19), where T0 is the Kelvin temperature of the coldest heat reservoir. In this case T0 = TC.
SOLUTION
a. From part a of the Concept Questions, the change in entropy of the universe is
∆Suniverse = −
QH
TH
+
QC
TC
Chapter 15 Problems
807
The magnitude QC of the heat rejected to the cold reservoir is related to the magnitude
QH of the heat supplied to the engine from the hot reservoir and the magnitude W of the
work done by the engine via QC = QH − W (Equation 15.1). Thus, ∆Suniverse becomes
∆Suniverse = −
QH
TH
+
QH − W
TC
=−
1285 J 1285 J − 264 J
+
= +1.74 J/K
852 K
314 K
As expected, the entropy of the universe increases when an irreversible process occurs.
b. The magnitude W of the work done by any engine depends on its efficiency e and input
heat QH via W = e QH (Equation 15.11). For a reversible engine, the efficiency is related
to the temperatures of its hot and cold reservoirs by e = 1 − (TC/TH), Equation 15.15. The
work done by the reversible engine is
⎛ T
Wreversible = e QH = ⎜⎜1 − C
⎝ TH
⎞
⎛ 314 K ⎞
⎟⎟ QH = ⎜ 1 −
⎟ (1285 J ) = 811 J
⎝ 852 K ⎠
⎠
c. According to the discussion in part d of the Concept Questions, the difference between
the work done by the reversible and irreversible engines is
Wreversible − Wirreversible = TC ∆Suniverse = ( 314 K )(1.74 J/K ) = 547 J
Wunavailable
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