ME12001 Thermodynamics T2
Exercise 17.26
In an effort to stay awake for an all­night study session, a student makes a cup of coffee by first placing a 200­W
electric immersion heater in 0.350 kg of water.
Part A
How much heat must be added to the water to raise its temperature from 21.5 ∘ C to 84.0 ∘ C ?
ANSWER:
Q
= 9.17×104 J Correct
Part B
How much time is required? Assume that all of the heater's power goes into heating the water.
ANSWER:
t
= 458 s Correct
Exercise 17.28
In very cold weather a significant mechanism for heat loss by the human body is energy expended in warming the
air taken into the lungs with each breath.
Part A
On a cold winter day when the temperature is −20 ∘ C , what amount of heat is needed to warm to body
temperature (37 ∘ C ) the 0.50 L of air exchanged with each breath? Assume that the specific heat of air is 1020
−3
J/kg ⋅ K and that 1.0 L of air has mass 1.3 × 10
kg .
Express your answer using two significant figures.
ANSWER:
Q
= 38 J Correct
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ME12001 Thermodynamics T2
Part B
How much heat is lost per hour if the respiration rate is 20 breaths per minute?
Express your answer using two significant figures.
ANSWER:
ΔQ
Δt
= 4.5×104 J/hour Correct
Exercise 17.29
You are given a sample of metal and asked to determine its specific heat. You weigh the sample and find that its
weight is 25.2 N . You carefully add 1.15×104 J of heat energy to the sample and find that its temperature rises
16.0 ∘ C .
Part A
What is the sample's specific heat?
ANSWER:
c
= 280 J/(kg ⋅ K) Correct
Exercise 17.30
Conventional hot­water heaters consist of a tank of water maintained at a fixed temperature. The hot water is to be
used when needed. The drawback is that energy is wasted because the tank loses heat when it is not in use, and
you can run out of hot water if you use too much. Some utility companies are encouraging the use of on­demand
water heaters (also known as flash heaters), which consist of heating units to heat the water as you use it. No water
tank is involved, so no heat is wasted. A typical household shower flow rate is 5.99×10−4 gal/min (9.49 L/min )
with the tap water being heated from 51.0 ∘ F (10.6 ∘ C ) to 120 ∘ F (48.9 ∘ C ) by the on­demand heater.
Part A
What rate of heat input (either electrical or from gas) is required to operate such a unit, assuming that all the
heat goes into the water?
ANSWER:
H
= 2.54×104 W Correct
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ME12001 Thermodynamics T2
Exercise 17.36
A technician measures the specific heat of an unidentified liquid by immersing an electrical resistor in it. Electrical
energy is converted to heat transferred to the liquid for 150 s at a constant rate of 65.0 W . The mass of the liquid
is 0.750 kg , and its temperature increases from 18.57 ∘ C to 22.55 ∘ C .
Part A
Find the average specific heat of the liquid in this temperature range. Assume that negligible heat is transferred
to the container that holds the liquid and that no heat is lost to the surroundings.
ANSWER:
c
= 3270 J/(kg ⋅ K) Correct
Part B
Suppose that in this experiment heat transfer from the liquid to the container or surroundings cannot be ignored.
Is the result calculated in part (a) an overestimate or an underestimate of the average specific heat? Explain.
ANSWER:
3664 Character(s) remaining
Overestimate, because this suggests that liquid's temperature would increase by the same value with less energy hence c has to be lower.
Submitted, grade pending
Adding Ice to Water
An insulated beaker with negligible mass contains liquid water with a mass of 0.215 kg and a temperature of 72.5 ∘
C .
Part A
How much ice at a temperature of ­15.4 ∘ C must be dropped into the water so that the final temperature of the
system will be 25.0 ∘ C ?
Take the specific heat of liquid water to be 4190 J/kg ⋅ K , the specific heat of ice to be 2100 J/kg ⋅ K ,
and the heat of fusion for water to be 3.34×105 J/kg .
Hint 1. How to approach the problem
Calculate the heat lost by the water when cooled to 25.0 ∘ C . Determine an expression for the heat
gained by the ice in terms of the mass of ice. Since there is no heat added to or removed from the
system, the total heat change of the system must be zero. Set up the appropriate equation and solve for
the mass of ice.
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ME12001 Thermodynamics T2
Hint 2. Calculate the heat lost by the water
Calculate Q water , the heat loss of the water when it is cooled to a final temperature of 25.0 ∘ C . Recall
that Q = mcΔT , where m is the mass, c is the specific heat, and ΔT is the temperature change of
the substance.
Take the specific heat of water to be 4190 J/kg ⋅ K .
ANSWER:
Q water
= −4.28×104 J Hint 3. How to calculate the heat gained by the ice
Since the ice will reach a final temperature of 25.0 ∘ C , which is above the melting point of ice, there will
be a phase change that occurs. To find the total heat gained by the ice, the heat gain must be split into
three parts: heating the ice to the melting point, melting the ice at a constant temperature of 0 ∘ C , and
heating the melted ice (now liquid water) to the final temperature.
Hint 4. Heat gained by the ice
The expression for the heat gained by the ice is mice [cice (0 − Tice ) + Lf
the value of the term in the square brackets and use it in your calculation.
+ c water (Tf − 0)]
. Calculate
ANSWER:
m ice
= 9.08×10−2 kg Correct
Calorimetry Conceptual Question
A 0.5 kg block of aluminum (caluminum = 900 J/kg ⋅ ∘ C ) is heated to 200∘ C . The block is then quickly placed in
an insulated tub of cold water at 0∘ C (cwater = 4186 J/kg ⋅ ∘ C ) and sealed. At equilibrium, the temperature of the
water and block are measured to be 20∘ C .
Part A
If the original experiment is repeated with a 1.0 kg aluminum block, what is the final temperature of the water
and block?
Hint 1. Heat energy and temperature
The amount of heat energy Q needed to change the temperature of a substance depends on the mass
of the substance m, the specific heat of the substance c, and the change in the temperature ΔT of the
substance. This is summarized mathematically as
.
Q = mc(ΔT )
Hint 2. Conservation of energy
The total amount of energy in the system must be conserved. Once the block is placed in contact with
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ME12001 Thermodynamics T2
the cold water, heat energy will flow from the hot test block into the cold water, lowering the temperature
of the block and raising the temperature of the water. A larger block can transfer a larger amount of heat
energy for a given change in temperature. This will result in a larger equilibrium temperature.
ANSWER:
less than 20∘ C
20∘ C
greater than 20∘ C
Correct
Part B
If the original experiment is repeated with a 1.0 kg copper (ccopper
temperature of the water and block?
= 387 J/kg ⋅
∘
C
) block, what is the final
ANSWER:
less than 20∘ C
20∘ C
greater than 20∘ C
Correct
Part C
If the original experiment is repeated but 100 g of the 0∘ C water is replaced with 100 g of 0∘ C ice, what is the
final temperature of the water and block?
Hint 1. Latent heat of fusion
The latent heat of fusion is the amount of heat energy needed to melt 1.0 kg of a substance. The latent
heat of fusion of ice is 334,000 J/kg .
ANSWER:
less than 20∘ C
20∘ C
greater than 20∘ C
Correct
Part D
∘
C
∘
C
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ME12001 Thermodynamics T2
If the original experiment is repeated but 100 g of the 0∘ C water is replaced with only 25 g of 0∘ C ice, what is
the final temperature of the water and block?
Hint 1. Find the equilibrium temperature
In the original experiment, the temperature of 100 g of 0∘ C water was raised by 20∘ C . The amount of
heat Q required to do this can be found as follows. For specific heat c and change in temperature ΔT ,
,
Q = mc(ΔT )
or
Q = (0.1kg)(4186 J/kg ⋅
Therefore, Q
= 8372 J
∘
∘
.
C)(20 C)
.
Is more or less energy than this required to melt and heat 25 g of ice to 20∘ C ?
Hint 1. How to approach the problem
The amount of energy required to melt and heat 25 g of ice to 20∘ C can be broken into two parts:
the energy required to melt the ice and the energy required to heat the water after the ice melts.
ANSWER:
more
less
ANSWER:
less than 20∘ C
20∘ C
greater than 20∘ C
Correct
Exercise 17.40
One suggested treatment for a person who has suffered a stroke is to immerse the patient in an ice­water bath at 0 ∘
C to lower the body temperature, which prevents damage to the brain. In one set of tests, patients were cooled
until their internal temperature reached 32.4 ∘ C .
Part A
To treat a 69.0 kg patient, what is the minimum amount of ice (at 0 ∘ C ) that you need in the bath so that its
temperature remains at 0 ∘ C ? The specific heat capacity of the human body is 3480 J/(kg ⋅ ∘ C) , and recall
that normal body temperature is 37.0 ∘ C .
ANSWER:
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ME12001 Thermodynamics T2
m ice
= 3.31 kg Correct
Exercise 17.50
An open container holds 0.550 kg of ice at ­ 15.0 ∘ C . The mass of the container can be ignored. Heat is supplied to
the container at the constant rate of 800 J/min for 500 min.
Part A
After how many minutes does the ice start to melt?
ANSWER:
t
= 21.7 min Correct
Part B
After how many minutes, from the time when the heating is first started, does the temperature begin to rise
above 0 ∘ C ?
ANSWER:
t
= 252 min Correct
Part C
Plot a curve showing the temperature as a function of the elapsed time.
ANSWER:
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ME12001 Thermodynamics T2
Correct
Steam vs. Hot­Water Burns
Just about everyone at one time or another has been burned by hot water or steam. This problem compares the
heat input to your skin from steam as opposed to hot water at the same temperature.
Assume that water and steam, initially at 100∘ C , are cooled down to skin temperature, 34∘ C , when they come in
contact with your skin. Assume that the steam condenses extremely fast. We will further assume a constant
specific heat capacity c = 4190 J/(kg ⋅ K) for both liquid water and steam.
Part A
Under these conditions, which of the following statements is true?
ANSWER:
Steam burns the skin worse than hot water because the thermal conductivity of steam is much higher
than that of liquid water.
Steam burns the skin worse than hot water because the latent heat of vaporization is released as
well.
Hot water burns the skin worse than steam because the thermal conductivity of hot water is much
higher than that of steam.
Hot water and steam both burn skin about equally badly.
Correct
The key point is that the latent heat of vaporization has to be taken into account for the steam.
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ME12001 Thermodynamics T2
Part B
How much heat H1 is transferred to the skin by 25.0 g of steam onto the skin? The latent heat of vaporization
for steam is L = 2.256 × 10 6 J/kg.
Express the heat transferred, in joules, to three significant figures.
Hint 1. Determine the heat transferred from steam to skin
Find an expression for the heat H1 transferred from the vapor to the skin.
Express your answer in terms of c, the mass of steam m, the latent heat of vaporization L, and
the temperature difference ΔT between the initial temperature of the steam and the skin
temperature.
ANSWER:
H1
= m(cΔT + L)
ANSWER:
H1
= 6.33×104 J Correct
Here we assumed that the skin continues to remain at 34∘ C . Actually the local temperature in the area
where the steam condenses can be raised quite significantly.
Part C
How much heat H2 is transferred by 25.0 g of water onto the skin? To compare this to the result in the
previous part, continue to assume that the skin temperature does not change.
Express the heat transferred, in joules, to three significant figures.
Hint 1. Determine the heat transferred from water to skin
Find an expression for the heat H2 transferred from the hot water to the skin.
Express your answer in terms of c, the mass of water m, and the temperature difference ΔT
between the initial temperature of the water and the skin temperature.
ANSWER:
H2
= mcΔT
ANSWER:
H2
= 6910 J 9/19
ME12001 Thermodynamics T2
Correct
The amount of heat transferred to your skin is almost 10 times greater when you are burned by steam
versus hot water. The temperature of steam can also potentially be much greater than 100∘ C . For these
reasons, steam burns are often far more severe than hot­water burns.
An Electric Water Heater
An engineer is developing an electric water heater to provide a continuous ("on demand") supply of hot water. One
trial design is shown in the figure. Water is flowing at the
rate F , the inlet thermometer registers T1 , the voltmeter
reads V , and the ammeter reads current I . Then the
power (i.e., the heat generated per unit time by the heating
element) is V I .
Assume that the heat capacity of water is C and that the
heat capacity of the heater apparatus is Ch .
Part A
When a steady state is finally reached, what is the temperature reading T2 of the outlet thermometer?
Express the outlet temperature in terms of T1 , F , C , and any other given quantities.
Hint 1. How to approach the problem
Determine the amount of heat input per unit mass of water that flows through the heater. Then use the
definition of heat capacity to compute how much the temperature of a unit mass of water will increase
for a given quantity of heat. Combine these two calculations to determine the increase in temperature of
the water as it flows through the heater.
Hint 2. Find heat input per unit mass of water
How much heat Q is delivered by the resistor per unit mass m of water flowing through the heater?
Express the heat per unit mass in terms of quanitities given in the problem introduction.
Hint 1. What relationships are involved
The power P dissipated in the resistor is equal to the energy dissipated per unit time. The flow
rate F is equal to the mass of the water that flows through the heater per unit time. Thus, if all of
the energy dissipated by the resistor goes into heating the water, then
heat
P
F
=
time
Mass
Time
=
heat
.
mass
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ME12001 Thermodynamics T2
Hint 2. Electrical power
The electrical power disspated in a device is given by the product of the voltage and the current:
P = V I.
ANSWER:
Q/m
= VI
F
Hint 3. Find the change in water temperature
If an amount of heat Q is added uniformly to a substance of mass m and heat capacity C , by how
much will the temperature of that substance increase?
Express the temperature increase ΔT in terms of Q , m, and C .
ANSWER:
ΔT
= Q
1
m
C
ANSWER:
T2
= VI
FC
+ T1
Correct
Part B
Why is it unnecessary to take into account the heat capacity of the apparatus itself?
Choose the best explanation.
ANSWER:
In steady state the temperature of the apparatus doesn't change. Hence its heat capacity is
irrelevant.
The heat capacity of any of the materials used in a water heater is much smaller than that of water.
Since the actual heating unit is immersed in the flow all the heat goes directly into the water.
Correct
Imagine that the input temperature of the water is T1 = 18 ∘ C, the ammeter reads I = 15.0 A, the voltmeter reads
V = 120 V , and the flow rate is F = 0.500 kg/min . The heat capacity of water C = 4200 J/(kg ⋅ K).
Part C
P
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ME12001 Thermodynamics T2
What is the power P at which the heater operates?
Express your answer numerically, in watts.
Hint 1. Electrical power
The electrical power disspated in a device is given by the product of the voltage and the current:
P = V I.
ANSWER:
P
= 1800 W Correct
This is a large, but reasonable, power requirement for a household appliance.
Part D
Calculate the temperature T2 of the water leaving the heater. Note that F is defined as the number of
kilograms of water flowing through the heater per minute, whereas the power is measured in watts
(joules/second).
Express your answer numerically, in degrees Celsius, to the nearest integer.
ANSWER:
T2
= 69 ∘ C Correct
Part E
Consider using this heater to generate "on demand" hot water for a home shower. The input water temperature
(during the winter) can be as low as T1 = 5.00∘ C and the output temperature should be at least T2 = 40.0∘ C
for a moderately warm shower. Assume a conservative flow rate of F = 8.00 kg/min (corresponding to about
2.5 gallons per minute).
If the heater is to operate on a US standard V
meet the design requirements?
= 120 V
wall plug, how much current I would it need to draw to
Express the current numerically, in amperes, to the nearest integer.
ANSWER:
I
= 163 A 12/19
ME12001 Thermodynamics T2
Correct
163 A is a tremendous amount of current for a home appliance. Most homes in the United States are wired
for a maximum of 200 A for the entire house, and only 20 A on any particular outlet. For the electric heater
to meet the modest design requirements given, it would consume almost 20 kW of power!
For this reason, most "on demand" water heaters generate heat by burning a fuel such as propane, rather
than using electricity.
Exercise 17.71
A picture window has dimensions of 1.40 m × 2.50 m and is made of glass 5.70 mm thick. On a winter day, the
outside temperature is ­18.0 ∘ C , while the inside temperature is a comfortable 21.0 ∘ C .
Part A
At what rate is heat being lost through the window by conduction?
Express your answer using three significant figures.
ANSWER:
H
= 1.92×104 W Correct
Part B
At what rate would heat be lost through the window if you covered it with a 0.750 mm­thick layer of paper
(thermal conductivity 0.0500 W/m ⋅ K )?
Express your answer using three significant figures.
ANSWER:
H
= 6170 W Correct
Heat Flow through Three Rods
Three identical rods are welded together to form a Y­shaped figure. The cross­sectional area of each rod is A, and
they have length L and thermal conductivity k .
The free end of rod 1 is maintained at T1 and the free ends of rods 2 and 3 are maintained at a lower temperature T0 . You may assume that there is no heat loss from the surfaces of the rods.
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ME12001 Thermodynamics T2
Part A
What is Tj , the temperature of the junction point?
Express your answer in terms of T1 and T0 .
Hint 1. Find a relationship among the heat currents
Heat is flowing from the free end of rod 1 to the free ends of rods 2 and 3, breaking into two flows at the
junction point. What is the relationship between the heat current in rod 1 ( H1 ) to the heat currents in
rods 2 and 3 ( H2 and H3 respectively)?
Express H1 in terms of H2 and H3 .
Hint 1. Use an analogy to electric currents
If you are familiar with electrical current, you can solve this problem by using the analog of
Kirchoff's junction rule, which states that the algebraic sum of the currents into any junction is
zero (any current leaving the junction must be subtracted from the total). Similarly, in steady
state, the algebraic sum of all heat currents into a junction is zero (any heat current leaving the
junction must be subtracted from the total).
ANSWER:
H1
= H2 + H3
Hint 2. Find H 1 in terms of the junction temperature
Now, if you can find expressions for the heat currents in the rods in terms of their physical properties
and temperature difference across them, you can substitute these into the formula H1 = H2 + H3 and
solve algebraically for Tj .
Derive a general expression for the heat current through rod 1 in terms of the temperature of the junction,
Tj , and other quantities given in the problem introduction.
ANSWER:
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ME12001 Thermodynamics T2
H1
kA(T 1 −T j )
= L
ANSWER:
Tj
= T 1 +2T 0
3
Correct
This is a weighted average of the temperatures at the two ends of the system and so always lies between
them as you would expect. Also, it is closer to T0 than to T1 . This is because the heat current in rods 2
and 3 separately must be less than the heat current in rod 1 (since H2 + H3 = H1 ).
Part B
What is the heat current H1 in rod 1?
Express the heat current in terms of any or all of k , L, A, and the temperatures T1 and T0 .
Hint 1. Determine the general formula for heat current
Given the difference in temperature between the ends of the rod ( ΔT ), the rod's thermal conductivity k ,
and its length L and cross­sectional area A, what is the general formula for the heat current H in the
rod?
Express your answer in terms of k , L, A, and ΔT (typed as DeltaT).
ANSWER:
H
= kAΔT
L
Hint 2. Find the temperature difference
What is the temperature difference ΔT between the ends of rod 1?
Express your answer in terms of T1 and T0 . Your answer should be positive.
Hint 1. Use your answer to Part A
The temperature difference between the ends of rod 1 is ΔT = T1 − Tj . To express this in
terms of T1 and T0 , use the relationship you found in Part A of this problem.
ANSWER:
ΔT
= 2
3
(T 1 − T 0 )
ANSWER:
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ME12001 Thermodynamics T2
T
H1
= T1 −
kA
1
+2T
0
3
L
Correct
This is greater than the heat current due to only 2 rods. You would expect this to be the case since the
effective surface area is increased by having a third rod. You can check that if you wanted to remove the
third rod while keeping the same heat current, you must increase the cross­sectional area of the remaining
two rods by a factor of 1.33.
Part C
By symmetry, the heat current in rods 2 and 3 will be the same. Find this heat current.
Express your answer in terms of k , L, A, and the temperatures T1 and T0 .
Hint 1. How to approach this problem
Recall that H1
and H3 .
= H2 + H3
. Also, by symmetry, H2
= H3
. This information is sufficient to find H2
ANSWER:
T
H2 = H3
= kA
1
+2T
0
3
−T 0
L
Correct
Heat Radiated by a Person
In this problem you will consider the balance of thermal energy radiated and absorbed by a person.
Assume that the person is wearing only a skimpy bathing suit of negligible area. As a rough approximation, the area
of a human body may be considered to be that of the sides of a cylinder of length L = 2.0 m and circumference C = 0.8 m .
For the Stefan­Boltzmann constant use σ
= 5.67 × 10
−8
2
W/m /K
4
.
Part A
If the surface temperature of the skin is taken to be Tbody
body described in the introduction radiate?
Take the emissivity to be e
= 0.6
∘
= 30 C
, how much thermal power Prb does the
.
Express the power radiated into the room by the body numerically, rounded to the nearest 10 W.
Hint 1. Area of person
Find the area A of the person, ignoring the ends of the cylinder that represents the person's area.
Express the area in terms of L, C , and any constants.
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ME12001 Thermodynamics T2
ANSWER:
A
= LC
Hint 2. Symbolic expression
Find Prb , the total power radiated into the room by the body of the person.
Express the total power radiated in terms of the Stefan­Boltzmann constant σ, the body area A,
the body temperature Tbody , and the emissivity e .
ANSWER:
P rb
= eσATbody
4
ANSWER:
P rb
= 460 W Correct
Part B
The basal metabolism of a human adult is the total rate of energy production when a person is not performing
significant physical activity. It has a value around 125 W, most of which is lost by heat conduction to the
surrounding air and especially to the exhaled air that was warmed while inside the lungs. Given this energy
production rate, it would seem impossible for a human body to radiate 460 W as you calculated in the previous
part.
Which of the following alternatives seems to best explain this conundrum?
ANSWER:
The human body is quite reflective in the infrared part of the spectrum (where it radiates) so e is in
fact less than 0.1.
The surrounding room is near the temperature of the body and radiates nearly the same power into the
body.
Correct
The human body contains significant amounts of water and organic compounds. These typically have
many absorption bands in the infrared part of the spectrum where room temperature objects radiate thermal
energy. Consequently, human skin has a high emissivity in this part of the spectrum regardless of the
emissivity of the skin in the visible part of the spectrum.
Part C
Now calculate Pbr , the thermal power absorbed by the person from the thermal radiation field in the room,
which is assumed to be at Troom = 20 ∘ C. If you do not understand the role played by the emissivities of room
and person, be sure to open the hint on that topic.
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ME12001 Thermodynamics T2
Express the thermal power numerically, giving your answer to the nearest 10 W.
Hint 1. Role of emissivities
The emissivity e is the complement of the reflectivity. The power incident on the person is partly
absorbed and partly reflected. The fraction absorbed is proportional to the emissivity of the person and
thus is less than or equal to the thermal power striking the person by the amount of this power that is
reflected.
Because the room is closed, no allowance has to be made for its emissivity. If the room has an
emissivity of 0.7, then when you look at the wall (with an infrared viewer) 0.7 of what you see is the
power radiated by that spot on the wall; the remaining 0.3 of what you see is what is reflected by the
wall. But in a closed container like a room, this reflection is simply another part of the wall (since the
room is closed). Being at the same temperature, this other part of the wall radiates power at the same
rate as the spot you are looking at. When all the emissions and reflections are added together, the net
effect is that the radiation from any part of the wall has the same total thermal power as if it were
perfectly black. This accounts for the remarkable fact that the thermal radiation field in any closed cavity
is independent of the material of the cavity and depends only on the temperature.
Such radiation is called cavity radiation or blackbody radiation. The fact that the intensity at each color
depended only on the temperature was so striking to physicists that the explanation of this phenomenom
was assiduously studied until it unlocked the key to quantum mechanics and revealed a fundamental
quantum of nature: Planck's constant h.
A good example of this is a kiln. Just after a part is placed in a kiln, the heaters are much hotter than
the cool part, and the part is visible through the peephole. As thermal equilibrium is reached everything
glows a uniform red or orange, and the part disappears into the background color, becomming invisible.
Hint 2. Area of person
Find the area A of the person.
Express the area in terms of L, C , and any constants.
ANSWER:
A
= LC
Hint 3. Symbolic expression
Find Pbr , the total power absorbed by the person.
Express your answer in terms of the Stefan­Boltzmann constant σ, the body area A, the room
temperature Troom , and the emissivity e .
ANSWER:
P br
= eσATroom
4
ANSWER:
P br
= 400 W Correct
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ME12001 Thermodynamics T2
Part D
Find Pnet , the net power radiated by the person when in a room with temperature Troom
∘
= 20 C
.
Express the net radiated power numerically, to the nearest 10 W.
Hint 1. How to set up the problem
The net power radiated is simply the power radiated minus the power absorbed.
ANSWER:
P net
= 60 W Correct
This net thermal power radiated is less than half of the basal metabolic rate. Under these circumstances a
person would feel cool, but not cold (if we neglect loss of heat by conduction). However, if the surrounding
temperature is much cooler than this, the net radiative loss will become a significant loss mechanism for
body heat. This explains why lightweight survival blankets have a shiny side: The low emissivity
dramatically reduces radiative heat loss (and the air trapped between blanket and body reduces heat loss
by conduction).
Exercise 17.41
A copper pot with a mass of 0.500 kg contains 0.170 kg of water, and both are at a temperature of 20.0 ∘ C . A
0.250 kg block of iron at 85.0 ∘ C is dropped into the pot.
Part A
Find the final temperature of the system, assuming no heat loss to the surroundings.
ANSWER:
T
= 27.5 ∘ C Correct
19/19