Heat
Heat is a form of energy. Any object has
internal energy in the form of kinetic energy
of the atoms or molecules (heat – vibration
or motion) and in potential energy connected
with the molecular structure and the
electromagnetic forces among the
constituents.
If we add energy to an object the internal
energy increases.
Energy can be added in many ways. For
example as: HEAT
™from frictional forces on a moving object.
™electromagnetic radiation from the sun.
™a source of heat like a hair dryer.
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First Law of Thermodynamics
The increase in the internal energy of a
system is equal to the amount of heat added
to the system minus the amount of work
done by the system.
U = internal energy
Q = heat that is added or removed
Wsystem = is the work done by the system
Looking at the picture you can see that IF
the force moves the piston inward it does
positive work and therefore is putting
energy into the gas. We call it Wexternal
The work done by the gas is negative
because the force the gas exerts is to the
right and the movement of the piston is to
the left. We call it Wsystem
Here heat MAY flow in or out thru the walls
ENERGY CONSERVATION
ΔU = Q – Wsystem or ΔU = Q + Wexternal
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Positive Q is heat inflow
Negative Q is heat flow
out from the system
2
Temperature
Heat flows from a hot body to a cold body until they reach thermal
equilibrium. This heat flow is partly “irreversible”, and ultimately
determines the direction of “time’s arrow” – a profound fact.
Temperature is the quantity that measures (among other things)
whether one body is hotter than another.
At thermal equilibrium both bodies have the same temperature.
One simple system is an ideal
gas of free molecules where the
energy is kinetic energy of translation.
A higher temperature means higher
velocities and more stored energy.
Increase U Æ T increases
http://www2.biglobe.ne.jp/~norimari/science/JavaApp/Mole/e-gas.html
http://jersey.uoregon.edu/vlab/Thermodynamics/index.html
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Physics 214
Spring 2009
m=mass, c=specific heat
capacity per kg per degree
3
Measuring temperature
Physical properties of an object change with
temperature. For example mercury expands as the
temperature increases and we can use that
expansion to measure T. Or, use the change in
electronic properties of Silicon with Temperature.
There are three temperature scales,
Fahrenheit, Celsius,
Kelvin
00C
273K
Mixture of ice and water 320F
Boiling point of water 2120F
1000C
373K
Annoyingly, the standard usage is just K, for Kelvins,
NOT degrees Kelvin. K measures ABSOLUTE temp.
and cannot be negative.
Tc = 5/9(Tf – 32)
Tf = 9/5Tc + 32
Celsius and Farenheit are the same number at -40o
TK = Tc + 273
Note: this water boiling point is for sea level –
The liquid expands
water boils cooler at higher altitudes where air
more than the glass
pressure is less
Week 9
Physics 214
Spring 2009
as the thermometer
4
warms up.
Temperature Scales
Tc = 5/9(Tf – 32)
Tf = 9/5Tc + 32
100 oC
0 oC
-40 oC
-40
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212 oF
32
Physics 214
Spring 2009
5
Questions Chapter 10
Q1 Is an object that has a temperature of 0°C hotter than, colder
than, or at the same temperature as one that has a temperature of
0°F?
Water freezes at 0o C and 32o F so 0o F is colder
Q2 Which spans a greater range in temperature, a change in
temperature of 10 Fahrenheit degrees or a change of 10
Celsius degrees?
There is 100o C between water freezing and boiling and
180o F so 1oC = 1.8oF so a change of 10oC is larger
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Ch 10 E 2
Temperature is 14° F.
What is the temperature in Celsius?
Remember 32oF is 0oC and a rise of 9oF is the same
as a rise of 5oC.
C = (5/9)(F-32)
A. 7.8°C
B. 14°C
C. 44°C
D. -10°C
E. -40oC
March 9, 2009
Degrees C = (5/9)(14-32) = (5/9)(-18) = -10
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Absolute zero (the Kelvin scale)
An ideal gas obeys the law
PV = constant x T
so if we keep V constant the pressure is
proportional to T.
If we measure P versus T at constant
volume we find a straight line graph,
which points to -273oC at zero pressure.
This is absolute zero where the stored
internal energy has its lowest possible
value.
This is Zero degrees Kelvin.
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Absolute zero (the Kelvin scale)
An ideal gas obeys the law
PV = constant x T
BUT THIS IS AN IDEALIZATION. ALL
REAL GASES CONDENSE INTO LIQUID
OR SOLID AT LOW ENOUGH
TEMPERATURE, EVEN HELIUM.
That’s why the graph stops short of
absolute zero, although we CAN get
almost straight-line data at much lower
temperatures than shown below, for many
gases.
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Q4 We sometimes attempt to determine whether another person
has a fever by placing a hand on their forehead. Is this a reliable
procedure? What assumptions do we make in this process?
Not too accurate, but a high fever is likely to cause the forehead
to be perceptibly warmer than “usual”. One assumption is that
your hand is at normal body temperature, so the fever feels
hotter.
Q5 Is it possible for a temperature to be lower than 0°C?
Yes. Absolute zero is – 2730 which one can think of as the place
where the molecules of an ideal gas have zero energy
Q6 Is it possible for a temperature to be lower than 0 K on the
Kelvin temperature scale?
NO. Absolute zero is 0 K, the lowest possible, where
there is no remaining molecuar kinetic energy
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Heat capacity
For a given object at a specific
temperature the total internal energy
depends on how massive the object is. If
we add a fixed amount of energy the
temperature will rise but the rise will be
smaller the larger the object is. So we
define a quantity called specific heat
c = is the quantity of heat required to raise
unit mass of a substance by one degree.
Q = mcΔT
New unit
1 calorie = heat required to heat 1 gram of
1calorie = 4.186 Joules
water 10C
Note to raise 1kg of water 10C requires
1000cal
which is 1 kcal or kilocalorie
Note that internal energy can change
without a temperature change. This is due
to internal rearrangement of the molecules
a “phase change”.
EXAMPLE: water freezing or thawing
Week 9
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Spring 2009
Ice
Water
11
Heat capacity
c = is the quantity of heat required to raise unit mass
of a substance by one degree.
Q = mcΔT
Example: 100 g water at 100 oC in contact with 50 g
water at 20 oC.
What is the final temperature T at thermal
equilibrium?
Heat lost from object 1 = heat gained by object 2
100g(1 cal/goC)(100 oC - T) = 50g(1cal/goC)(T – 20 oC)
100 oC - T = (50/100)(T - 20 oC) = 0.5T – 10 oC
Ice
110 oC = 1.5 T
Water
T = 73.3 oC
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Q25 Heat is added to a hot-air balloon causing the air to expand.
Will this increased volume of air cause the balloon to fall?
Archimedes principle states that the buoyant force is equal to the
weight of fluid displaced. Assume the balloon stays the same size
– buoyancy does not change. Some expanding air leaves the
balloon. The remaining air is less dense, so balloon weighs less,
& rises. If no air escapes but the balloon increases in size it will
also rise because the buoyant force is larger (more displaced air)
Q27 A block of wood and a block of metal have been sitting on a
table for a long time. The block of metal feels colder to the touch
than the block of wood. Does this mean that the metal is actually
at a lower temperature than the wood?
No. What you feel is heat flow and the thermal conductivity of metal is much
bigger than that of wood. Also the heat capacity of the metal is greater. Of
course, this requires the room temperature to be lower than body
temperature
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Solids and liquids
Temperature depends on the internal energy
Put in heat, T rises except at a change of state
When not at a melting or boiling point, adding heat
changes T according to the specific heat (for liquids or
solids).
When at a “phase change” temperature, a lot of heat needs
to be added (or subtracted) to change from one phase to
another. During this process, the temperature remains
constant (if all the stuff remains in thermal equilibrium.)
Change of state
ice
water
steam
Requires 80 cal/gm to melt ice at 0C to water at 0C
Requires 540cal/gm to make steam at 1000C
To condense steam or freeze water requires the removal of
540 cal/gm or 80cal/gm.
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Change of state
We are used to solids, liquids and gases and that most substances
can be in any of the three states depending on temperature and
pressure. Solids have more orderly structure and the
atoms/molecules are tightly joined. In a liquid the atoms/molecules
are loosely joined. In a gas the atoms/molecules are “free”. When
we tear apart something that is joined, energy is required and this is
true as we go from solid to liquid and liquid to gas. Taking water as
an example:
80 calories/gram are required to melt ice at 0oC to water at 0oC and
540 calories/gram to convert water to steam at 100oC. When water
freezes 80cal/gm has to be removed and when water vapor
condenses 540 cals/gm is released.
http://jersey.uoregon.edu/vlab/Balloon/
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Change of state
Example: How much (mass) of steam at 100 oC is
needed to just melt 1 kg of ice at 0 oC ?
80 cals/gm is the latent heat of fusion (of water)
540 cals/gm is water’s latent heat of vaporization
heat from condensation = heat from melting:
Then 1000 g x 80 cal/ g = 540 cal/ g x Msteam so
Msteam = 148.15 g
We can assume all the water comes to a common temperature -- since
they must be in thermal contact in any case. What is that temperature?
Use same technique as on Monday, or a shortcut is to just compute the
average temperature (would not work for two DIFFERENT substances
coming into thermal equilibrium)
Tfinal = (1000 * 0 + 148.16 * 100)/1148.16 = 12.90 oC
http://jersey.uoregon.edu/vlab/Balloon/
Week 9
Physics 214
Spring 2009
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Water and Ice
Water has an unusual behavior as the
temperature is changed. As we lower the
temperature from room temperature it
shrinks but at 40C it starts to expand
and that is why ice floats.
Remember density ρ = mass/volume so
below 40C the volume increases and the
density decreases.
Above 40C the molecules are maximally
tight-packed. Below 40C the molecules
begin to rearrange themselves
somewhat into the less-dense solid (ice)
form they will have at and below 0oC : a
crystalline structure that has holes.
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Water and Ice
Water is VERY unusual, and if ice
didn’t float, lake bottoms would
freeze first and maybe never thaw
in the summer. The world would
be a much different place.
Also, the expansion of freezing
water accelerates the breakup of
rock formations, and the process
of “weathering”.
It’ll also burst the water pipes in
your house if allowed to freeze.
Week 9
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Geysers (Old Faithful)
Geysers produce a jet of water very
often at equal time intervals. Old
Faithful in Yellowstone erupts every
90 minutes or so. What is required is
a source of heat, a source of water
and a constricted vertical channel
that the water flows down into.
As water flows in and the column of
water increases in height the
pressure increases at the bottom of
the water column where the heat is.
The increase in pressure raises the
boiling point. But eventually the
water at the bottom is heated so
much it starts to boil at very high T.
This causes an expansion of the
column which reduces the pressure P
and immediately lowers the boiling
point of all the superheated water.
Then there is a massive explosive
boiling resulting in the eruption.
Week 9
= ρgh
Physics 214
Spring 2009
Heat
19
Q8 Two objects at different temperatures are placed in contact with
one another but are insulated from the surroundings. Will the
temperature of either object change?
They will exchange heat until they both reach the same (intermediate) temperature
Q10 Two objects of the same mass, but made of different
materials, are initially at the same temperature. Equal amounts of
heat are added to each object. Will the final temperature of the
two objects necessarily be the same?
No. The specific heat, which is the heat energy required to raise
the temperature one degree, may be different for the two materials
Q13 What happens if we add heat to water that is at the
temperature of 100°C? Does the temperature change? Explain.
First, all of the water turns into steam at 100o C
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Gases
Now piston moves to the right
As gas expands, piston moves d
Work done by gas = Fd = PAd
ΔV = Ad
Wgas = PΔV
External work done = Wext = - P ΔV
Here ΔV is the + volume swept out by
the piston. Gas volume increases.
A quantity of gas can be described by three
variables. Pressure, volume and temperature
U = internal energy (kinetic energy of the molecules)
U determines the temperature
Q = heat energy that is put in or taken out
Wgas = work done by the gas when it expands
or is compressed or contracts
Wgas is + if it expands, – if it is compressed
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Physical Laws for a gas
Energy conservation
ΔU = Q – Wgas = heat inflow – work gas does on surroundings
Ideal gas law
PV = NkT (T in degrees Kelvin)
N is the number of molecules, k is Boltzmann’s constant
1 gram of Hydrogen atoms has ~6 x 1023 atoms
Boltzmann constant = 1.3806503 × 10-23 J/K ( sort of “per molecule”)
So at a given T, NkT is in units of Joules
PV is also in units of Joules, as we have seen before.
Each molecule of gas carries, on average, 3/2kT of kinetic energy
Actually, there is a random distribution of speeds and energies, the KE
above is the average over this distribution
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Physical Laws for a gas
As a gas volume is changed, or as it is allowed to change,
various conditions might hold:
Adiabatic
no heat in or out
Q=0
Compression: work on gas is + (U, T increase)
Expansion:
work on gas is – (U, T decrease)
Isothermal T does not change ΔU = 0 = Q – Wsystem = gas
Q = Wsystem = PΔV
put in heat gas expands take out heat gas will shrink or be
compressed
Isobaric pressure is kept constant
Put in heat
T increases gas expands (hot air balloon)
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Heat engines
The use of energy and the conversion of energy is essential not
only in our practical everyday life but is a requirement for life to
exist. The sun puts energy into the earth – warmth, growth of
plants for food, evaporation of sea-water to make rain, etc.
Humans use such energy sources as gasoline, coal, nuclear
power, wind power, solar electricity, geothermal energy.
The conversion of one form of energy into another form of energy
is governed by physical laws. The extraction of mechanical energy
from a heat source requires some kind of device, generically
called a heat engine.
A car engine is a practical example. Also steam engines, turbines,
jet engines, rockets. Heat is a “low-grade” form of energy, as
we’ll soon see – not all the heat can be turned back into
mechanical energy.
Note that solar cells, which turn sunlight directly into electrical
energy, do NOT suffer from the above “thermal inefficiency”. (but
generally don’t convert all the light)
An electric battery, or a fuel cell, turns chemical energy dirctly
into electical energy, also avoiding thermal inefficiency.
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2nd Law of Thermodynamics
Heat engine uses energy QH and produces
work W and releases Qc (H for hot, c for cold)
Efficiency = ε = W/QH
W = QH – Qc
(c = the “cold reservoir” = the environment)
2nd law
No engine working in a continuous cycle can
take heat at a single temperature and convert
that heat completely to work.
An absolute zero cold reservoir would give
100% efficiency, but can not be realized in
practice OR in principle.
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Carnot Cycle
We can imagine a “perfect engine”
without friction. The Carnot cycle
gives an ideal engine and an
analysis reveals that
ε=
(TH –Tc)/TH
(T is in oK)
For example in a gasoline engine TH is the temperature of combustion
and Tc is the outside temperature. If we take TH as 1200oC and Tc as
300o then ε = 75%. In practice the efficiency is maybe 35% because of
heat leakage, friction, exhaust back-pressure, non-ideal cycle, etc.
An ideal Carnot engine would have to run infinitely slowly, because to
make it “reversible” the temperature difference between each reservoir
and the working fluid has to be close to zero (in parts A and C of the
cycle above) To maximize the POWER (rate of converting heat to
mechanical work) there has to be a significant Temperature difference in
stages A and C, resulting in a much lower thermal efficiency.
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Carnot Cycle
We can imagine a “perfect engine”
without friction. The Carnot cycle
gives an ideal engine and an
analysis reveals that
ε=
(TH –Tc)/TH
(T is in oK)
Turbines can run hotter and do a bit better.
Note that in a “fuel cell” which could convert the chemical energy of the
gasoline directly into electricity, we could evade the Carnot inefficiency.
Practical engineering difficulties exist, but people are working very hard
to perfect fuel cells for cars, laptops, portable appliances, etc.
Week 9
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Car engines
In an engine with spark plugs a gasoline-air mixture is injected
when the cylinder volume is maximal (the piston is farthest
withdrawn). Then the piston moves in and compresses it, and a
spark ignites the mixture when the piston is farthest in. The
expanding explosion drives the piston, i.e. it does work on the
piston.
Diesel engines use a thicker grade of petroleum and also run at a
higher temperature. Instead of a spark plug, the ignition comes
from very high compression heating of the fuel-air mixture. The
bigger compression ratio and the hotter burning lead to somewhat
improved Carnot efficiency of diesel vs. gasoline engines.
There is actually more energy per kilogram of diesel fuel than of
gasoline – but unfortunately it also contains a higher ratio of
Carbon to Hydrogen (Hydrogen burns to harmless water, Carbon
burns to C02 which is the predominant greenhouse gas currently
frying our planet.)
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Review Chapters 10 and 11
Wgas = PΔV for the expanding piston
Wexternal is negative
ΔU = Q – Wsystem or ΔU = Q + Wexternal
Work done by gas = Fd = PAd
ΔV = Ad
W = PΔV
Work done on gas is - PΔV
Temperature scales Celsius, Fahrenheit Kelvin
Mixture of ice and water 00C
320F 2730K
Boiling point of water 1000C 2120F 3730K
Tc = 5/9(Tf – 32)
Tf = 9/5Tc + 32 TK = Tc + 273
c = is the quantity of heat required to raise unit mass of a substance by one degree.
Q = mcΔT
1 calorie = heat required to raise 1 gram of water 10C = 4.186 joules
Change of state (internal energy changes temperature is constant)
80 cals/gm is the latent heat of fusion
540 cals/gm is the latent heat of vaporization
Week 9
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Heat engines
Efficiency = ε = W/QH
Change in internal energy in one cycle
is zero
W = QH – Qc (c = environment)
Ideal Carnot ε = (TH –Tc)/TH = 1 - Tc/TH
2nd law
No engine working in a continuous
cycle can take heat at a single
temperature and convert that heat
completely to work.
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3E-03 Fire Syringe
Compression and rise in air temperature
What will happen
to the
combustible
material when the
plunger is rapidly
pushed down ?
Can you guess
the every-day
application of this
phenomenon ?
This system is analogous
to the combustion cycle
within a diesel engine.
RAPID COMPRESSION IS ADIABATIC GIVING RAPID RISE OF
AIR TEMPERATURE IN THE CHAMBER WHICH EXCEEDS THE
IGNITION TEMPERATURE OF THE FLAMMABLE MATERIAL.
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Q21 An ideal gas is compressed without allowing any heat to flow
into or out of the gas. Will the temperature of the gas increase,
decrease, or remain the same in this process?
The temperature will increase: ΔU = W
Q23 Heat is added to an ideal gas, and the gas expands in the
process. Is it possible for the temperature to remain constant
in this situation?
Yes. IF the work done by the gas equals the heat inflow, then
this is an isothermal expansion and W = Q
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Ch 10 E 2
Temperature is 14° F.
What is the temperature in Celsius?
Tc= 5/9 (TF – 32) = 5/9 (14 - 32)=5/9 (-18) = -10° C
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Ch 10 E6
How much heat does it take to raise the temperature
of 70 g of H2O from 20°C to 80°C? C= 1 cal/gram/oC.
Q = mcΔT=(70)(1)(80-20) = 4200 cal
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Ch 10 E 12
Add 600 J to 50 g of H2O initially at 20°C
a) How many calories?
b) What is the final temperature of the H2O
a) 600 /4.186 = 143.3 cal = Q
b) Q=mcΔT
ΔT=Q/mc=143.3/(50)(1)=2.87°C
TF=22.87°C
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Ch 10 E 16
Add 500 cal of heat to gas. Gas does 500 J of
work on surroundings.
What is the change in internal energy of gas?
ΔU = Q-W
Q = 500/ 4.186 = 2093 J
W=500 J
ΔU = 2093 – 500 = 1593 J
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Ch 10 CP 2
A student’s temperature scale = 0°s = ice point of H2O; 50° s = boiling
point of H2O. The student then measures a temperature of 15°s.
a) What is this temperature in degrees Celsius?
b) What is this temperature in degrees Farenheit?
c) What is this temperature in Kelvins?
d) Is the temperature range spanned by 1°s larger or smaller than spanned by
1°C?
a) Ice point H2O = 0°s=0°C
Boiling point H2O = 50°s=100°C
Obvious relationship: Ts=1/2 Tc, Tc=2Ts
Tc=2Ts=2(15) = 30°C
b) TF = 9/5 Tc+ 32 = 9/5(30) + 32 = 86°F
c) Tk = Tc + 273.2 = 30 + 273.2 = 303.2K
Ts
50°s
0°s
Tc
100°c
0°c
d) 1°s spans 2°C, so that the range spanned by 1°S is LARGER than that
spanned by 1°C.
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Ch 10 CP 4
150 g of metal at 120°C is dropped in a beaker containing 100 g H2O
at 20°C. (Ignore the beaker). The final temperature of metal and water
is 35°C.
a) How much heat is transferred to the H2O?
b) What is the specific heat capacity of metal?
c) Use the same experimental setup – how much metal at 120°C to have final
temperature of metal and water = 70°C?
a) Q = mcΔT = (100)(1)(35-20) = 1500 cal
b) Q = mcΔT → c = Q/mΔT =
c = -1500/(150)(35-120) = 0.12 cal/gram/°C
c) Q H2O = mcΔT = (100)(1)(70-20) = 5000 cal
Q metal = - Q H2O = - 5000
Q metal = mcΔT; m = Q metal/cΔT = -5000/(0.12)(70-120)
m = 833 g
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Ch 11 E 6
A Carnot engine takes in heat at 650 K and releases
heat to a reservoir at 350K.
What is the efficiency?
A. 0.46
B. 0.26
C. 0.33
D. 0.75
E. 0.90
March 2009
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Ch 11 E 6
A Carnot engine takes in heat at 650 K and releases
heat to a reservoir at 350K.
What is the efficiency?
ec = (TH – Tc) / TH = 650-350 / 650 = 0.46
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Ch 11 E 8
Carnot engine operates b/w 600 K and 400 K and
does 200 J of work in each cycle.
a) What is the efficiency?
b) How much heat does it take in from the high-temp
reservoir during each cycle?
a) εc= (TH- Tc)/TH = (600-400)/600 = 0.33
b) ε = W/QH , QH = W/ε = 200/0.33 = 600J
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Ch 11 CP 2
Carnot engine operates b/w 500° C and 150° C and does
30 J of work in each cycle.
a) What is the efficiency?
b) How much heat is taking in from the high-temp reservoir each cycle?
c) How much heat is released to low-temp reservoir each cycle?
d) What is the change, if any, in the internal energy of gas each cycle?
a) εc= (TH-Tc)/ TH= (500+273.3)-(150+273.3)/(500+273.3) = .045
b) ε = W/QH , QH = W/ε = 30/0.45 = 66.3 J
c) W = QH-Qc, Qc = QH-W = 66.3J - 30J = 36.3J
d) ΔU = Q-W, W = 30J
Q = QH - Qc = 66.3 – 36.3 = 30 J
ΔU = 30-30 = 0
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