Chapter 18: The Laws of Thermodynamics

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18 – 1
Chapter 18: The Laws of Thermodynamics

Answers to Even-Numbered Conceptual Questions


2. (a) Yes. Heat can flow into the system if at the same time the system expands, as in an isothermal expansion
of a gas. (b) Yes. Heat can flow out of the system if at the same time the system is compressed, as in an
isothermal compression of a gas.
4. No. The heat might be added to a gas undergoing an isothermal expansion. In this case, there is no change
in the temperature.
6. Yes. In an isothermal expansion, all the heat added to the system to keep its temperature constant appears as
work done by the system.
8. The final temperature of an ideal gas in this situation is T; that is, there is no change in temperature. The
reason is that as the gas expands into the vacuum, it does no work because it has nothing to push against.
The gas is also insulated, so no heat can flow into or out of the system. It follows that the internal energy of
the gas is unchanged, which means that its temperature is unchanged as well.
10. This would be a violation of the second law of thermodynamics, which states that heat always flows from a
high-temperature object to a low-temperature object. If heat were to flow spontaneously between objects of
equal temperature, the result would be objects at different temperatures. These objects could then be used to
run a heat engine until they were again at the same temperature, after which the process could be repeated
indefinitely.
12. Yes. In fact, the heat delivered to a room is typically 3 to 4 times the work done by the heat pump.
14. The law of thermodynamics most pertinent to this situation is the second law, which states that physical
processes move in the direction of increasing disorder. To decrease the disorder in one region of space
requires work to be done, and a larger increase in disorder in another region of space.


Solutions to Problems and Conceptual Exercises



1. Picture the Problem: Thermodynamic systems change their internal energy when heat flows and when work is done.
Strategy: Use equation 18-3 to find the change in internal energy for each system.

Solution: 1. (a) Calculate ΔU:
50 J 50 J 0 JU Q WΔ = − = − =


2. (b) Calculate ΔU:
( )
50 J 50 J 0 JU Q WΔ = − = − − − =


3. (c) Calculate ΔU:
( )
50 J 50 J 100 JU Q WΔ = − = − − = −

Insight: In part (c) 50 J of heat energy flows out of the system, and the system does work on the external world,
removing an additional 50 J of energy from the system. The net effect is 100 J of energy is removed from the system.




2. Picture the Problem: A gas expands, doing 100 J of work while its internal energy increases by 200 J.
Strategy: Use equation 18-3 to find the heat flow into the system.

Solution: Solve equation 18-3 for Q:
100 J 200 J 300 JQ W U= + Δ = + =

Insight: The work is positive because the system is doing work on the external world. The heat flow into the system
provides the energy to do this work plus an additional 200 J for the system to store as internal energy.

Chapter 18: The Laws of Thermodynamics James S. Walker, Physics, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 2

3. Picture the Problem: A swimmer does work and gives off heat during a workout.
Strategy: The heat and work are given in the problem. Use equation 18-3 to find the change in internal energy.

Solution: 1. Write down the heat:
5
4.1 10 JQ = − ×


2. Write down the work:
5
6.7 10 JW = ×


3. Find the change in internal energy:
5 5 5
4.1 10 J 6.7 10 J 10.8 10 JU Q WΔ = − = − × − × = − ×

Insight: The heat is negative because the system (the swimmer) loses heat. Q is positive when the system gains heat.


4. Picture the Problem: The temperature of one mole of an ideal gas increases as heat is added.
Strategy: Use the first law of thermodynamics to determine the work, recognizing that the change in internal energy of
a monatomic gas is given by
3
2
.U n R TΔ = Δ


Solution: Solve equation 18-3 for W:
( ) ( ) ( )
3
2
3
2
1210 J 1mol 8.31 J/mol K 276K 272K 1160 J
W Q U Q nR T= −Δ = − Δ
= − ⋅ − =
⎡ ⎤
⎣ ⎦

Insight: Of the 1210 J of heat flow into the gas, only 50 J was converted to internal energy. The rest was energy for
the gas to do work on the external world.


5. Picture the Problem: Three different processes act on a system, resulting in different final states.
Strategy: Use the first law of thermodynamics (equation 18-3) to solve for the unknown quantity in each process.

Solution: 1. (a) Apply equation 18-3 directly:
77 J ( 42 J) 119 JU Q WΔ = − = − − =


2. (b) Apply equation 18-3 directly:
77 J 42 J 35 JU Q WΔ = − = − =


3. (c) Solve equation 18-3 for Q:
120 J 120 J 0Q U W
=
Δ + = − + =

Insight: Conservation of energy requires that the net energy flow (Q and −W) into the system must equal the change in
internal energy, irregardless of the process by which the energy exchange occurs.


6. Picture the Problem: A cycle of four processes is shown on the pressure–
volume diagram at right.
Strategy: Set the sum of the changes in internal energy equal to zero. Then
solve for
CD

.
Solution: 1. Sum the
changes in internal energy:
AB BC CD DA
0U U U UΔ + Δ + Δ + Δ =


2. Solve for
CD

:
CD AB BC DA
U U U UΔ = −Δ −Δ −Δ


3. Insert the numeric values:
(
)
CD
82 J 15 J 56 J 41 JUΔ = − − − − = −

Insight:
In a complete cycle the system returns to its original state, which means that the internal energy must return to
its initial value. Therefore the net change in internal energy must be zero.

Chapter 18
: The Laws of Thermodynamics James S. Walker, Physics, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 3

7.
Picture the Problem
: In a basketball game, the player does work and gives off heat in the form of perspiration.

Strategy:
Use the first law of thermodynamics (equation 18-3) to solve for the change in internal energy. Calculate the
heat loss from the latent heat of vaporization of the perspiration.

Solution:

1. (a)
Write equation 18-3
using the latent heat of vaporization:
U Q W mL W
Δ
= − = − −


2.
Enter the given values:
(
)
(
)
6 5
0.110 kg 2.26 10 J/kg 2.43 10 J 492 kJUΔ = − × − × = −

3. (b)
Convert the change in energy
from Joules to Calories:
(
)
(
)
492 kJ 0.239 kcal/kJ 117 kcal 117 CalUΔ = = =


Insight:
There are about 150 Cal in a 1.0 oz bag of potato chips.

8.
Picture the Problem
: A monatomic ideal gas undergoes a process in which heat is absorbed and work is done by the
gas. The process results in a change in temperature for the gas.

Strategy:
Use the change in energy for a monatomic gas
(
)
3
f i
2
U n R T T
Δ
= −
with the first law of thermodynamics to
solve for the final temperature.

Solution:

1. (a)
Replace ΔU in the first law with
the change in internal energy for a monatomic gas:
( )
3
f i
2
U Q W
nR T T Q W
Δ
= −

=−


2.
Solve for the final temperature:
(
)
f i
2
3
Q W
T T
nR

=
+


3.
Insert the numeric values:

(
)
( ) ( )
f
2 3280 J 722 J
263 K 468 K
3 1mol 8.31 J/mol K
T

= + =

⎡ ⎤
⎣ ⎦

4. (b)
When there are more molecules to share the energy, the average energy gained per molecule is smaller, resulting
in a decrease in the final temperature found in part (a).

Insight:
Because the heat added was greater than the work done, the net change in internal energy was positive. A
positive change in internal energy results in a temperature increase. If the internal energy had decreased, the
temperature would have dropped.

9.
Picture the Problem
: As a car operates, its engine converts the internal energy of the gasoline into both work and heat.

Strategy:
Use the first law of thermodynamics to calculate the heat given off,
rel
.Q


Solution:

1. (a)
Solve the first
law for the heat released.
(
)
rel
8
5
rel
1.19 10 J/gal
5.20 10 J/mi 4.24 MJ mi
25.0 mi/gal
U Q W
Q U W
Δ = − −
⎛ ⎞
− ×
= −Δ − = − − × =
⎜ ⎟
⎝ ⎠


2.

(b)
Increasing miles per gallon improves efficiency, resulting in a decrease of heat released to the atmosphere.

Insight:
The gas mileage of a car is a measure of how much work (miles transported) the car can do for a given amount
of available energy (in gallons of gas). The maximum amount of work would occur if all of the energy was converted
to work and no heat was ejected. However, the Second Law of Thermodynamics says this is not possible.

10.
Picture the Problem
: A monatomic gas undergoes a process in which work is done on the gas, resulting in an increase
in temperature. During the process heat may enter or leave the gas.

Strategy:
Combine the change in energy for a monatomic gas
(
)
3
f i
2
U n R T T
Δ
= −
with the first law of
thermodynamics (equation 18-3) to solve for the heat flow.

Solution:
Solve equation 18-3 for Q:
(
)
( ) ( ) ( )
3
f i
2
3
2
560 J 4 mol 8.31 J/mol K 130 C 5.9 kJ
Q W U W n R T T= + Δ = + −
= − + ⋅ ° =⎡ ⎤
⎣ ⎦


Insight:
It was not necessary to convert the temperature difference from Celsius degrees to kelvins in this problem
because temperature differences are the same in both scales.

Chapter 18
: The Laws of Thermodynamics James S. Walker, Physics, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 4

11.
Picture the Problem
: The three-process cycle for an ideal gas is shown by
the pressure–volume plot at right.


Strategy:
For each process in which two of the parameters Q, W, or
Δ
U
are known, use the first law of thermodynamics to solve for the third
parameter. Process
A
B→
is at constant volume, so use
AB
0W
=
.
Because the three processes form a complete cycle, we can find the change
in internal energy for Process
C A→
by setting the net change in internal
energy equal to zero.

Solution:

1. (a)
Find the constant volume work:
AB AB
(0) 0W P V P= Δ = =


2.

(b)
Use the first law to solve for
the change in internal energy:
AB AB AB
53 J 0 53 JU Q WΔ = − = − − = −


3. (c)
Use the first law to solve for
the change in internal energy:
( )
BC BC BC
280 J 130 J 150 JU Q WΔ = − = − − − = −


4. (d)
Set the net change in
internal energy equal to zero:

AB BC CA
0U U U
Δ
+ Δ + Δ =


5.
Solve for

CA

:

( ) ( )
CA AB BC
53 J 150 J 200 JU U UΔ = −Δ −Δ = − − − − =


6.

(e)
Solve the first law for the heat transfer:
CA CA CA
200 J 150 J 350 JQ U W= Δ + = + =


Insight:
This three-step process converts 17 J of heat
(
)
AB BC CA
Q Q Q+ +
into 17 J of work
( )
AB BC CA
W W W+ +
. Any
time a cycle forms a clockwise loop on a pressure–volume diagram, heat will be transformed into work. If the cycle is
counterclockwise, the net result will be work transformed into heat.

12.
Picture the Problem
: The PV plots at right show three different multi-step
processes, labeled A, B, and C.

Strategy:
The work done by a gas is equal to the area under the PV plot as
long as the gas is expanding (the volume increases). In each of the
depicted processes the gas expands so the work is positive in each case.
Find the area under each plot to determine the ranking of the work done.

Solution:

1.
Find W
A
:
( )
( )
A 1 1 2 2 3 3
3
0 4 kPa 1 m 0 4 kJ
W P V P V P V= Δ + Δ + Δ
= + + =


2.

(b)
Find W
B
:
(
)
(
)
(
)
( )
3 3
1
B 1 1 2 2 3 3
2
1 kPa 1 m 1 3 kPa 2 m 0 5 kJW P V P V P V= Δ + Δ + Δ = + + + =


3. (c)
Find W
C
:
(
)
(
)
3
C 1 1 2 2 3 3
0 1 kPa 3 m 0 3 kJW P V P V P V= Δ + Δ + Δ = + + =

4. (d)
By comparing the values of the work done in each process we arrive at the ranking W
C
< W
A
< W
B
.

Insight:
Even if process B had begun at 0 kPa of pressure and increased linearly to the point (5 m
3
, 3 kPa) to form a
triangle shape, the work done would be
( )
(
)
3
1
2
3 kPa 3 m 4.5 kJ=
and the ranking would remain the same.

13.
Picture the Problem
: An ideal gas absorbs the same amount of heat during two different constant pressure processes.

Strate
gy
: Use the first law of thermodynamics (equation 18-3) and the work at constant pressure (equation 18-4) to
solve for the change in volume.

Solution:

1.
Solve equation 18-3 for the work:


U Q W W Q U
Δ
= − ⇒ = −Δ


2.
Replace the work with equation 18-4
and solve for the change in volume:

Q U
P V Q U V
P
−Δ
Δ = −Δ ⇒ Δ =


3. (a)
Insert the numeric values with
920 J:UΔ =

3
920 J 920 J
0
110 10 Pa
V

Δ
= =
×

Chapter 18
: The Laws of Thermodynamics James S. Walker, Physics, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 5

4.

(b)
Insert the numeric values
360 J:UΔ =

3 3
3
920 J 360 J
5.1 10 m
110 10 Pa
V


Δ = = ×
×


Insight:
When the heat absorbed and change in internal energy were equal, no work was done. Therefore the volume
remained constant. If the change in internal energy were greater than the heat absorbed, then the volume would have
decreased, as work was done on the gas.

14.
Picture the Problem
: An ideal gas is compressed at constant pressure to one-half of its initial volume.

Strategy:
Because this is a constant pressure process, solve equation 18-4 for the initial volume.

Solution:

1.
Write equation 18-4 in
terms of initial and final volumes:
( )
( )
1
f i i i i
2
2

W
P V V W P V V W V
P

− = ⇒ − = ⇒ =


2.
Insert the numeric values:
(
)
3
i
3
2 790 J
0.013 m
120 10 Pa
V
− −
= =
×


Insight:
If the initial volume were larger than 0.013 m
3
, more work would be needed to compress the gas to half its
volume at the same pressure. For instance, it would require 1200 J to compress a gas with an initial volume of 0.020 m
3

to 0.010 m
3
at a constant pressure of 120 kPa.

15.
Picture the Problem
: The ideal gas expands at constant pressure.

Strategy:
Because this is a constant pressure process, solve equation 18-4 for the pressure.

Solution:

1.
Solve equation 18-4 for the pressure:
( )
f i
f i

W
P V V W P
V V
− = ⇒ =



2.
Insert the given values:
3 3
93 J
60 Pa
2.3 m 0.74 m
P = =



Insight:
The pressure is proportional to the amount of work done. For example, compressing a gas at ten times the
pressure (or 600 Pa) between the same two volumes requires ten times the work (or 930 J).

16.
Picture the Problem
: A monatomic ideal gas expands to twice its original volume, while the temperature is held
constant.

Strategy:
An ideal gas undergoing an isothermal process obeys the relation
constant.PV
=
Use this relation to calculate
the ratio of final to initial pressure.

Solution:

1.
Write the isothermal relation in terms
of initial and final conditions and solve for
f i
P P
:
f i
i i f f
i f

P V
PV PV
P V
= ⇒ =


2.
Set
f i
2V V=
and solve:
f i
i i
1
2 2
P V
P V
= =


Insight:
Because the pressure and volume are inversely proportional to each other, increasing one by a given factor will
reduce the other by the same factor. For example, to decrease the volume to one-third its original volume, the pressure
would have to be increased by a factor of three.

17.
Picture the Problem
: A system that is thermally isolated from its surroundings undergoes a process in which its
internal energy increases.

Strategy:
Use the first law of thermodynamics to calculate the work done during an adiabatic process (
0Q
=
).



Solution:

1.
Solve the first law for W, setting Q = 0:
U Q W
W U
Δ
= −
= −Δ


2. (a)
Since the internal energy increased (
0U
Δ
>
) the work must be negative, which means it is done on the system.

3.

(b)
Solve numerically:
( )
670 J 670 JW U= −Δ = − = −


Insight:
When no heat can enter or leave a system, any change in internal energy is equal to the work done on the
system.

Chapter 18
: The Laws of Thermodynamics James S. Walker, Physics, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 6

18.
Picture the Problem
: The pressure–volume diagram shows five
processes through which a system passes in going from state A to
state C. The area under the curve is divided into four geometric
shapes to simplify the area measurement.


Strategy:
The area under the line on the pressure–volume diagram
is the work. Break the area into rectangles and triangles and add up
their areas to find the work. Use the ideal gas law to find the final
temperature and use the first law of thermodynamics to find the
heat.

Solution:

1. (a)
Break the area into four
regions and sum the areas to get the work:
AC rect tri,1 tri,2 square
W P V A A A A= Δ = + + +



2.
Find each area:

(
)
(
)
3
rect
10 m 2 m 200 kPa 1600 kJA lw
3
= = − =
(
)
(
)
3 3
1 1
tri,1
2 2
6 m 4 m 600 kPa – 200 kPa 400 kJ
A bh
= = − =

(
)
(
)
3 3
1 1
tri,2
2 2
8 m 6 m 600 kPa – 400 kPa 200 kJ
A bh
= = − =

(
)
(
)
3 3
square
8 m 6 m 400 kPa – 200 kPa 400 kJA wh= = − =


3.
Sum all four areas:

AC
1600 kJ 400 kJ 200 kJ 400 kJ 2.6 MJ
W
= + + + =

4. (b)
Use the ideal gas law to find a ratio
between volumes and temperatures:

i f
i i f f
i f
,
V V
nR
PV nRT PV nRT
T P T
= = ⇒ = =


5.
Solve the ratio for the final temperature:

( )
3
f
f i
3
i
10 m
220 K 1100 K
2 m
V
T T
V
= = =

6. (c)
Solve the first law for the heat:
3
2
Q U W nR T W
=
Δ + = Δ +


7.
Use the ideal gas law to eliminate
nR
:
i i
i
3
2
PV
Q T W
T
⎛ ⎞
=
Δ +
⎜ ⎟
⎝ ⎠


8.
Solve numerically:

( )
(
)
( )
3
6
200 kPa 2 m
3
880 K 2.60 10 J 5.0 MJ
2 220 K
Q
⎡ ⎤
⎢ ⎥
= + × =
⎢ ⎥
⎣ ⎦


Insight:
The work done in an isobaric process between points A and C would be the area of the rectangle only.
Increasing the pressure as the gas expands increases the work done by the gas.




19.
Picture the Problem
: The pressure–volume diagram shows the two processes
through which a system passes in going from state A to state B. The area under
the curve is divided into a rectangle and triangle to simplify the area
measurement.

Strategy:
Find the area under the curve in the pressure–volume diagram,
because this is equal to the work.

Solution:

1. (a)
Set the work equal
to the area under the curve:
AB rect tri
W P V A A
= Δ = +



2.
Find the area of the rectangle:
( )
( )
rect
3 3
rect
6 m 2 m 200 kPa
800 kJ
A lw
A
=
= −
=

Chapter 18
: The Laws of Thermodynamics James S. Walker, Physics, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 7

3.
Find the area of the triangle:
(
)
(
)
3 3
1 1
tri
2 2
6 m 4 m 600 kPa – 200 kPa 400 kJA bh= = − =


4.
Add the two areas:
AB
800 kJ 400 kJ 1200 kJW = + =


5.

(b)
No, the work only depends on the area under the PV plot.

Insight:
The area under a PV plot is always equal to the work done by the system. It is independent of the type of
system.




20.
Picture the Problem
: A monatomic ideal gas expands at constant temperature.

Strategy:
Use equation 18-5 to find the work done in the isothermal process. Then
solve the first law of thermodynamics (equation 18-3) for the heat.

Solution:

1. (a)
Apply equation 18-5 directly:
f
i
ln
V
W nRT
V
⎛ ⎞
=
⎜ ⎟
⎝ ⎠


2.
Insert the given data:

( ) ( )
4.33L
8.00 mol 8.31 J/mol K 245 K ln 22.0 kJ
1.12L
W
⎛ ⎞
= ⋅ =
⎡ ⎤
⎜ ⎟
⎣ ⎦
⎝ ⎠

3.

(b)
Solve equation 18-3for Q:
0 22.0 kJ 22.0 kJQ U W= Δ + = + =
The heat flow is positive
so this is heat flow into the gas.

4. (c)
Both answers increase by a factor of 2 because the work done is proportional to number of moles of gas.

Insight:
The internal energy of a monatomic ideal gas depends only on the temperature of the gas. In an isothermal
process the temperature remains constant, and therefore so does the internal energy.




21.
Picture the Problem
: A monatomic ideal gas expands at constant
temperature.


Strategy:


Use the Ideal Gas Law to solve for the constant temperature.
Then use equation 18-5 to solve for the work.

Solution:

1. (a)
Solve the Ideal
Gas Law for the temperature:


PV
PV nRT T
nR
= ⇒ =


2.
Insert the numeric values:
(
)
( )
3 3
i f
100 10 Pa 4.00 m
145 mol 8.31 J/mol K
332 K
T T
×
= =





=


3.

(b)
Write equation 18-5,
using the Ideal Gas Law to
replace
nRT PV=
:
( )
3
3
f f
i i
3
i i
4.00 m
ln ln 400 kPa 1.00 m ln 555 kJ
1.00 m
V V
W nRT PV
V V
⎛ ⎞ ⎛ ⎞
⎛ ⎞
= = = =
⎜ ⎟ ⎜ ⎟
⎜ ⎟
⎝ ⎠
⎝ ⎠ ⎝ ⎠


Insight:
If the gas had expanded at constant pressure to the same final volume, and then cooled at constant volume to
the final pressure, the total work done would have been 1200 kJ. Isothermal expansion does less work than isobaric
expansion.

Chapter 18
: The Laws of Thermodynamics James S. Walker, Physics, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 8

22.
Picture the Problem
: A monatomic ideal gas expands at constant
temperature.


Strategy:
Use equation 18-5 and the first law of thermodynamics to
calculate the heat between any two volumes of the isothermal expansion.

Solution 1
. Solve the first law for heat:

Q U W
=
Δ +

2.
Set
0UΔ =
and use equation 18-5 for W:

f
i
0 ln
V
Q nRT
V
⎛ ⎞
= +
⎜ ⎟
⎝ ⎠


3 (a)
Because the gas expands (
f i
V V>
) heat must be positive, so heat
enters the system.


4.

(b)
Heat input equals the work done; work done is equal to the area under the PV plot. The area from
3
1.00 m
to
3
2.00 m
is greater than the area from
3
3.00 m
to
3
4.00 m.


5. (c)
Insert the numeric values for
the expansion from
3
1.00 m
to
3
2.00 m
:
( )
3
3
3
2.00 m
400 kPa 1.00 m ln 277 kJ
1.00 m
Q
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠


6.

(d)
Insert the numeric values for
the expansion from
3
3.00 m
to
3
4.00 m
:
( )
3
3
3
4.00 m
100 kPa 4.00 m ln 115 kJ
3.00 m
Q
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠


Insight:
The heat absorbed in the expansion from
3
1.00 m
to
3
2.00 m
is equal to the heat absorbed in the expansion
from
3
2.00 m
to
3
4.00 m,
because the heat absorbed is a function of the relative volume expansion. In both of these
cases the volume doubles.


23.
Picture the Problem
: A monatomic ideal gas expands under constant pressure.


Strategy:
Use the first law of thermodynamics to solve for the heat transfer. The
change in internal energy can be obtained from equation 17-16, and the work from
the area under the P–V curve.

Solution:

1.
Solve the first law for Q:

3
2
Q U W nR T P V
=
Δ + = Δ + Δ


2.
Use the ideal gas law to eliminate
nR TΔ
:
( )
( )
3 5
2 2
Q P V P V P V
=
Δ + Δ = Δ


3. (a)
From the heat equation we see that heat must be added to the system for the volume to expand.

4.

(b)
Insert given values for the heat:
( )
( )
3 3
5
130 kPa 0.93 m – 0.76 m 55 kJ
2
Q = =


Insight:
When a monatomic ideal gas expands, the density of particles within the gas decreases. In order for the
pressure to remain constant, the temperature of the gas must then increase. This increase in internal energy of the gas
necessitates a heat flow into the gas.


24.
Picture the Problem
: A monatomic ideal gas is thermally isolated from its surroundings. As the gas expands its
temperature decreases.

Strategy:
Use the first law of thermodynamics (equation 18-3) to solve for the work done by the gas. Use
equation 17-16 to find the change in internal energy.

Solution:

1.
Solve the first law for W:
W Q U
=
−Δ


2.

(a)
Because the process is adiabatic, Q = 0.

3. (b)
Substitute
3
2
U n R TΔ = Δ
(equation 17-16):
3
2
0W nR T
=
− Δ

Chapter 18
: The Laws of Thermodynamics James S. Walker, Physics, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 9

4.
Insert the given values:
(
)
(
)( )
3
2
3.92 mol 8.31 J mol K 205 C– 485 C
13.7 kJ
W
=
− ⋅ ° °
=


5. (c)
Solve the first law for the
change in internal energy:
13.7 kJU WΔ = − = −


Insight:
When a monatomic ideal gas expands adiabatically it does work on the external world. The energy to do this
work cannot come from the absorption of heat, so the internal energy, and therefore the temperature, must decrease.


25.
Picture the Problem
: An ideal gas completes a three-process cycle.

Strategy:
Find the net work done by the system by measuring the area
enclosed by the PV plot. Then use the first law of thermodynamics to
calculate the heat absorbed.


Solution:

1. (a)
Find the area of
the triangle on the PV-plot:
(
)
(
)
3
1
2
4 1 m 150 50 kPa
= 150 kJ
W = − −


2.

(b)
0UΔ =
over a complete cycle.

3. (c)
Solve the first law for the heat:
150 kJQ W U= + Δ =


Insight:
Because the change in energy for a complete cycle is zero, the work done is equal to the net heat absorbed.


26.
Picture the Problem
: A monatomic ideal gas undergoes a process in which its
volume increases while the pressure remains constant.


Strategy:
Use the area under the PV plot to find the work done by the gas.
Calculate the initial and final temperatures from the ideal gas law. Use the
change in temperature to calculate the change in internal energy. Finally,
calculate the heat added to the system using the first law of thermodynamics.

Solution:

1. (a)
Multiply P by
:VΔ


(
)
3 3 5
210 kPa 1.9 m 0.75 m 2.4 10 JW P V= Δ = − = ×


2.

(b)
Solve the ideal gas law for T:

PV
T
nR
=


3.
Insert initial conditions:

(
)
( )
3 3
2
i
210 10 Pa 0.75 m
3.9 10 K
49 mol 8.31 J/mol K
T
×
= = ×

⎡ ⎤
⎣ ⎦


4.
Insert the final conditions:
(
)
( )
3 3
2
f
210 10 Pa 1.9 m
9.8 10 K
49 mol 8.31 J/mol K
T
×
= = ×

⎡ ⎤
⎣ ⎦


5. (c)
Use equation 17-15 to solve for
:UΔ

(
) ( ) ( )
3 3
2 2
5
49 mol 8.31 J/mol K 980 K 390 K
3.6 10 J
U n R TΔ = Δ = ⋅ −
⎡ ⎤
⎣ ⎦
= ×


6.

(d)
Solve the first law for heat:
5
360 kJ 240 kJ 6.0 10 JQ U W= Δ + = + = ×


Insight:
The work done by the gas is proportional to the pressure. The temperature, and thus the change in internal
energy, is also proportional to the pressure. Therefore the heat absorbed must also be proportional to the pressure. If
the initial pressure was 420 kPa (double that given in the problem) then the work done would be 480 kJ, the change in
internal energy 720 kJ, and the heat 1200 J.

Chapter 18
: The Laws of Thermodynamics James S. Walker,
Physics
, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 10

27.
Picture the Problem
: An ideal gas completes a three-process cycle.

Strategy:
Use the ideal gas law to calculate the temperature at each state.
Solve the first law of thermodynamics for the heat absorbed in each
process.

Solution:

1. (a)
Solve the ideal
gas law for temperature:

PV
T
nR
=


2.
Solve for
A
T
:
(
)
(
)
( )
( )
3 3
A
150 10 Pa 1.00 m
67.5 mol 8.31 J mol K
267 K
T
×
=

=


3.
Solve for
B
T
:
(
)
(
)
( )
( )
3 3
B
50 10 Pa 4.00 m
357 K
67.5 mol 8.31 J mol K
T
×
= =



4.
Solve for
C
T
:
(
)
(
)
( )
( )
3 3
C
50 10 Pa 1.00 m
89.1 K
67.5 mol 8.31 J mol K
T
×
= =



5.

(b)
A B:→
The temperature rises and the gas does work, so heat enters the system.

B C:→
The temperature drops and work is done on the gas, so heat leaves the system.

C A:→
The temperature rises and no work is done on or by the gas, so heat enters the system.

6. (c)
Solve the first law for
Q
:

3
2
Q U W nR T W= Δ + = Δ +


7.
Insert values for
A B:→

(
)
(
)
(
)
( )
( )
3
2
3
1
2
67.5 mol 8.31 J mol K 357 K 267 K
150 kPa 50 kPa 3.00 m 376 kJ
Q = ⋅ −
+ + =


8.
Insert values for
B C:→

(
)
(
)
(
)
( )
3
2
3
67.5 mol 8.31 J mol K 89.1 K 357 K
50 kPa 3.00 m 375 kJ
Q = ⋅ −
+ − = −


9.
Insert values for
C:
A


(
)
(
)
(
)
3
2
67.5 mol 8.31 J mol K 267 K 89.1 K 0 150 kJQ = ⋅ − + =


Insight:
The net work done during this cycle is equal to the net heat absorbed, or W

=

150 kJ. Although the sum of the
heats in part (d) appears to be 151 kJ, that is an artifact of significant digits and rounding issues, it’s exactly 150 kJ.


28.
Picture the Problem
: A gas with initial volume
i
V
either expands at
constant pressure to twice its initial volume, or contracts to 1/3 its original
volume.

Strategy:
Use the area under the PV plot to find the work done by the gas.

Solution:

1. (a)
Set the work equal
to the pressure times volume:
(
)
( )
f i
i i
i
2
W P V V
P V V
W PV
= −
= −
=


2.
Insert the numeric values:

(
)
3
140 kPa 0.66 m 92 kJW = =


3.

(b)
Set the work equal to the
pressure times volume:
( )
i
f i i
i
2
3 3
V
W P V V P V PV
⎛ ⎞
= − = − = −
⎜ ⎟
⎝ ⎠


4.
Insert the numeric values:
( )
( )
3
2
140 kPa 0.66 m 62 kJ
3
W = − = −


Insight:
The work is positive when the volume increases and negative when the volume decreases.

Chapter 18
: The Laws of Thermodynamics James S. Walker, Physics, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 11

29.
Picture the Problem
: A system expands at constant pressure.

Strategy:
Use the first law of thermodynamics to find the heat absorbed

Solution:

1.
Solve the first law for heat:

Q U W U P V
=
Δ + = Δ + Δ


2. (a)
Find Q for
65 J:UΔ =

(
)
(
)
3 3
65 J 125 10 Pa 0.75 m
94 kJ; into the gas
Q = + ×
=


3.

(b)
Find Q for
1850 J:UΔ = −

(
)
3 3
1850 J 125 10 Pa 0.75 m 92 kJ;into the gasQ = − + × =


Insight:
For heat to flow out of the gas, the internal energy would need to decrease by at least 93.8 kJ, which is equal to
the amount of work done by the gas.

30.
Picture the Problem
: A monatomic ideal gas is adiabatically compressed.

Strategy:
Use the ideal gas law to relate the initial and final conditions of the gas. Then use equation 18-9 to eliminate
the unknown volumes from the equation.

Solution:

1 (a)
Doing work on the system must increase the internal energy and therefore the temperature when no heat
flows into or out of the system.

2. (b)
Write the ideal gas law in
terms of the initial and final states:

i i f f
i f

PV PV
PV nRT nR
T T
= ⇒ = =


3.
Solve for the final temperature:

f f
f i
i i
P V
T T
P V
⎛ ⎞⎛ ⎞
=
⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠


4.
Now solve equation 18-9
for the ratio of volumes:
1 1
f i f
i f i

i i f f
V P P
PV P V
V P P
γ
γ
γ γ

⎛ ⎞ ⎛ ⎞
= ⇒ = =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
, where
5
3
γ
=


5.
Insert this ratio into the equation for
f
T
:
( )
(
)
1 1
3
5
1
1
f f f
f i i
i i i
140 kPa
280 K 310 K
110 kPa
P P P
T T T
P P P
γ γ
− −

⎛ ⎞⎛ ⎞ ⎛ ⎞
⎛ ⎞
= = = =
⎜ ⎟⎜ ⎟ ⎜ ⎟
⎜ ⎟
⎝ ⎠
⎝ ⎠⎝ ⎠ ⎝ ⎠

Insight:
As predicted, when work was done on the gas in an adiabatic process the temperature increased.

31.
Picture the Problem
: A monatomic ideal gas expands where the increase
in pressure is proportional to the increase in volume.

Strategy:
Calculate the work done from the area under the PV plot. Use
equations 17-16 and the ideal gas law to find the change in internal energy.
Lastly, use the first law to calculate the heat flow into the gas.


Solution:

1. (a)
Determine the work
from the area under the graph:
(
)
(
)
1
i i i i
2
i i
3 2
3
W V V P P
PV
= − +
=

2.

(b)
Write the change in internal
energy in terms of equation 17-16:
f i
3 3
f i
2 2
U U U
nRT nRT
Δ
= −
= −


3.
Use the ideal gas law to write in
terms of pressure and volume:

(
)
3 3 3
f f i i f f i i
2 2 2
U P V PV P V PVΔ = − = −

4.
Write in terms of initial conditions:

( )( )
3 15
i i i i i i
2 2
2 3U P V PV PVΔ = − =⎡ ⎤
⎣ ⎦

5. (c)
Solve the first law for heat:
15
21
i i i i i i
2 2
3Q U W PV PV PV= Δ + = + =


Insight:
Suppose the pressure were to triple as the volume triples. In this case, the work would increase to
i i
4PV
, the
change in internal energy would increase to
i i
12PV
, and the heat would increase to
i i
16PV
.

Chapter 18
: The Laws of Thermodynamics James S. Walker, Physics, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 12

32.
Picture the Problem
: A gas expands to double its initial volume in three different
processes: constant pressure, isothermal, and adiabatic.


Strate
gy
:
Use the P–V diagram to rate the work because the work is the area under
t
he curve. Determine the ranking of final temperatures from the ideal gas law.

Solution:

1. (a)
The area under the constant-pressure curve is the greatest, hence this
p
rocess does the most work.


2.

(b)
The area under the adiabatic curve is the smallest, hence this process does the
least work.


3.
Solve the ideal gas law for temperature and note that
the final temperature is proportional to the final pressure:

0
f f
2V
PV
T P
nR nR
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠


4.

(c)
The constant-pressure expansion has the highest final temperature because it has the highest final pressure.

5.

(d)
The adiabatic expansion has the lowest final temperature, because at final volume, it has the lowest pressure.

Insight:
In the isothermal process the temperature remained constant. From the graph we see that the final temperature
in the constant pressure is higher than the temperature in the isothermal process, therefore the temperature increased
during the constant pressure process. In the adiabatic process the temperature decreased.




33.
Picture the Problem
: You plan to add a certain amount of heat to a gas in order to raise its temperature.

Strategy:
Use the principles concerned with constant volume and constant pressure processes to answer the question.

Solution:

1. (a)
At constant pressure, the gas expands and does work as the heat is added. Hence, only part of the heat
goes into increasing the internal energy. When heat is added at constant volume no work is done. In this case, all the
heat goes into increasing the internal energy. We conclude that if you add the heat at constant volume, the increase in
temperature is greater than the increase in temperature if you add the heat at constant pressure.

2.

(b)
The best explanation is
II
. All the heat goes into raising the temperature when added at constant volume; none
goes into mechanical work. Statement I ignores the work done during gas expansion at constant pressure, and
statement III is false.

Insight:
More heat is required to increase the temperature at constant pressure because part of the input heat is
converted to the work of the expanding gas.




34.
Picture the Problem
: Heat is added to monatomic ideal gas to increase its temperature. The heat is added either at
constant volume or constant pressure.

Strategy:
Use the molar specific heat at constant volume
V
C
(equation 18-6) and at constant pressure
P
C
(equation
18-7) to solve for heat.

Solution:

1. (a)
Write Q in terms of
P
C
:
( ) ( ) ( )
5 5
p
2 2
3.5 mol 8.31 J/mol K 23 K 1.7 kJQ nR T= Δ = ⋅ =
⎡ ⎤
⎣ ⎦

2.

(b)
Write Q in terms of
V
C
:
( ) ( ) ( )
3 3
v
2 2
3.5 mol 8.31 J/mol K 23 K 1.0 kJQ nR T= Δ = ⋅ =⎡ ⎤
⎣ ⎦

Insight:
More heat is required to increase the temperature at constant pressure because part of the input heat is
converted to the work of the expanding gas.

Chapter 18
: The Laws of Thermodynamics James S. Walker, Physics, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 13

35.
Picture the Problem
: Heat is added to a monatomic gas resulting in a temperature increase. This process is performed
at constant volume and then again at constant pressure.

Strategy:
Use the molar specific heat at constant volume (equation 18-6) and at constant pressure (equation 18-7) to
solve for the temperature change for a given heat input.

Solution:

1. (a)
Write Q in terms
of
V
C
and solve for
:TΔ

( )
( ) ( )
3
v
2
v
2 535 J
2
0.95 K
3
3 45 mol 8.31 J/mol K
Q nR T
Q
T
nR
= Δ
Δ = = =

⎡ ⎤
⎣ ⎦


2.

(b)
Q in terms of
P
C
and solve for
:TΔ

( )
( ) ( )
5
p
2
p
2
2 535 J
0.57 K
5
5 45 mol 8.31 J/mol K
Q nR T
Q
T
nR
= Δ
Δ = = =

⎡ ⎤
⎣ ⎦


Insight:
A greater temperature change occurs at constant volume, because all of the heat goes into increasing the
temperature. In the constant pressure case, some of the heat is converted to work as the gas expands.


36.
Picture the Problem
: Heat is added to a monatomic ideal gas until the internal energy has doubled. This process is
first done at constant pressure and then at constant volume.

Strategy:
Use equation 17-15 to determine the relationship between initial and final temperatures when the internal
energy doubles. Then use the molar heat at constant volume (equation 18-6) and at constant pressure (equation 18-7) to
calculate the heat added.


Solution:

1. (a)
Divide equation 17-15 by
temperature and equate initial and final terms:
i f
i f
3
2
U U
nR
T T
= =


2.


Set
f i
2U U=
and solve for T
f

:
f i
f i i i
i i
2
2
U U
T T T T
U U
⎛ ⎞ ⎛ ⎞
= = =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠


3.
Write the heat in terms of molar
heat capacity at constant pressure:
(
)
5
P f i
2
Q nR T T= −


4.
Write final temperature in
terms of initial temperature:

(
)
5 5
P i i i
2 2
2Q nR T T nRT= − =


5.
Solve numerically:

( ) ( ) ( )
5
P
2
2.5 mol 8.31 J/mol K 325 K 17 kJQ = ⋅ =
⎡ ⎤
⎣ ⎦


6.

(b)
Write the heat in terms of molar
heat capacity at constant volume:
3 3
V i
2 2
Q nR T nRT= Δ =


7.
Solve numerically:

( ) ( ) ( )
3
V
2
2.5 mol 8.31 J/mol K 325 K 10 kJQ = ⋅ =⎡ ⎤
⎣ ⎦

Insight:
At 325 K the internal energy of the gas was 10 kJ. Adding 10kJ at constant volume is sufficient to double the
internal energy, but it is not sufficient at constant pressure.


37.
Picture the Problem
: Heat is added to monatomic ideal gas to increase its temperature. The heat is added at constant
pressure and then at constant volume.

Strategy:
Use the molar specific heat at constant volume (equation 18-6) and at constant pressure (equation 18-7) to
solve for heat.

Solution:

1. (a)
Solve
5
P
2
Q nR T= Δ
for
:TΔ

(
)
( )
( )
P
2 170 J
2
2.9 K
5 5 2.8 mol 8.31 J mol K
Q
T
n R
Δ = = =



2.

(b)
Solve
3
V
2
Q nR T= Δ
for
:TΔ

(
)
( )
( )
V
2 170 J
2
4.9 K
3 3 2.8 mol 8.31 J mol K
Q
T
n R
Δ = = =



Insight:
A greater temperature change occurs at constant volume, because all of the heat goes into increasing the
temperature. In the constant pressure case some of the heat is converted to work as the gas expands.

Chapter 18
: The Laws of Thermodynamics James S. Walker, Physics, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 14

38.
Picture the Problem
: A monatomic ideal gas is expanded at constant
pressure by a fixed change in volume. This process is done twice with
differing initial volumes.

Strategy:
Use the area under the PV plot to calculate the work done in
each process.

Solution:

1. (a)
Write the work
as area under the PV plot:

W P V
=
Δ


2.
Solve numerically:
( )( )
6 3
3 3 3
3
1 10 m
160 10 Pa 8600 cm 5400 cm 0.51 kJ
cm
W

⎛ ⎞
×
= × − =
⎜ ⎟
⎝ ⎠


3.

(b)
Work is directly proportional to the change in volume. Therefore, the work done by the gas in the second
expansion is equal to that done in the first expansion.

4. (c)
Solve numerically:
( )( )
6 3
3 3 3
3
1 10 m
160 10 Pa 5400 cm 2200 cm 0.51 kJ
cm
W

⎛ ⎞
×
= × − =
⎜ ⎟
⎝ ⎠


Insight:
Because the work is proportional to the change in volume, increasing the volume by 3200 cm
3
from any initial
volume will produce the same amount of work.


39.
Picture the Problem
: A monatomic ideal gas that is thermally isolated from its surroundings expands to double its
initial volume.

Strategy:
Use equation 18-9 to calculate the ratio of the initial and final pressures. Use that result and the ideal gas law
to determine the ratio of the initial and final temperatures.

Solution:

1. (a)
Solve equation 18-9 for
the ratio of final to initial pressure:
f i
f f i i
i f

P V
P V PV
P V
γ
γ γ
⎛ ⎞
= ⇒ =
⎜ ⎟
⎝ ⎠


2.
Set final volume as twice the initial:

5 3
5 3
f i
i i
1
0.315
2 2
P V
P V
⎛ ⎞
⎛ ⎞
= = =
⎜ ⎟
⎜ ⎟
⎝ ⎠
⎝ ⎠


3.

(b)
Calculate the ratio of initial to final
temperatures using the ideal gas law:

( )( )
f f f f f
i i i i i
0.315 2 0.630
T P V nR P V
T PV nR P V
⎛ ⎞⎛ ⎞
= = = =
⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠


4 (c)
Calculate
f
P
from equation 18-9:
5/3
3
i
f i
3
f
1.2 m
330 kPa 100 kPa
2.4 m
V
P P
V
γ
⎛ ⎞
⎛ ⎞
= = =
⎜ ⎟
⎜ ⎟
⎝ ⎠
⎝ ⎠


5.
Calculate the ratio of pressures:
f
i
104 kPa
0.315
330 kPa
P
P
= =
, which matches step 2.

6.
Calculate
i
T
from the ideal gas law:
(
)
( )
3 3
i i
i
330 10 Pa 1.2 m
353 K
135 mol 8.31 J/mol K
PV
T
nR
×
= = =

⎡ ⎤
⎣ ⎦


7.
Calculate
f
T
from the ideal gas law:

(
)
( )
3 3
f f
f
104 10 Pa 2.4 m
222 K
135 mol 8.31 J/mol K
PV
T
nR
×
= = =

⎡ ⎤
⎣ ⎦


8.
Calculate the ratio of temperatures:

f
i
222 K
0.630
353 K
T
T
= =
, which matches step 3.

Insight:
When a monatomic ideal gas expands to double its volume, its pressure decreases rapidly to less than a third of
its initial pressure while its temperature decreases to a little under two-thirds of the initial Kelvin temperature.

Chapter 18
: The Laws of Thermodynamics James S. Walker, Physics, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 15

40.
Picture the Problem
: A thermally isolated monatomic ideal gas is compressed causing its pressure and temperature to
increase.

Strategy:
Use equation 18-9 to calculate the final volume in terms of the initial and final pressures. Then combine
equation 18-9 and the ideal gas law to derive an equation for the final volume in terms of the temperatures.

Solution:

1. (a)
Solve equa-
tion 18-9 for the final volume:
( )
1
3/5
3 3
i
i i f f f i
f
105 kPa
0.0750 m 0.0618 m
145 kPa
P
PV P V V V
P
γ
γ γ
⎛ ⎞
⎛ ⎞
= ⇒ = = =
⎜ ⎟
⎜ ⎟
⎝ ⎠
⎝ ⎠


2.

(b)
Combine the Ideal
Gas Law with equation 18-9
and solve for final volume:
( )
( )
5
3
i i f f
i f
i f
i f
1
1/1
1
3 3
f
f i
i
295 K
0.0750 m 0.0835 m
317 K
PV P V
nRT nRT
V V
V V
T
V V
T
γ γ
γ γ
γ


=
⎛ ⎞ ⎛ ⎞
=
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
⎛ ⎞
⎛ ⎞
= = =
⎜ ⎟
⎜ ⎟
⎝ ⎠
⎝ ⎠


Insight:
When a monatomic ideal gas is compressed adiabatically, its volume decreases while its temperature and
pressure increase. Because the temperature in part (b) is less than the initial temperature, the gas had to expand as it
cooled. This is seen in the final volume being greater that the initial volume.



41.
Picture the Problem
: A monatomic ideal gas is heated at constant
volume and then expanded at constant pressure.

Strategy:
Calculate the three temperatures (
i 1 f
, and T T T
) using the
Ideal Gas Law. Use these temperatures together with
V
C
and
P
C
to
calculate the heat transferred during each process. Calculate the work
during the constant pressure process from the area under the PV plot.
Finally, use the first law of thermodynamics to find the change in
internal energy.

Solution:

1. (a)
Solve the Ideal Gas Law for
i
T
:
(
)
( )
3 3
i i
i
106 10 Pa 1.00 m
212.6 K
60.0 mol 8.31 J/mol K
PV
T
nR
×
= = =

⎡ ⎤
⎣ ⎦


2.
Solve the Ideal Gas Law for
1
T
:
(
)
( )
3 3
1 1
1
212 10 Pa 1.00 m
425.2 K
60.0 mol 8.31 J/mol K
PV
T
nR
×
= = =

⎡ ⎤
⎣ ⎦


3.
Solve the Ideal Gas Law for
f
T
:
(
)
( )
3 3
f f
f
212 10 Pa 3.00 m
1275.6 K
60.0 mol 8.31 J/mol K
P V
T
nR
×
= = =

⎡ ⎤
⎣ ⎦


4.
Find the heat for process 1:

(
)
( ) ( ) ( )
3
v 1 i
2
3
2
60.0 mol 8.31 J/mol K 212.6 K 159 kJ
Q nR T T= −
= ⋅ =
⎡ ⎤
⎣ ⎦


5.
Find the heat for process 2:

(
)
( ) ( ) ( )
5
P f 1
2
5
2
60.0 mol 8.31 J/mol K 850.4 K 1060 kJ
Q nR T T= −
= ⋅ =⎡ ⎤
⎣ ⎦


6. (b)
Find the work from the area under
the PV plot:
(
)
(
)
3 3 3
212 10 Pa 3.00 m 1.00 m 424 kJW P V= Δ = × − =


7. (c)
Use the first law to determine the total
change in internal energy:
159 kJ 1060 kJ – 424 kJ 795 kJU Q WΔ = − = + =


Insight:
The change in internal energy is independent of the processes between the initial and final states. If the initial
process had been at constant pressure and the final at constant volume, the net work and heat absorbed would be less
than in this problem, but the change in internal energy would still be the same.

Chapter 18
: The Laws of Thermodynamics James S. Walker, Physics, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 16

42.
Picture the Problem
: A monatomic ideal gas expands at constant
pressure and then is heated at constant volume.

Strategy:
Calculate the three temperatures (
i 2 f
, and T T T
) using the
Ideal Gas Law. Use these temperatures together with
V
C
and
P
C
to
calculate the heat transferred during each process. Calculate the work
during the constant pressure process from the area under the PV plot.
Finally, use the first law of thermodynamics to find the change in
internal energy.

Solution:

1. (a)
Solve the Ideal Gas Law for
i
T
:
(
)
( )
3 3
i i
i
106 10 Pa 1.00 m
212.6 K
60.0 mol 8.31 J/mol K
PV
T
nR
×
= = =

⎡ ⎤
⎣ ⎦


2.
Solve the Ideal Gas Law for
2
T
:
(
)
( )
3 3
2 2
2
106 10 Pa 3.00 m
637.8 K
60.0 mol 8.31 J/mol K
P V
T
nR
×
= = =

⎡ ⎤
⎣ ⎦


3.
Solve the Ideal Gas Law for
f
T
:
(
)
( )
3 3
f f
f
212 10 Pa 3.00 m
1275.6 K
60.0 mol 8.31 J/mol K
P V
T
nR
×
= = =

⎡ ⎤
⎣ ⎦


4.
Find the heat for process 3:

(
)
( ) ( ) ( )
5
p 2 i
2
5
2
60.0 mol 8.31 J/mol K 425.2 K 530 kJ
Q nR T T= −
= ⋅ =⎡ ⎤
⎣ ⎦


5.
Find the heat for process 4:

(
)
( ) ( ) ( )
3
v f 2
2
3
2
60.0 mol 8.31 J/mol K 637.8 K 477 kJ
Q nR T T= −
= ⋅ =
⎡ ⎤
⎣ ⎦


6. (b)
Find the work from the
area under the PV plot:
(
)
(
)
3 3 3
106 10 Pa 3.00 m 1.00 m 212 kJW P V= Δ = × − =


7. (c)
Use the first law to determine
the total change in internal energy:
530 kJ 477 kJ – 212 kJ 795 kJU Q WΔ = − = + =


Insight:
The change in internal energy is independent of the processes between the initial and final states. If the initial
process had been at constant volume and the final at constant pressure, the net work and heat absorbed would be greater
than in this problem, but the change in internal energy would still be the same.




43.
Picture the Problem
: A Carnot engine operates between a hot reservoir at the Kelvin temperature T
h
and a cold
reservoir at the Kelvin temperature T
c
.

Strategy:
Use the expression for the efficiency of a Carnot heat engine to answer the conceptual questions.

Solution:

1. (a)
The efficiency of a Carnot heat engine (e
max
= 1 – T
c
/T
h
) (equation 18-13) depends on the ratio of T
c
to
T
h
. Doubling both temperatures leaves this ratio unchanged. We conclude that if both temperatures are doubled, the
efficiency of the engine will stay the same.

2.

(b)
Adding the same temperature to both T
c
and T
h
means that the ratio T
c
/T
h
will have a value that is closer to 1.
Therefore, the efficiency of the engine will decrease.

Insight:
The best way to increase the efficiency of a Carnot engine is to increase the temperature difference between
the two reservoirs. In real life the cold reservoir is usually the environment (near 300 K) and cannot be changed, so the
best option is to increase the temperature of the hot reservoir.

Chapter 18
: The Laws of Thermodynamics James S. Walker, Physics, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 17

44.
Picture the Problem
: A Carnot engine can be operated with one of four sets of reservoir temperatures

Strategy:
Use the expression for the efficiency of a Carnot heat engine (equation 18-13) to determine the ranking of the
efficiencies when the engine is operated between the stipulated temperature reservoirs.

Solution:

1.
Find e
A
:
c, A
A
h, A
400 K
1 1 0.50
800 K
T
e
T
= − = − =

2.
Find e
B
:
c, B
B
h, B
400 K
1 1 0.33
600 K
T
e
T
= − = − =


3.
Find e
C
:
c, C
C
h, C
800 K
1 1 0.33
1200 K
T
e
T
= − = − =

4.
Find e
D
:
c, D
D
h, D
800 K
1 1 0.20
1000 K
T
e
T
= − = − =


5.
By comparing the efficiencies we arrive at the ranking e
D
< e
B
= e
C
< e
A
.

Insight:
The larger the temperature difference between the two reservoirs, the greater the efficiency of a Carnot heat
engine. A Carnot engine operating between 300 K and 1200 K would have an efficiency of 75%.

45.
Picture the Problem
: A heat engine does work as it extracts heat from the hot reservoir and
rejects heat to the cold reservoir.

Strategy:
Use equation 18-10 to calculate
h
Q
and then equation 18-11 to calculate the
efficiency:

Solution:

1.
Solve equation 18-10 for
h
:Q

h c
h c
W Q Q
Q W Q
=

= +


2.
Insert the expression from
step 1 into equation 18-11:

h c
340 J
0.28
340 J 870 J
W W
e
Q W Q
= = = =
+ +


Insight:
Because the work W is always less than
c
W Q
+
the efficiency must always be less than one.

46.
Picture the Problem
: A heat engine does work as it extracts heat from the hot reservoir and
rejects heat to the cold reservoir.

Strategy:
Use equation 18-10 to calculate
W
and then equation 18-11 to calculate the
efficiency.

Solution:

1. (a)
Calculate work
from equation 18-10:
h c
690 J 430 J 260 JW Q Q= − = − =


2.

(b)
Calculate e from equation 18-11:
h
260 J
0.38
690 J
W
e
Q
= = =


Insight:
Decreasing
c
Q
while keeping
h
Q
constant will always increase the work done and increase the efficiency.

47.
Picture the Problem
: A Carnot heat engine does 2500 J of work as it extracts heat from the
hot reservoir and rejects heat to the cold reservoir.


Strategy:
Use equation 18-14 to solve for the heat input
h
,Q
and then calculate the heat
rejected
c
Q
from equation 18-10.

Solution:

1. (a)
Solve
equation 18-14 for
h
:Q

h
c h
2500 J
8.5 kJ
1 1 290 K 410 K
W
Q
T T
= = =
− −


2.

(b)
Solve equation 18-10 for
c
:Q

3
c h
8.5 10 J 2500 J 6.0 kJQ Q W= − = × − =


Insight:
Equation 18-14 shows that for two fixed reservoirs the work done by a Carnot
engine is proportional to the heat absorbed. Equation 18-10 shows that the heat rejected is
also proportional to the work.

Chapter 18
: The Laws of Thermodynamics James S. Walker, Physics, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 18

48.
Picture the Problem
: A heat engine does work at a rate of 250 MW as it extracts heat from
the hot reservoir at a rate of 820 MW and rejects heat to the cold reservoir.


Strategy:
Solve equation 18-10 for
c
Q
and divide by time to determine the rate that heat is
discarded from the heat engine. Then find the efficiency from equation 18-11.


Solution:

1. (a)
Solve eq. 18-10
for
c
Q
and divide by time:
c
h
c h

Q
Q
W
Q Q W
t t t
=
− ⇒ = −


2.
Insert the numeric values:

c
838 MW 253 MW 585 MW
Q
t
= − =


3.

(b)
Write equation 18-11 in terms of rates:
h h
253 MW
0.302
838 MW
W W t
e
Q Q t
= = = =


Insight:
Typically in power plants it is the rate of heat flow and power output that is important. As shown in this
problem, the same equations used for heat and work can be used to determine heat flow rates and power output.

49.
Picture the Problem
: A heat engine produces power as it extracts heat from a hot temperature reservoir and rejects
some heat to a cold temperature reservoir.

Strategy:
Combine equations 18-10 and 18-11 to solve for the rate of heat rejected to the cold reservoir in terms of the
efficiency and power output. Then use equation 18-11 to solve for the rate that heat must be supplied.


Solution:

1. (a)
Combine equations
18-10 and 18-11 to find
c
Q
:

h c
c c
1
1
W W
e
Q W Q
W
W Q Q W
e e
= =
+
⎛ ⎞
+
= ⇒ = −
⎜ ⎟
⎝ ⎠

2.
Divide by the time to find the exhaust rate:

( )
c
1 1
1 548 MW –1 1.16 GW
0.320
Q
W
t t e
⎛ ⎞ ⎛ ⎞
= − = =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠

3.

(b)
Write equation 18-11 as a
rate equation and solve for
h
Q t
:
h
548 MW
1.71 GW
0.32
Q
W t
t e
= = =


Insight:
Typically in power plants it is the rate of heat flow and power output that is important. As shown in this
problem, the same equations used for heat and work can be used to determine heat flow rates and power output.

50.
Picture the Problem
: A heat engine produces 2700 J of work as it extracts heat from a hot
temperature reservoir and rejects heat to a cold temperature reservoir.

Strategy:
Use equation 18-11 to calculate
h
Q
and then equation 18-10 to calculate
c
Q
:

Solution:

1. (a)
Solve equation 18-11 for
h
Q
:
h
W
Q
e
=


2.


Insert the given values:
h
2700 J
15 kJ
0.18
Q = =


3. (b):
Solve equation 18-19 for
c
Q
:
4
c h
1.50 10 J 2700 J 12 kJQ Q W= − = × − =


4. (c)
Higher efficiency means less heat input is needed to produce the same work. Consequently, less heat is lost to the
surroundings. The answers in parts (a) and (b) will decrease.

Insight:
As an example of higher efficiency, if
0.24e
=
and
W
= 2700 J, then
h
10.0 kJQ
=
and
c
7.6 kJ.Q =


51.
Picture the Problem
: The efficiency of a Carnot engine is increased by lowering the cold temperature reservoir.

Strategy:
Use equation 18-13 to solve for the temperature of the cold temperature reservoir using the initial efficiency,
and then the new efficiency.

Solution:

1. (a)
Solve equation 18-13 for
c
T
:
(
) ( )
c h max
1 545 K 1 0.300 382 KT T e= − = − =

Chapter 18
: The Laws of Thermodynamics James S. Walker,
Physics
, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 19

2.

(b)
The efficiency of a heat engine increases as the difference in temperature of the hot and cold reservoirs increases.
Therefore, the temperature of the low temperature reservoir must be decreased.

3. (c)
Insert the new efficiency:
(
)
c
545 K 1 0.400 327 KT = − =


Insight:
The efficiency of a Carnot engine is increased when the temperature difference between the two reservoirs
increases. When the hot temperature reservoir is fixed, the efficiency can be increased by lowering the temperature of
the cold reservoir.

52.
Picture the Problem
: A Carnot engine produces 2200 J of work as it extracts 2500 J of heat
from a hot temperature reservoir and rejects heat to a cold temperature reservoir.


Strategy:
Use equation 18-11 to determine the efficiency. Solve equation 18-10 for the heat
exhausted at the low temperature. Finally, solve equation 18-13 for the ratio of reservoir
temperatures.

Solution:

1. (a)
Calculate the
efficiency from equation 18-11:
h
2200 J
0.88
2500 J
W
e
Q
= = =


2.

(b)
Solve equation 18-10 for
c
Q
:
c h
2500 J 2200 J 300 JQ Q W= − = − =


3. (c)
Solve equation 18-13
for temperature ratio:
c
h
max
h c max
1 1
1 8.3
1 1 0.88
T
T
e
T T e
= − ⇒ = = =
− −


Insight:
A heat engine with 88% efficiency would need to have the hot temperature reservoir at least 8.3 times hotter
than the cold temperature reservoir. For instance, if the cold reservoir is at room temperature (300 K) the hot
temperature reservoir would have to be at least 2500 K, plenty hot enough to melt steel!

53.
Picture the Problem
: A Carnot engine operates between two temperature reservoirs where the hot reservoir is 55 K
warmer than the cold reservoir.

Strategy:
Insert the efficiency and reservoir temperatures into equation 18-13 and solve for the cold temperature.

Solution:

1.
Write the hot temperature
as
h c
55T T= +
in equation 18-13.
c
max
c
1
55
T
e
T
= −
+


2.
Solve for
c
:T


2
c
max
55 K 55 K
55 K 55 K 445 K 4.5 10 K
0.11
T
e
= − = − = = ×


3.
Now find
h
T
from the given relation:
2
h c
55 K = 445 K + 55 K 500 K 5.0 10 K
T T= + = = ×


Insight:
The efficiency is determined by the ratio of the temperatures, not by their differences. For example, a second
set of temperatures
c
= 200 KT
and
h
= 255 KT
are separated by 55 K, but a Carnot engine operating between these
two temperatures has an efficiency of 22%.

54.
Picture the Problem
: A Carnot engine produces work in a cycle where it exhausts 2/3 of the
input heat to the cold reservoir (
2
c h
3
Q Q=
).

Strategy:
Calculate the efficiency from equation 18-12 and the fact that
2
c h
3
.Q Q=
Then
use equation 18-13 to determine the reservoir temperature ratio.

Solution:

1. (a)
Write equation 18-12:
2
h
c 3
h h
1
1 1
3
Q
Q
e
Q Q
= − = − =


2. (b)
Solve equation 18-13
for the temperature ratio:

c
max
h
1 2
1 1
3 3
T
e
T
= − = − =


Insight:
In a Carnot engine, the ratio of heat exhausted to heat input will always be equal to the ratio of the temperature
of cold reservoir to the temperature of the hot reservoir, as in this case
c h c h
2 3Q Q T T
=
=
.

Chapter 18
: The Laws of Thermodynamics James S. Walker,
Physics
, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 20

55.
Picture the Problem
: A dozen ice cubes are frozen in a freezer that is located in a kitchen with a variable temperature.

Strategy:
Determine the temperature dependence of the coefficient of performance of an ideal refrigerator in order to
answer the conceptual question.

Solution:

1. (a)
The coefficient of performance of an ideal refrigerator is given by
c
Q W
(equation 18-15). Use
h c
W Q Q= −
(equation 18-10) together with the Carnot relationship
h h
c c
Q T
Q T
=
to determine the temperature dependence
of the COP of an ideal refrigerator.

2.
Substitute for
W
and
h c
Q Q

in equation 18-15:
c c c
h c h c h c h c
1 1
COP
1 1
Q Q T
W Q Q Q Q T T T T
= = = = =

− − −


3.
From this expression we can see that the smaller the temperature difference
T
h

T
c
, the greater the coefficient of
p
erformance. Cooling the kitchen reduces the temperature difference between the freezer compartment and the kitchen.
We conclude that if the temperature in the kitchen is decreased, the cost (work needed) to freeze a dozen ice cubes is
less than it was before the kitchen was cooled

4. (b)
The best explanation is
I.
The difference in temperature between the inside and the outside of the refrigerator is
decreased, and hence less work is required to freeze the ice. Statements II and III are each false.

Insight:
The derived result suggests that air conditioners become less efficient the larger the temperature difference
between indoors and outdoors. This is one reason why electrical demand is so high on very hot days. Many air
conditioners are turned on during such a heat wave, and each of them operates less efficiently than they would on a
cooler day.


56.
Picture the Problem
: A refrigerator extracts heat from the cold reservoir and ejects the heat
to the hot reservoir. This process requires work to be input into the system.

Strategy:
Solve equation 18-10 for the heat exhausted to the hot reservoir. Then use
equation 18-15 to solve for the coefficient of performance of the refrigerator.

Solution:

1. (a)
Solve equation 18-10 for
h
:Q

h c
Q Q W
=
+


2.
Insert the given values:
h
110 J 480 J 0.59 kJQ = + =


3.

(b)
Insert the values into equation 18-15:
c
110 J
COP 0.23
480 J
Q
W
= = =


Insight:
This rather inefficient refrigerator exhausts to the room over five times the amount of heat extracted from
inside the refrigerator. The additional heat comes from the work done by the refrigerator.


57.
Picture the Problem
: A refrigerator extracts heat from the cold reservoir and ejects the heat
to the hot reservoir. This process requires work to be input into the system.

Strategy:
Solve equation 18-15 for the work needed to run the refrigerator. Then use
equation 18-10 to determine the heat exhausted to the hot reservoir.



Solution:

1. (a)
Solve equation 18-15 for
W
:
4
c
3.45 10 J
19.7 kJ
COP 1.75
Q
W
×
= = =


2. (b)
Solve equation 18-10 for
h
:Q
:

h c
4 4
3.45 10 J 1.97 10 J 54.2 kJ
Q Q W= +
= × + × =


Insight:
Increasing the coefficient of performance decreases work required to extract the
same heat from the inside of the refrigerator. For example, if the COP were increased to
2.75, only 12.5 kJ of work would be required to extract the 34.5 kJ.

Chapter 18
: The Laws of Thermodynamics James S. Walker,
Physics
, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 21

58.
Picture the Problem
: A heat pump injects 3240 J of heat into a warm room by pulling heat
from the cold outside. This process requires 345 J of work.


Strategy:
Use equation 18-10 to calculate the heat removed from the outside air. Then use
the Carnot relation,
h c h c
=,Q Q T T
to determine the outside temperature.

Solution:

1. (a)
Solve
equation 18-10 for
c
Q
:
c h
3240 J 345 J 2.90 kJ
Q Q W
=

= − =


2.
Solve the Carnot relation
for cold temperature:
[ ]
c
c h
h
2895 J
273.15 21.0 K 262.8 K
3240 J
Q
T T
Q
⎛ ⎞
= = + =
⎜ ⎟
⎝ ⎠


3.
Convert to Celsius:
c
262.8 K 273.15 K 10 C
T
=
− = − °


Insight:
Decreasing the outside temperature decreases the coefficient of performance, thereby increasing the work
necessary to produce the same heat output. When the external temperature drops to −20
°
C, the same 3240 J of heat
output requires 455 J of work.


59.
Picture the Problem
: An air conditioner extracts heat from a room at a rate of 11 kW.


Strategy:
Modify equation 18-10 and the Carnot relation to be rate equations by dividing
each energy term by time. Then use the resulting equations to determine the mechanical
power.

Solution:

1.
Solve equation 18-10 for
h
Q t
:
h c
Q t W t Q t
=
+


2.
Write the Carnot relation as a rate equation:
h h
c c
=
Q t T
Q t T


3.
Substitute the expression for
h
Q t
from step 1:
c
h
c c
=
W t Q t
T
Q t T
+


4
Solve for mechanical power:
( )
h
c
c
1
T
P W t Q t
T
⎛ ⎞
=
= −
⎜ ⎟
⎝ ⎠


5.
Insert numerical values:
273.15 32°C
11 kW 1 0.41 kW
273.15 21°C
P
⎛ ⎞
+
= − =
⎜ ⎟
+
⎝ ⎠


Insight:
The coefficient of performance for the air conditioner can be found by dividing the rate of heat flow by the
mechanical power giving
26.8.COP =



60.
Picture the Problem
: A reversible refrigerator can operate as a refrigerator or as a heat engine.

Strategy:
Combine equations 18-10, 18-11, and 18-15 to solve for efficiency in terms of the coefficient of
performance.

Solution:

1.
Combine equations 18-10 and
18-11, then divide top and bottom by
W
:
h c c
1
1
W W
e
Q W Q Q W
= = =
+ +


2.
Replace
c
Q W
using equation 18-15:
1 1
0.0909
1 COP 1 10.0
e = = =
+ +


Insight:
Writing the efficiency as
1
1 COP
e =
+
shows that the larger the coefficient of performance, the smaller the
efficiency. It also shows that while the COP can range from zero to infinity, the efficiency ranges from zero to one.

Chapter 18
: The Laws of Thermodynamics James S. Walker,
Physics
, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 22

61.
Picture the Problem
: A freezer (refrigerator) extracts heat from water to freeze liquid water into ice cubes. Electrical
energy must be supplied to the freezer to extract the heat.

Strategy:
Calculate the heat that must be extracted from the water to cool the water to freezing, freeze the water into
ice, and cool the ice to −5
°
C from the specific heats and latent heat of fusion. Then use equation 18-15 to calculate the
required electrical work.

Solution:

1
. Calculate the heat
necessary to freeze the water:
(
)
(
)
(
) ( )
c w f ice w f ice
1 2 1 2
Q mc T mL mc T m c T L c T


= Δ + + Δ = Δ + + Δ




2.
Now substitute the expression
from step 1 into equation 18-15:

( ) ( )
( ) ( )
( ) ( )
c
w f ice
1 2
4
5
COP COP
4186 J/kg K 15 C 33.5 10 J/kg
1.5 kg
1.5 10 J
4.0
2090 J/kg K 5.0 C
Q
m
W c T L c T
⎡ ⎤
= = Δ + + Δ
⎣ ⎦
⎧ ⎫
⋅ ° + ×
⎡ ⎤
⎪⎣ ⎦ ⎪
= = ×
⎨ ⎬
+ ⋅ °
⎡ ⎤
⎪ ⎪
⎣ ⎦
⎩ ⎭


Insight:
The heat extracted from the water is equal to the electrical power times the COP or 600 kJ.

62.
Picture the Problem
: A reversible engine can operate in reverse as a heat pump.

Strategy:
Combine equations 18-11 and 18-17 to solve for efficiency in terms of the coefficient of performance.

Solution:
Combine equations 18-11 and 18-17:
1 1
COP 4.3
0.23
h
Q
W e
= = = =


Insight:
The coefficient of performance of the heat pump is the inverse of the reversible engine’s efficiency. Because
the efficiency ranges from zero to one, the coefficient of performance can range from one to infinity.

63.
Picture the Problem
: You rub your hands together and friction converts the mechanical energy into thermal energy.

Strategy:
Use the definition of entropy change to answer the conceptual question.

Solution:

1. (a)
The definition of entropy change
S Q T
Δ
=
(equation 18-18) indicates that whenever heat is transferred
there is a corresponding entropy change. In this case the mechanical energy of the hands is converted into thermal
energy
Q
that is deposited into the hands, increasing their entropy. No heat is subtracted from anywhere, so we
conclude that the entropy of the universe will increase if you rub your hands together.

2.

(b)
The best explanation is
III
. The heat produced by rubbing raises the temperature of your hands and the air, which
increases the entropy. Statements I and II are each false.

Insight:
All such frictional processes produce an increase in the entropy of the universe, because low-entropy
mechanical energy is converted into high-entropy thermal energy.

64.
Picture the Problem
: An ideal gas is expanded slowly while the temperature is held constant.

Strategy:
Use the definition of entropy change to answer the conceptual question.

Solution:

1. (a)
The definition of entropy change
S Q T
Δ
=
(equation 18-18) indicates that whenever heat is transferred
there is a corresponding entropy change. In this case heat is added to the gas so that it might maintain a constant
internal energy while it expands and does work on the external world. We conclude that the entropy of the gas will
increase as it is expanded slowly and isothermally.

2.

(b)
The best explanation is
I
. Heat must be added to the gas to maintain a constant temperature, and this increases the
entropy of the gas. Statement II mistakenly supposes entropy is a function of temperature only, and statement III is
partially true (the temperature will decrease) but ignores the increase in disorder and entropy that results from the
expansion.

Insight:
In this case the entropy change of the
universe
is zero because the process is reversible. Any entropy gained by
the gas is lost by the reservoir when the heat is extracted.

65.
Picture the Problem
: A gas is expanded reversibly and adiabatically.

Strategy:
Use the definition of entropy change to answer the conceptual question.

Solution:

1. (a)
The definition of entropy change
S Q T
Δ
=
(equation 18-18) indicates that whenever heat is transferred
there is a corresponding entropy change. In this case no heat is exchanged because the expansion process is adiabatic.
We conclude that the entropy of the gas will stay the same as it is expanded reversibly and adiabatically.
Chapter 18
: The Laws of Thermodynamics James S. Walker,
Physics
, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 23

2.

(b)
The best explanation is
I
. The process is reversible, and no heat is added to the gas. Therefore, the entropy of the
gas remains the same. Statement II is partly true (when randomly moving molecules occupy a larger volume they
become more disordered) but the work required for an adiabatic expansion comes at the expense of internal energy, and
the temperature of the gas will decrease, making the system more ordered. Statement III is partially true (the
temperature will decrease) but entropy is not a function of temperature only.

Insight:
One type of
irreversible
adiabatic expansion is the free expansion of a gas into a vacuum. In this case no work
is done on the external world (because the vacuum offers no resistance to the expansion) so the internal energy and the
temperature of the gas remain the same. The gas now occupies a larger volume, however, and has become more
disordered, so the entropy increases during this process. Entropy always increases during irreversible processes.

66.
Picture the Problem
: Heat is added to water at its boiling point, converting the water entirely to steam.

Strategy:
Use the latent heat of vaporization of water to calculate the heat added to the water. Then calculate the
change in entropy from equation 18-18.

Solution:
Combine equa-
tions 17-20 and 18-18:
(
)
5
4
v
1.85 kg 22.6 10 J/kg
1.12 10 J/K 11.2 kJ/K
100 273.15 K
mL
Q
S
T T
×
Δ = = = = × =
+


Insight:
As heat is added to the water, the entropy of the water increases.

67.
Picture the Problem
: Heat is removed from water at its freezing point, converting the water to ice.

Strategy:
Use the latent heat of fusion of water to calculate the heat extracted from the water. Then calculate the
change in entropy from equation 18-18.

Solution:
Combine equations 17-20 and 18-18:
(
)
4
f
3.1 kg 33.5 10 J/kg
3.8 kJ K
273.15 0 K
mL
Q
S
T T
− ×

Δ = = = = −
+


Insight:
As heat is taken from the water, the entropy of the water decreases.

68.
Picture the Problem
: You heat a pan of water on the stove.

Strategy:
Use
Q mc T= Δ
(equation 16-13) to determine the amount of heat added to the water for each temperature
change, and then use
av
S Q TΔ =
(equation 18-18) to determine the entropy change, where
T
av
is the average
temperature of the water during the heating process.

Solution:
Combining equations 16-13 and 18-18 we see that the entropy change,
av
,S mc T T
Δ
= Δ
increases linearly
with the temperature change
ΔT
and is inversely proportional to the average temperature,
T
av
. Noting that processes C
and D involve only a 5°C temperature change, we arrive at the following ranking: D < C < B < A.

Insight:
A 10°C temperature increase near 30°C temperature (process A) produces a larger entropy increase than a
10°C increase near 40°C (process B) because the average temperature
T
av
is smaller for process A. Using
T
av
is only an
approximation; the exact solution requires calculus and reveals that
(
)
f i
ln.S mc T TΔ =
A comparison of the two
approaches reveals that they differ by less than 0.01% over the temperature intervals given in this question.

69.
Picture the Problem
: As heat flows out of a house, the entropy of the house decreases and the entropy of the outside
increases.

Strategy:
Use equation 18-18 to sum the entropy changes for inside and outside the house. The resulting equation can
be divided by time in order to determine the rate of entropy change.

Solution:
Sum the inside and outside changes
in entropy, and divide each term by the time:
( )
inside outside
20.0 kW 20.0 kW
9.6W K
273.15 22 K 273.15 14.5 K
S Q t Q t
t T T
Δ
⎛ ⎞ ⎛ ⎞
= +
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠

= + =
+ + −


Insight:
Although the change in entropy of the inside was negative, the net change in entropy is positive. This will be
true for all physical situations.

Chapter 18
: The Laws of Thermodynamics James S. Walker,
Physics
, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 24

70.
Picture the Problem
: While descending at constant speed, the parachutist’s gravitational potential energy is converted
to heat, resulting in an increase in entropy.

Strategy:
Use equation 18-18 to calculate the change in entropy with the heat given by the change in gravitational
potential energy.

Solution:

1
. Set the heat in equation 18-18 equal
to the change in gravitational potential energy:


Q U mgh
S
T T T
Δ
Δ = = =


2.
Solve numerically for the change in entropy:
(
)
(
)
2
88 kg 9.81 m/s 380 m
1100 J/K 0.11 kJ/K
21 273.15 K
SΔ = = =
+


Insight
: In the absence of air resistance, the parachutist’s gravitational potential energy would have been converted to
kinetic energy, not heat, with no increase in entropy.




71.
Picture the Problem
: Imperfect insulation in a house allows heat
Q
c
to leak into the house from the outside. An air
conditioner extracts that same heat from the house and expels slightly more heat outside, where the temperature is
warmer. The entropy change of the universe takes is the sum of the changes in entropy due to the heat entering the
house and the air conditioner expelling the heat outside.

Strategy:
Use equation 18-18 to sum the entropy changes for inside and outside the house. The resulting equation can
be divided by time to determine the rate of entropy change. Use equation 18-10 to determine the rate that the air
conditioner expels heat to the external world.

Solution:

1. (a)
The entropy of the universe will increase. The air conditioner has the efficiency of a Carnot engine, so
it does not increase the entropy of the universe, but the heat leaking into the house does produce a net entropy increase.

2. (b)
Sum the inside and outside changes
in entropy due to the air conditioner and the
leaking insulation:
in out in out
AC Leak
c h c c
c h c h
Q Q Q Q
S
T T T T

⎤ ⎡ ⎤
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
− −
Δ = + + +

⎥ ⎢ ⎥
⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎥ ⎢ ⎥
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎦ ⎣ ⎦


3.


Eliminate the indoor changes in entropy
because they are equal and opposite:

c h c
h
h h h
Q Q Q
Q
S
T T T
⎛ ⎞ ⎛ ⎞
− −
Δ = + =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠


4.
Use equation 18-10 to eliminate
h
,Q

and divide by time to obtain the rate
of entropy increase:

h c
h h h
0.41 kW
1.3 W/K
32 273.15 K
Q Q
W S W t
S
T T t T

Δ
Δ = = ⇒ = = =
+


Insight:
The heat exchange through the air conditioner does not change the entropy of the universe, but it does increase
due to the heat flow of 11 kW into the house through the windows and doors (see problem 69).


72.
Picture the Problem
: A heat engine extracts 6400 J from a hot reservoir and performs
2200 J of work. The remainder of the heat is exhausted to the cold reservoir.


Strategy:
Calculate the net change in entropy by summing the changes in entropy at the hot
and cold reservoirs. Use equation 18-18 to write the entropy change in terms of the heat
transfers and temperature. Finally, use equation 18-10 to find the heat exchange at the cold
reservoir.

Solution:

1.
Sum the entropy changes:

h c
Q Q
S
T T
⎛ ⎞ ⎛ ⎞
Δ = +
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠


2.
Write in terms from the hot and cold reservoirs:

c
h
h c
Q
Q
S
T T
⎛ ⎞
⎛ ⎞

Δ = +
⎜ ⎟
⎜ ⎟
⎝ ⎠
⎝ ⎠


3.
Use equation 18-10 to eliminate
c
:Q

h h
h c
Q Q W
S
T T
⎛ ⎞
⎛ ⎞
− −
Δ = +
⎜ ⎟
⎜ ⎟
⎝ ⎠
⎝ ⎠

Chapter 18
: The Laws of Thermodynamics James S. Walker,
Physics
, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 25

4.
Substitute numerical values:
6400 J 6400 J 2200 J
2.6 J K
610 K 320 K
S
− −
Δ = + =


Insight:
This engine operates at less than maximum efficiency. The Carnot efficiency of this engine is 0.475 and the
actual efficiency is 0.344. If the engine were running at maximum efficiency the net change in entropy would be zero.

73.
Picture the Problem
: An ideal gas is held in an insulated container at the temperature
T
. All the gas is initially in one-
half of the container, with a partition separating the gas from the other half of the container, which is a vacuum. The
partition then ruptures, and the gas expands to fill the entire container.

Strategy:
Use the first law of thermodynamics to determine the change in the internal energy of the gas. For an ideal
gas the internal energy depends only upon the temperature, so any change in the internal energy will produce a
corresponding change in temperature.

Solution:
The change in the internal energy of the ideal gas is given by
U Q W
Δ
= −
(equation 18-3). The container is
insulated so no heat is exchanged with the external world and
Q
= 0. Furthermore, no work is done on the external
world (because the vacuum offers no resistance to the expansion) so
W
= 0. We conclude that the internal energy does
not change and neither does the temperature. The final temperature is therefore equal to
T
.

Insight:
This process can be considered an irreversible adiabatic expansion of the gas into a vacuum. Although the
internal energy and the temperature of the gas remain the same, the gas now occupies a larger volume and has become
more disordered, so the entropy increases during this process. Entropy always increases during irreversible processes.

74.
Picture the Problem
: A system is taken through the three-process cycle shown in
the
PV
plot shown at right.

Strategy:
Whenever a system expands (volume is increasing) at nonzero pressure it
does positive work on the external world. Likewise, if the system is compressed,
work is done on the system and the work has a negative value. Use these principles
to determine the correct sign of the work for each of the indicated processes.

Solution:

1. (a)
In the process
A B→
the volume decreases, work is done on the
system, and the work is negative.

2. (b)
In the process
B C→
the volume increases, work is done by the system, and the work is positive.

3. (c)
In the process
C A→
the volume decreases, work is done on the system, and the work is negative.

Insight:
The two negative works occur at higher pressure than the positive work, so the net work done by the entire
cycle is negative. Another way to remember this is to note that for complete cycles depicted on a
PV
plot,
counterclockwise cycles correspond to negative work and clockwise cycles correspond to positive work.

75.
Picture the Problem
: An ideal gas has the pressure and volume indicated by point I
in the
PV
plot at right. At this point its temperature is
1
T
.
The temperature of the gas
can be increased to
2
T
by using the constant-volume process,
I II→
,
or the
constant-pressure process,
I III.→


Strategy:
The change in entropy for each process equals
av
,S Q T
Δ
=
where
Q
is
the heat transferred and
T
av
is the average temperature.
T
av
is the same for the
constant-volume process as it is for the constant pressure process, so the process
with the largest heat transfer
Q
results in the largest increase in entropy.

Solution:
In the process
I II→
the volume remains constant so no work is done.
In the process
I III→
the system does positive work on the external world, so the
heat flow must both increase the internal energy (because of the increase in
temperature) and provide the energy to do the work. We conclude that the heat flow
and the entropy change for the process
I II→
is less than the entropy change for
the process
I III.→


Insight:
More heat flow is always required for such a constant pressure process than is required for a constant volume
process because of the work done on the external world during the expansion.

Chapter 18
: The Laws of Thermodynamics James S. Walker,
Physics
, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 26

76.
Picture the Problem
: Note the three processes on the
P–V
plot.


Strategy:
Equate the work done in the process with the area under
the
PV
plot.

Solution:

1. (a)
Find area of
the rectangle under process A:

3
(4.0 kPa)(1.0 m )
4.0 kJ
W =
=


2. (b)
Sum the areas of
the rectangle and triangle
under process B:

(
)
( )
( )
3
3
1
2
1.0 kPa 3.0 m
2.0 kPa 2.0 m
5.0 kJ
W
W
=
+
=

3.

(c)
Find the area of the rectangle
under process C:
(
)
3
1.0 kPa 3.0 m 3.0 kJW = =


Insight:
In each of the processes shown on the
P–V
plot, the vertical (constant volume) segments did not contribute to
the work done.


77.
Picture the Problem
: Heat is added to an ice cube at 0°C, increasing its entropy.

Strategy:
Use equation 18-18 to calculate the amount of heat absorbed by the ice. Then use equation 17-20 and the
latent heat of fusion for water to determine the amount of ice that melts.

Solution:
Solve equation 17-20 for the mass
and then write
Q T S= Δ
(equation 18-18):

( )
4
f f
273.15 K 87 J/K
0.071 kg
33.5 10 J/kg
Q T S
m
L L
Δ
= = = =
×


Insight:
About one-half of the 0.14-kg ice cube has melted. Melting the entire ice cube would result in an increase in
entropy of 172 J/K.


78.
Picture the Problem
: A heat engine does work while extracting heat from a hot reservoir and exhausting heat to a cold
reservoir.

Strategy:
Use equation 18-11 to determine the efficiency.

Solution:
Apply equation 18-11 directly:

h
1
0.250
4 4
W W
e
Q W
= = = =


Insight:
Efficiency is a ratio of the work done to the input heat.


79.
Picture the Problem
: A monatomic ideal gas expands at constant
temperature.


Strategy:
Use the Ideal Gas Law and the initial state to determine the
temperature. Since this is an isothermal process, you can calculate the
work during the process from equation 18-5. Use the first law of
thermodynamics (equation 18-3) to relate the work to heat absorbed during
the process, noting that
0UΔ =
for an isothermal process of an ideal gas.
Then use equation 18-18 to calculate the change in entropy.

Solution: 1.
(a) Solve the Ideal
Gas Law for temperature:

PV
T
nR
=


2.
Insert given data from the initial state:
(
)
( )
3
400 kPa 1.00 m
365 K
132 mol 8.31 J/mol K
T = =

⎡ ⎤
⎣ ⎦


3.

(b)
Heat is added to the gas, while the temperature remains constant. Therefore the entropy must increase.
Chapter 18
: The Laws of Thermodynamics James S. Walker,
Physics
, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 27

4. (c)
Combine equations 18-3 and 18-18:

0Q U W W
S
T T T
Δ
+ +
Δ = = =


5.
Insert equation 18-5 for the work:

(
)
( )
f i
f
i
3
3
ln
ln
4.00 m
132 mol 8.31 J/mol K ln 1.52 kJ/K
1.00 m
nRT V V
V
S nR
T V
⎛ ⎞
Δ = =
⎜ ⎟
⎝ ⎠
⎛ ⎞
= ⋅ =⎡ ⎤
⎜ ⎟
⎣ ⎦
⎝ ⎠


6.

(d)
Find
W
with equation 18-5:
(
)
( ) ( )
f i
3
3
ln
4.00 m
132 mol 8.31 J/mol K 365 K ln 555 kJ
1.00 m
W nRT V V=
⎛ ⎞
= ⋅ =
⎡ ⎤
⎜ ⎟
⎣ ⎦
⎝ ⎠


7.
Now compare with
T SΔ
:

( )
365 K 1.52 kJ/K 555 kJT SΔ = =


Insight:
The area under a
PV
plot is equivalent to the work done in the process represented by the graph. Similar to a
P
V
plot, a
T–S
diagram plots the temperature as a function of the entropy of the system. The area under a
T–S
plot is
equal to the heat transfer during the process. In this problem, the work and heat (
T S
Δ
) are equal because the change in
internal energy is zero.




80.
Picture the Problem
: A monatomic ideal gas undergoes four processes
represented by the accompanying
P–V
plot.


Strategy:


A gas does work when it expands. Examine the
P–V
diagram
for each process to determine whether the gas is expanding (positive
work), contracting (negative work), or at constant volume (no work).
Use the first law of thermodynamics to calculate the work done in
process AB. Use equation 17-15 and the Ideal Gas Law to relate the
change in internal energy to the work.

Solution:

1. (a)
The volume is expanding so the work is positive.

2.

(b)
The volume is constant so the work is zero.

3. (c)
The volume is contracting so the work is negative.

4.

(d)
The volume is contracting so the work is negative.

5.
Solve the first law for work:

W Q U
=
−Δ


6.
Use equation 17-15 to eliminate

and
use the Ideal Gas Law to set
:n R T P VΔ = Δ

3 3
2 2
W Q n R T Q P V
=
− Δ = − Δ


7.
Write set pressure times change in volume
equal to the work and gather like terms:

3 5
2 2
W Q W W Q
=
− ⇒ =


8.
Solve for the work:

( )
2 2
5 5
27 J 11 JW Q= = =


Insight:
For a monatomic ideal gas undergoing a constant pressure process, the work, heat, and change in internal
energy are all proportional to each other. Therefore given one of the three quantities it is possible to solve for the other
two. In this problem the change in internal energy is 16 J.

Chapter 18
: The Laws of Thermodynamics James S. Walker,
Physics
, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 28

81.
Picture the Problem
: A monatomic ideal gas is expanded such that the
change in volume is proportional to the change in pressure.

Strategy:
Calculate the work by measuring the area under the
PV
plot.
Use the Ideal Gas Law to determine the initial and final temperatures of
the gas. Then use equation 17-15 to calculate the change in internal
energy. Determine the heat from the first law of thermodynamics.

Solution:

1. (a)
Divide the
area under the curve into a
rectangle and triangle:
rectangle triangle
1
2
W A A
lw bh
=
+
= +


2.
Insert the numeric values:

( )
( )
( )
3 3
1
2.00 m 106 kPa + 2.00 m 106 kPa 318 kJ
2
W = =


3.

(b)
Use the Ideal Gas Law to find
the initial and final temperatures:
f f i i
f i
,
PV PV
T T
nR nR
= =


4.
Insert these temperatures into
equation 17-15 and simplify:
( )
( )
f f i i
f i f f i i
3 3 3
2 2 2
PV PV
U nR T T nR PV PV
nR nR
⎛ ⎞
Δ = − = − = −
⎜ ⎟
⎝ ⎠


5.
Substitute numerical values:

(
) ( )
3 3 3 3
3
2
212 10 Pa 3.00 m 106 10 Pa 1.00 m 795 kJU
⎡ ⎤
Δ = × − × =
⎣ ⎦


6.

(c)
Solve the first law for heat:
795 kJ 318 kJ 1113 kJQ U W= Δ + = + =


Insight:
Compare the result of this problem with the results of problems 40 and 41, in which different processes were
used to move between the same initial and final states. In all three problems the change in internal energy is the same.
Internal energy is a function of the initial and final states. The change in internal energy is independent of the path
between the states. On the other hand, the heat absorbed and the work done is dependent upon the process.

82.
Picture the Problem
: Two engines absorb the same amount of heat from a hot reservoir. Engine B exhausts twice as
much heat to the cold reservoir as does Engine A.

Strategy:
Use equation 18-12 to determine a relationship between exhausted heat and the efficiency for each engine.
Set the exhausted heat of Engine B equal to twice the exhausted heat of Engine A and combine the equations to solve
for the efficiency of Engine B.


Solution:

1. (a)
Engine B does less work than engine A because it exhausts more heat to the low temperature reservoir.
Therefore, Engine A has the greater efficiency.

2. (b)
Write equation 18-12 for Engine A
and solve for the heat ratio:

A,c A,c
A A
h h
1 1
Q Q
e e
Q Q
=
− ⇒ = −

3.
Write equation 18-12 for Engine B
and
s
et
B,c A,c
2:Q Q=

B,c A,c
B
h h
2
1 1
Q Q
e
Q Q
= − = −


4.
Write the heat ratio in terms
of the efficiency of Engine A:
(
)
(
)
B A A
1 2 1 2 1 2 0.66 1 0.32e e e= − − = − = − =


Insight:
As expected, Engine B has a lower efficiency because it exhausts more heat to the cold reservoir.

83.
Picture the Problem
: A freezer (refrigerator) extracts heat from water, thereby cooling the water, freezing the water,
and cooling the ice. During this process the freezer uses electrical work and exhausts heat to the surrounding room.

Strategy:
Determine the heat extracted from the water using specific heats and latent heat of fusion. Then use equation
18-15 to calculate the work and equation 18-10 to find the heat exhausted to the room.

Solution:

1. (a)
Sum the heat needed to cool
the water, freeze it to ice, and cool the ice:
(
)
(
)
w f ice
1 2
Q mc T mL mc T= Δ + + Δ


2.
Insert the numeric values to solve for the heat:

( ) ( )
( ) ( )
4
4186 J/kg K 20.0 C
1.75 kg 33.5 10 J/kg 751 kJ
2090 J/kg K 5.00 C
Q
⎧ ⎫
⎡ ⎤
⋅ °
⎡ ⎤
⎣ ⎦
⎪ ⎪
⎢ ⎥
⎪ ⎪
= + × =
⎢ ⎥
⎨ ⎬
⎢ ⎥
⎪ ⎪
+ ⋅ °
⎡ ⎤
⎢ ⎥
⎣ ⎦
⎪ ⎪
⎣ ⎦
⎩ ⎭

Chapter 18
: The Laws of Thermodynamics James S. Walker, Physics, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 29

3. (b)
Solve equation 18-15 for work:

c
751 kJ
194 kJ
COP 3.88
Q
W = = =

4. (c)
Solve equation 18-10 for heat
exhausted to the room:
h c
194 kJ 751 kJ 945 kJQ W Q= + = + =


Insight:
The amount of heat that must be extracted to freeze the ice is independent of the coefficient of performance of
the freezer. However, the greater the coefficient of performance, the less work is needed to freeze the ice and therefore
the less heat is exhausted into the room.


84.
Picture the Problem
: As heat is added to a monatomic ideal gas (argon) it expands at constant pressure.

Strategy:
Use the molar specific heat at constant pressure (equation 18-7) and the internal energy of a monatomic ideal
gas (equation 17-15) to write the change in internal energy as a function of the heat and to calculate the temperature
difference. Use the first law of thermodynamics to solve for the work in terms of the change in internal energy and
heat. Then calculate the change in volume from the work at constant pressure.

Solution:

1. (a)
Solve the molar heat
at constant pressure for
nR T
Δ
:
5
2
p p
2 5
Q n R T n R T Q= Δ ⇒ Δ =


2.
Eliminate
nR TΔ
from equation 17-15
for the change in internal energy:
(
)
3 3 3
2
p p
2 2 5 5
U n R T Q QΔ = Δ = =


3.
Solve numerically:

(
)
3
5
1800 J 1080 J 0.11 kJUΔ = = =


4. (b)
Solve the equation for molar heat at
constant pressure for the temperature change:
( )
( )
( )
p
2
2 1800 J
24 K
5 5 3.6 mol 8.31 J mol K
Q
T
n R
Δ = = =



5. (c)
Solve the first law for work,
set
W P V= Δ
and solve for

:

3 3
3
1800 J 1080 J
6.0 10 m
120 10 Pa
W Q U P V
Q U
V
P

=
−Δ = Δ
−Δ −
Δ = = = ×
×

Insight:
When a monatomic ideal gas expands at constant pressure the heat, work, and internal energy are all
proportional to each other. If the amount of heat were to be doubled to 3600 J, each of the other terms in this problem
would also be doubled:
2160 J,UΔ = 48 K,TΔ =
and
3 3
12 10 m.V

Δ = ×



85.
Picture the Problem
: The hot Sun transfers heat to cold space. This process decreases the entropy of the Sun and
increases the entropy of space.

Strategy:
Multiply the radiation rate by one day to determine the heat transferred per day. Then sum the changes in
entropy (equation 18-18) for the Sun and space. Use equation 18-14 to find the work that could have been done.

Solution:

1. (a)
Multiply the
radiation rate by one day:

(
)
(
)
(
)
26 31
3600 s
24 h
h day
3.80 10 J/s 3.28 10 JQ Pt= = × = ×


2.
Sum the changes in entropy:

Sun space
31 31
31
3.28 10 J 3.28 10 J
1.1 10 J/K in one day
5500 273.15 K 3.0 K
Q Q
S
T T
⎛ ⎞ ⎛ ⎞
Δ = +
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
− × ×
= + = ×
+


3.

(b)
Write equation 18-14:
( )
( )
c
max h h
h
31 31
1
3.0 K
1 3.28 10 J 3.28 10 J
5500 273 K
T
W e Q Q
T
⎛ ⎞
= = −
⎜ ⎟
⎝ ⎠
⎡ ⎤
= − × = ×
⎢ ⎥
+
⎢ ⎥
⎣ ⎦


Insight:
A small fraction of that available work is captured by the Earth each day and used to grow plants, warm the
Earth, etc., and in some way is the source of almost all of the energy used on Earth.

Chapter 18
: The Laws of Thermodynamics James S. Walker, Physics, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 30

86.
Picture the Problem
: The table lists the heat, work, and change in internal energy of a monatomic ideal gas for each o
f

the processes covered in the chapter.

Strategy:
Use the first law of thermodynamics to relate the heat, work, and change in internal energy. Use equation 18-
5 for the work in an isothermal process.

Solution:

1. (a)
Solve the first law for the work:
5 3
2 2
W Q U nR T nR T nR T
=
−Δ = Δ − Δ = Δ


2.

(b)
Adiabatic means no heat transfer
:

Q = 0

3. (c)
Solve the first law for
:UΔ

(
)
3 3
2 2
0U Q W n R T n R T
Δ
= − = − − Δ = Δ


4.

(d)
Work requires a change in volume:
W = 0

5. (e)
Solve the first law for
:UΔ

3 3
2 2
0U Q W n R T n R T
Δ
= − = Δ − = Δ


6. (f)
Solve the first law for heat
using equation 18-5 for work:

( ) ( )
f i f i
0 ln/ln/Q U W n RT V V nRT V V= Δ + = + =


7. (g)
Change in internal energy
requires a change in temperature:

0UΔ =


Insight:
This table provides a quick reference for each of the processes covered in this chapter. It would be a good idea
to memorize the table, or keep a copy in your notes.




87.
Picture the Problem
: A monatomic ideal gas is compressed isothermally.

Strategy:
Calculate the final volume of the gas from the Ideal Gas Law. Then use
equation 18-5 to calculate the work done. Finally determine the heat added from the
first law of thermodynamics.

Solution:

1. (a)
Solve the Ideal
Gas Law for the final volume:
( )( )
f
f
3
3 2 3
f
2.75 mol 8.31 J mol K 295 K
121 10 Pa
0.0557 m 5.57 10 m
nRT
V
P
V

=

=
×
= = ×


2. (b)
Use the Ideal Gas Law to relate
initial and final pressures and volumes
and solve for the volume ratio:

f i
i i f f
i f

V P
PV PV
V P
= ⇒ =

3.
Write equation 18-5 and replace the
volume ratio with the pressure ratio:

f i
i f
ln ln
V P
W n RT n RT
V P
⎛ ⎞ ⎛ ⎞
= =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠

4.


Solve numerically:
( )( )
101 kPa
2.75 mol 8.31 J mol K 295 K ln 1.22 kJ
121 kPa
W
⎛ ⎞
= ⋅ = −
⎜ ⎟
⎝ ⎠


5. (c)
Use the first law to determine Q:
(
)
0 1.22 kJ 1.22 kJQ U W= Δ + = + − = −


Insight:
Compressing a monatomic ideal gas isothermally requires work to be done on the gas and heat to be exhausted
from the gas. This process converts mechanical work into heat.

Chapter 18
: The Laws of Thermodynamics James S. Walker, Physics, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 31

88.
Picture the Problem
: A heat engine operates between reservoirs at 1010 K and 302 K. The
inventor claims that the engine can do 1120 J by extracting 1250 J from the hot reservoir.


Strate
gy
:
Use equation 18-11 to calculate the claimed efficiency. Then use equation 18-13
t
o calculate the maximum possible efficiency.

Solution:

1. (a)
Solve equation
18-11 for the efficiency:
h
1120 J
0.896
1250 J
W
e
Q
= = =


2.

(b)
Solve equation 18-13 for the
maximum efficiency:
c
max
h
302 K
1 1 0.701
1010 K
T
e
T
= − = − =


Insight:
Do not invest in this machine. The claimed efficiency exceeds the maximum
possible efficiency.

89.
Picture the Problem
: A heat engine operates between reservoirs at 810 K and 320 K.
During its cycle it extracts 660 J of heat from the hot reservoir and performs 250 J of work.

Strategy:
Sum the entropy changes at the hot and cold reservoirs using equation 18-18 and
equation 18-10. Determine the maximum work possible from equation 18-14.
Solution:

1. (a)
Sum the entropy changes:
c
h
tot
h c
h c
Q
Q
Q Q
S
T T T T

⎛ ⎞ ⎛ ⎞
Δ = + = +
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠

2.
Use equation 18-10 to eliminate
c
Q

h h
tot
h c
Q Q W
S
T T


Δ = +

3.
Substitute numerical values:
tot
660 J 660 J 250 J
0.47 J K
810 K 320 K
S
− −
Δ = + =


4. (b)
Write equation 18-14 for the
maximum possible work:

( )
c
max h
h
320 K
1 1 660 J 0.40 kJ
810 K
T
W Q
T
⎛ ⎞
⎛ ⎞
= − = − =
⎜ ⎟
⎜ ⎟
⎝ ⎠
⎝ ⎠

5. (c)
Use equation 18-14 and eq.18-10
to write the difference between
the maximum and the actual work:

( )
c c
max actual h h c h c
h h
1
T T
W W W Q Q Q Q Q
T T
⎛ ⎞
Δ
= − = − − − = − +
⎜ ⎟
⎝ ⎠

6.
Write in terms of change in entropy:
c
h
c c tot
h c
Q
Q
W T T S
T T
⎛ ⎞

Δ
= + = Δ
⎜ ⎟
⎝ ⎠


Insight:
The net change in entropy is related to the difference between the work done by a real engine and the work
done by a Carnot engine operating between the same two temperatures. An engine operating at maximum efficiency
does not cause a net change in entropy.

90.
Picture the Problem
: The water in a dish cools and freezes solid.

Strategy
: Use equation 18-18 to calculate the change in entropy, where the heat leaving the water is the latent heat of
fusion. We concentrate only on the entropy change during the freezing process at 0°C.

Solution:

1. (a)
The entropy of the water decreases because heat flows out of the water. You can also visualize the
entropy change by realizing that the water goes from a disordered liquid state to an ordered crystalline state.

2. (b)
Write equation 18-18 in terms of
f
:L

(
)
( )
4
f
0.53 kg 33.5 10 J/kg

0.65 kJ K
0 273.15 K
mL
Q
S
T T
− ×

Δ = = = =
+


3. (c)
The entropy will increase in the object that absorbed the heat given off by the ice.

Insight:
If the surrounding air (at −5°C) absorbs the heat, its entropy will increase by
0.66 kJ K,S
Δ
=
producing a
positive net change in the entropy of the universe.

Chapter 18
: The Laws of Thermodynamics James S. Walker, Physics, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 32

91.
Picture the Problem
: An ideal gas undergoes a three-process cycle, as
shown in the PV plot.


Strategy:
Solve for the missing terms using the first law of thermodynamics
and the properties of a constant volume process (W = 0), a constant pressure
process (
W P V= Δ
), and a cycle (
0UΔ =
).


Solution:

1. (a)
For process
A B,→

solve the first law with W
= 0:
38 J + 0 38 J
Q U W
=
Δ +
= − = −


2. (b)
For process
A B,→
set the work
equal to the area under the
PV
plot:
(
)
0 0
W P V P
=
Δ = =


3.

(c
) For process B C,→ solve
the first law for heat:
(
)
82 J 89 J 171 J
Q U W
= Δ + = − + − = −


4. (d)
For process
C A,→
sum the
changes in internal energy for the cycle:

( )
C A A B B C
C A A B B C
0
U U U
U U U
→ → →
→ → →
Δ
+ Δ + Δ =
Δ = − Δ + Δ


5.
Solve first law for work:

(
)
C A
332 J 38 J 82 J 212 J
W Q U

= −Δ = + − + − =⎡ ⎤
⎣ ⎦


6. (e)
For process
C A,→
numerically
solve for change in internal energy:

(
)
( )
C A A B B C
= 38 J 82 J 120 J
U U U
→ → →
Δ = − Δ + Δ
− − + − =
⎡ ⎤
⎣ ⎦


Insight:
Because the change in internal energy for the entire cycle is zero, the first law requires that net heat equals the
net work for the cycle. The total work is
(
)
(
)
0 89 J 212 J 123 J
W
= + − + =
and the total heat transfer is
( ) ( )
38 J 171J 332 J 123 J.
Q
= − + − + =
These are equal as expected.




92.
Picture the Problem
: A Carnot engine operates between two temperature reservoirs, with an efficiency determined by
the ratio of the reservoir temperatures.

Strategy:
Because
c
h
1,
T
e
T
= −
we need to make
c
h
T
T
smaller to increase the efficiency. We can determine which is
smaller,
c
h
T
T T
+ Δ
or
c
h
T T
T
−Δ
, by checking the sign of their difference.

Solution: 1.
Subtract the two ratios, find
the common denominator and simplify:
( )( )
( )
( ) ( )
( )
2
c h c h h c
c c
h h h h h h
T T T T T T T T T T
T T T
T T T T T T T T T
− −Δ + Δ Δ − + Δ
−Δ
− = =
+ Δ + Δ + Δ


2.
Because
0,T
Δ >
and
h c
( ) 0,T T
− >
the difference is greater than zero. Therefore, lowering the temperature of the
low temperature reservoir would result in the greater change in efficiency.

Insight:
To show this numerically, we can let
h c
500 K, 300 K, and Δ 20 K.T T T
=
= =
The initial efficiency is 0.40.
Increasing the hot reservoir temperature creates an efficiency of 0.42, while decreasing the cold temperature creates an
efficiency of 0.44. Clearly, lowering the cold reservoir temperature has the greatest effect on the efficiency.

Chapter 18
: The Laws of Thermodynamics James S. Walker,
Physics
, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 33

93.
Picture the Problem
: A monatomic ideal gas completes a three-process
cycle as shown in the
PV
plot.

Strategy:
Calculate the work, heat, and change in internal energy for each
process using the first law of thermodynamics and the properties of a
constant temperature process (Δ
U
= 0, and equation 18-5), a constant
pressure process (
W
=
P
Δ
V
), a constant volume process (
W
= 0), and a
complete cycle (Δ
U
= 0).

Solution:

1 (a) For process

A B:


Write equation 18-5 for the work, using
the Ideal Gas Law to eliminate
A
nRT
:
B B
AB A A A
A A
ln ln
V V
W nRT P V
V V
⎛ ⎞ ⎛ ⎞
= =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠


2.
Substitute numerical values:

( )( )
3 3
3 3 3
AB
3 3
4.50 10 m
600 10 Pa 0.750 10 m ln 806 J
0.750 10 m
W



⎛ ⎞
×
= × × =
⎜ ⎟
×
⎝ ⎠


3.
Set change in internal energy
equal to zero (isothermal process):
AB
0U
Δ
=


4.
Solve the first law for heat:
AB AB AB
0 806 J 806 JQ U W
= Δ + = + =


5. For process
B C:

Use molar specific
heat at constant pressure to solve for heat:

BC
5 5
2 2
Q nR T P V
=
Δ = Δ


6
. Substitute numerical values:

(
)
(
)
3 3 3 3 3
5
BC
2
100 10 Pa 0.750 10 m 4.50 10 m 938 JQ
− −
= × × − × = −


7.
The area under
PV
plot is
W
:

(
)
(
)
3 3 3 3
BC
100 10 Pa 0.750 10 4.50 10 m 375 JW P V
− −
= Δ = × × − × = −


8.
Use the first law to solve for
:U
Δ

(
)
BC BC BC
938 J 375 J 563 JU Q W
Δ = − = − − − = −


9. For process
C A:


Set the work
equal to zero (constant volume):

CA
0W P V
=
Δ =


10.
Set the sum of changes in internal
energy equal to zero and solve for
CA
:U
Δ

(
)
CA AB BC
0 563 563 JU U U
Δ = −Δ −Δ = − − − =


11.
Use the first law to solve for heat:

CA CA CA
563 J 0 563 JQ U W
= Δ + = + =


12. (b)
Calculate
e
from equation 18-11:

h
806 J 375 J
0.31
806 J 563 J
W
e
Q

= = =
+


Insight:
In calculating the efficiency only the two positive heat terms are included in
h
.Q
. The heat exhausted in
process
B C

is
c
Q
.

94.
Picture the Problem
: A monatomic ideal gas under goes a constant pressure process. During the process heat enters
the system and the system does work.

Strategy:
Calculate the ratio of work over heat. Use the area under the
PV
plot for the work and the molar specific heat
to calculate the heat.

Solution:

1.
Set
W
equal to the area under the
PV
plot,
and use the Ideal Gas Law to write in terms of
:T
Δ

W P V n R T
=
Δ = Δ


2.
Use equation 18-7 find the heat:
5
2
Q nR T
=
Δ


3.
Calculate the ratio of work to heat:
5
2
p
2
5
W nR T
Q nR T
⎛ ⎞
Δ
= =
⎜ ⎟
Δ
⎝ ⎠


Insight:
Two-fifths of the heat absorbed by a monatomic ideal gas during a constant pressure process is converted to
work as the gas expands. The remaining 3/5 of the heat becomes internal energy.

Chapter 18
: The Laws of Thermodynamics James S. Walker,
Physics
, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 34

95.
Picture the Problem
: A Carnot engine cycle consists of four processes:
isothermal expansion, adiabatic expansion, isothermal compression, and
adiabatic compression.

Strategy
: Use equation 18-12 to write the efficiency in terms of heat
absorbed and exhausted during the cycle. Use equation 18-5 to calculate
the heat transfers during the isothermal processes, where the volume
ratios have been calculated from equation 18-9 and the Ideal Gas Law.

Solution:

1.
Use equation 18-9
to relate the initial and final states
of the adiabatic processes:
i i f f
PV PV
γ
γ
=

2.
Use the Ideal Gas Law to eliminate
the pressure and simplify:

1 1
i f
i f i i f f
i f

nRT nRT
V V TV T V
V V
γ
γ γ γ
− −
⎛ ⎞ ⎛ ⎞
= ⇒ =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠


3.
Write this explicitly for
process 2 and process 4:
1 1 1 1
h 2 c 3 h 1 c 4
and T V TV T V TV
γ
γ γ γ

− − −
= =


4.
Divide these relations to eliminate the
temperatures and take the
1
γ

-root:

1 1
1
1
c 3 3 3
h 2 2 2
1 1
1 4 1 4
h 1 c 4

TV V V
T V V V
V V V V
T V TV
γ γ
γ
γ
γ γ
− −


− −
⎛ ⎞ ⎛ ⎞
= ⇒ = ⇒ =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠


5.
Solve the first law for heat during the
isothermal processes, setting
0:UΔ =

Q U W W
=
Δ + =


6.
Use equation 18-5 to write the
heat for each isothermal process:

2
h 1 h
1
ln
V
Q W n RT
V
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠
and
3
c 3 c
4
ln
V
Q W n RT
V
⎛ ⎞
= =
⎜ ⎟
⎝ ⎠


7.
Use the volume ratios
from step 4 to rewrite
Q
c
:
2
c c
1
ln
V
Q n RT
V
⎛ ⎞
=
⎜ ⎟
⎝ ⎠


8.
Divide the two heat equations:
(
)
( )
h 2 1
h h
c c 2 1 c
ln
ln
nRT V V
Q T
Q nRT V V T
= =


9.
Insert the ratio into equation 18-12:
c c
h h
1 1
Q T
e
Q T
= − = −

Insight:
As expected, the ratio of the heat from the hot and cold reservoir is equal to the ratio of the temperatures of
those two reservoirs.

Chapter 18
: The Laws of Thermodynamics James S. Walker, Physics, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 35

96.
Picture the Problem
: A Carnot refrigerator and a Carnot engine operate between the same two temperature reservoirs.

Strategy:
Solve equation 18-13 to obtain the ratio of reservoir temperatures in terms of the efficiency. Then use
equation 18-15 and equation 18-10 to write the coefficient of performance in terms of the ratio of heats. Finally employ
the Carnot relation
c h c h
Q Q T T
=
to write the coefficient of performance in terms of the temperature rations and then
in terms of the efficiency.

Solution:

1.
Solve equation 18-13 for
the ratio of reservoir temperatures:
c c
h h
1 1
T T
e e
T T
=
− ⇒ = −


2.
Combine equations 18-15 and 18-10:
c c
h c
COP
Q Q
W Q Q
= =



3.
Divide top and bottom by
h
Q
and
substitute in the Carnot relation:
c h c h
c h c h
COP
1 1
Q Q T T
Q Q T T
= =
− −


4.
Write the temperature ratio as (1-e) and simplify:

( )
1 1
COP
1 1
e e
e e


= =
− −


Insight:
The efficiency can range from zero to one. This range of efficiencies gives a range to the coefficient of
performance of zero to infinity. The efficiency is a minimum when the COP is a maximum.




97.
Picture the Problem
: An OTEC (Ocean Thermal Energy Conversion) heat engine produces work by extracting heat
from the surface of the ocean and rejecting heat to the colder, deep-water of the ocean.

Strategy:
Use equation 18-13 to determine the maximum efficiency. Remember to convert the temperature from
Celsius to Kelvin.

Solution
Insert the given temperatures
(in kelvins) into equation 18-13:
(
)
( )
c
max
h
4.0 273.15 K
1 1 0.0610 6.10%
22 273.15
T
e
T
+
= − = − = =
+


Insight:
Six percent efficiency means that for every 100 Joules of heat extracted from the water only 6 Joules of work
can be produced.




98.
Picture the Problem
: During the operation of an OTEC heat engine, 1500 kg of water at 22°C is cooled to 4.0°C.

Strategy:
Use
Q mc T= Δ
(equation 16-13) to determine the heat released by the warm water.

Solution:
Apply equation 16-13:
( )( )( )
8
1500 kg 4186 J/kg/C 4.0 22°C 1.1 10 JQ mc T= Δ = ° − = − ×


Insight:
If this much energy were transferred every second, and 6.1% of the energy flow were converted into work, the
output of this heat engine would be 6.9 MW.

Chapter 18
: The Laws of Thermodynamics James S. Walker, Physics, 4
th
Edition

Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No
portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
18 – 36

99.
Picture the Problem
: An OTEC (Ocean Thermal Energy Conversion) heat engine produces work by extracting heat
from the surface of the ocean and rejecting heat to the colder, deep-water of the ocean.

Strategy:
Use equation 18-13 to determine the maximum efficiency. Remember to convert the temperature from
Celsius to Kelvin.

Solution
Insert the given temperatures
(in kelvins) into equation 18-13:
(
)
( )
c
max
h
2.0 273.15 K
1 1 0.0678 6.78%
22 273.15
T
e
T
+
= − = − = =
+


Insight:
If the rate of cooling were the same as in problem 98 (1500 kg of water changing from 22°C to 2.0°C every
second) the power output would then be 8.5 MW. A detailed analysis would be required to decide if the extra 1.6 MW
of output would adequately compensate for the cost required to get down to the 2.0°C water.




100.
Picture the Problem
: A heat engine operates between a hot and cold reservoir. The hot
reservoir is the same as in Active Example 18-3, but the cold reservoir is 10 degrees cooler.


Strategy:
Solve for the efficiency using equation 18-13. Calculate the work from the
efficiency and input heat using equation 18-11.

Solution:

1.
The efficiency increases when the temperature difference between the two
reservoirs increases. Lowering the temperature of the cold reservoir increases the efficiency,
so the new efficiency is greater than 0.47.


2.

(b)
Use equation 18-13
to solve for the efficiency:
295
1 1 0.488
576
C
H
T
K
e
T K
= − = − =


3. (c)
Solve equation 18-11 for W:
(
)
(
)
h
0.488 1050 J 512 JW eQ= = =


Insight:
By increasing the efficiency, it is possible to extract more work from the same available heat.




101.
Picture the Problem
: A heat engine operates as in Active Example 18-3. However, the
temperatures of the hot and cold reservoirs have both increased by 16 K.


Strategy:
Use equation 18-13 to solve for the efficiency and equation 18-18 to solve for
the entropy change. To calculate the heat exhausted to the cold reservoir, combine equations
18-10 and 18-11.

Solution:

1. (a)
The efficiency increases when the ratio of the temperatures of the two
reservoirs increases. Raising both temperatures does not affect the temperature difference,
but it does decrease their ratio. Increasing both temperatures decreases the efficiency, so the
new efficiency is less than 0.47.


2.

(b)
Use equation 18-13 to solve for e:
321
1 1 0.46
592
C
H
T
K
e
T K
= − = − =


3. (c)
Use equation 18-18 to solve for
:SΔ

h
h
h
1050 J
= 1.8 J/K
592 K
Q
S
T

Δ = = −


4.

(d)
Combine equations 18-10
and 18-11 to solve for Q
c
:
(
)
c h h h h
1Q Q W Q eQ Q e
=
− = − = −


5.
Use equation 18-18 to solve
for entropy change:

(
)
(
)( )
h
c
c
c c
1 1050 J 1 0.458
1.8 J/K
321 K
Q e
Q
S
T T
− −
Δ = = = =


Insight:
Because this is a Carnot engine, the net change in entropy is zero, that is,
h c
0S S
Δ
+ Δ =
.