19
HEAT ENGINES AND REFRIGERATORS
Conceptual Questions
19.1. Ws =−W .
(a) W < 0, Ws > 0. Work is done by the system; the area under the curve is positive.
(b) W > 0, Ws < 0. Work is done on the system to compress it to a smaller volume.
(c) W > 0, Ws < 0. More work is done on the system than by the system.
19.2. W3 > W1 = W2 > W4 . The amount of work done by the gas is the area inside the closed cycle loop (traversed in
a clockwise direction).
W
QH
out
19.3. No.
h =
=
what you get
. You cannot get out more than you put in.
what you had to pay
19.4. (a) 1
(b) 3
(c) 3
(d) 2
In stage 1 the volume is fixed (so no work is done) but the temperature increases, so heat was added. In stage 2 on the
isotherm work is done by the gas. In stage 3 work is done on the gas and heat is removed.
19.5. ηηηη
3 > 1 = 2 > 4. η =
4
6
40
6
Wout
η1 =
η3 =
η4 =
η2 =
10
10
100
100
QH
19.6. The thermal efficiency is larger for engine 1; the same amount of heat is added per cycle in both engines, but
the cycle for engine 1 has a larger Wout due to the larger enclosed area:=
η
Wout
.
QH
19.7. It is an isothermal process with equal amounts of heat added to the system and work done by the system.
19.8. (a) No; cannot have 15 J out > 10 Jin .
(b) Yes, this is a heat engine with η =
Wout 4
=
= 0.4, which is less than ηCarnot = 0.5.
QH 10
(c) No, it isn’t possible to have ηη
> Carnot : η =
Wout 6
=
= 0.6, but ηCarnot = 0.5.
QH 10
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19-1
19-2
Chapter 19
19.9. (a) No, the purpose of a refrigerator is to remove heat from the cold reservoir, and this diagram dumps heat
into the cold reservoir.
QC 10
(b) Yes, K =
=
= 0.5, which is less than K Carnot = 1.
Win 20
(c) No, =
K
QC 20
= = 2, which is less than K Carnot = 1.
Win 10
19.10. No. The first law of thermodynamics (energy conservation) for a refrigerator or air conditioner requires
Q=
H QC + W . There are no perfect refrigerators (second law), so W > 0 (work must be done by the compressor) and
thus QH > QC . Because the “air conditioner” exhausts more heat into the room than it extracts from the room, the net
effect is to increase the room temperature, not decrease it.
19.11. Yes, the first law says that energy is conserved, so we will never get more work out of the heat engine than
heat energy is transferred to the system. In fact, the second law (informal statement #4) says that there are no perfect
heat engines with η = 1, so there is always some waste heat exhausted to the cold reservoir.
Exercises and Problems
Section 19.1 Turning Heat into Work
Section 19.2 Heat Engines and Refrigerators
19.1. Solve: (a) The engine has a thermal efficiency of η = 40% = 0.40 and a work output of 100 J per cycle. The
heat input is calculated as follows:
η
=
Wout
QH
40
⇒ 0.=
100 J
QH
⇒ Q
=
H 250 J
(b) Because Wout
= QH − QC , the heat exhausted is
QC =QH − Wout =250 J − 100 J =150 J
19.2. Solve: During each cycle, the work done by the engine is Wout = 200 J and the engine exhausts QC = 400 J
of heat energy. By conservation of energy,
QH = Wout + QC = 200 J + 400 J = 600 J
Thus, the efficiency of the engine is
η=
Wout 400 J
=
= 0.33
QH 600 J
19.3. Solve: (a) During each cycle, the heat transferred into the engine is QH = 55 kJ, and the heat exhausted is
QC = 40 kJ. The thermal efficiency of the heat engine is
η =1 −
QC
40 kJ
=1 −
=0.27 =27%
55 kJ
QH
(b) The work done by the engine per cycle is
Wout = QH − QC = 55 kJ − 40 kJ = 15 kJ
19.4. Solve: The coefficient of performance of the refrigerator is
K=
QC QH − Win 600 J − 200 J
=
=
= 2.0
Win
Win
200 J
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Heat Engines and Refrigerators
19-3
19.5. Solve: (a) The heat extracted from the cold reservoir is calculated as follows:
K
=
QC
Win
⇒ 4=
.0
QC
50 J
⇒ Q
=
C 200 J
(b) The heat exhausted to the hot reservoir is
QH =QC + Win = 200 J + 50 J = 250 J
19.6. Model: Assume that the car engine follows a closed cycle.
Solve: (a) Since 2400 rpm is 40 cycles per second, the work output of the car engine per cycle is

kJ   1 s
kJ
kJ

Wout =  500
≈ 13
 = 12.5

s   40 cycles 
cycle
cycle

(b) The heat input per cycle is calculated as follows:
12.5 kJ
⇒ QH =
=.
62 5 kJ
0.20
W
QH
η =out
The heat exhausted per cycle is
QC = QH − Win = 62.5 kJ − 12.5 kJ = 50 kJ
19.7. Solve: The amount of heat discharged per second is calculated as follows:
Wout
W
= out
QH QC + Wout
η=
1 
 1

1 913 × 109 W
⇒ QC =Wout  − 1 =(900 MW) 
− 1 =.
 0.32 
η 
That is, each second the electric power plant discharges 1.913 × 109 J of energy into the ocean. Since a typical American house needs 2.0 × 104 J of energy per second for heating, the number of houses that could be heated with the
waste heat is (1.913 × 109 J)/(2.0 × 104 J) =96,000.
19.8. Solve: The amount of heat removed from the water in cooling it down in 1 hour
is QC mwater cwater ∆T . The
=
mass of the water is
mwater = r waterVwater = (1000 kg/m3 )(1 L) = (100 kg/m3 )(10−3 m3 ) = 1.0 kg
QC = (1.0 kg)(4190 J/kg K)(20°C − 5°C) = 6.285 × 104 J
The rate of heat removal from the refrigerator is
6.285 × 104 J
= 17.46 J/s
3600 s
QC=
The refrigerator does work W = 8.0 W = 8.0 J/s to remove this heat. Thus the performance coefficient of the refrigerator is
K=
17.46 J/s
= 2.2
8.0 J/s
Section 19.3 Ideal-Gas Heat Engines
Section 19.4 Ideal-Gas Refrigerators
19.9. Model: Process A is isochoric, process B is isothermal, process C is adiabatic, and process D is isobaric.
Solve: Process A is isochoric, so the increase in pressure increases the temperature and hence the thermal energy.
Because ∆Eth =Q − Ws and Ws =0 J, Q increases for process A. Process B is adiabatic, so Q = 0 J. Ws is positive
because of the increase in volume. Since Q= 0 =
J Ws + ∆Eth , ∆Eth is negative for process B. Process C is isothermal,
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19-4
Chapter 19
so T is constant and hence ∆Eth =
0 J. The work done Ws is positive because the gas expands. Because
Q= Ws + ∆Eth , Q is positive for process B. Process D is isobaric, so the decrease in volume leads to a decrease in
temperature and hence a decrease in the thermal energy. Due to the decrease in volume, Ws is negative. Because
Q= Ws + ∆Eth , Q also decreases for process D.
A
B
C
D
∆Eth
WS
+
−
0
−
0
+
+
−
Q
+
0
+
−
19.10. Model: Process A is adiabatic, process B is isochoric, and process C is isothermal.
Solve: Process A is adiabatic, so Q = 0 J. Work Ws is positive as the gas expands. Since Q= Ws + ∆Eth= 0 J, ∆Eth
must be negative. The temperature falls during an adiabatic expansion. Process B is isochoric. No work is done
(Ws = 0 J), and Q is positive as heat energy is added to raise the temperature (∆Eth positive). Process C is isothermal so ∆T =
0 and ∆Eth =
0 J. The gas is compressed, so Ws is negative. Q = Ws for an isothermal process, so Q is
negative. Heat energy is withdrawn during the compression to keep the temperature constant.
∆Eth
Ws
Q
−
+
0
+
0
−
0
+
−
A
B
C
19.11. Solve: The work done by the gas per cycle is the area inside the closed p-versus-V curve. The area inside the
triangle is
Wout
=
1 (3
2
atm − 1 atm)(600 × 10−6 m3 − 200 × 10−6 m3 )

1.013 × 105 Pa 
3
−6
= 12  2 atm ×
 (400 × 10 m ) =40 J

1 atm


19.12. Solve: The work done by the gas per cycle is the area enclosed within the pV curve. We have
60=
J
1(p
max
2
− 100 kPa)(800 cm3 − 200 cm3 ) ⇒
2(60 J)
= pmax − 1.0 × 105 Pa
600 × 10−6 m3
pmax = 3.0 × 105 Pa = 300 kPa
19.13. Model: The heat engine follows a closed cycle, which consists of four individual processes.
Solve: (a) The work done by the heat engine per cycle is the area enclosed by the p-versus-V graph. We get
Wout = (400 kPa − 100 kPa)(100 × 10−6 m3 ) = 30 J
The heat energy leaving the engine is QC = 90 J + 25 J = 115 J. The heat input is calculated as follows:
Wout =QH − QC
⇒ QH =QC + Wout =115 J + 30 J =145 J ≈ 0.15 kJ
(b) The thermal efficiency of the engine is
η=
Wout 30 J
=
= 0.21
QH 145 J
Assess: Practical engines have thermal efficiencies in the range η ≈ 0.1 − 0.4.
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Heat Engines and Refrigerators
19-5
19.14. Model: The heat engine follows a closed cycle, starting and ending in the original state. The cycle consists
of three individual processes.
Solve: (a) The work done by the heat engine per cycle is the area enclosed by the p-versus-V graph. We get
=
Wout
1 (200
2
kPa)(100 × 10−6=
m3 ) 10 J
The heat energy transferred into the engine is QH = 30 J + 84 J = 114 J. Because Wout
= QH − QC , the heat energy
exhausted is
QC= QH − Wout= 114 J − 10 J= 104 J ≈ 0.10 kJ
(b) The thermal efficiency of the engine is
η=
Wout 10 J
=
= 0.088
QH 114 J
Assess: Practical engines have thermal efficiencies in the range η ≈ 0.1 − 0.4.
19.15. Model: The heat engine follows a closed cycle.
Solve: The work done by the gas per cycle is the area inside the closed p-versus-V curve. We get
1 (300 kPa − 100 kPa)(600 cm3 − 300 cm3 ) =
1 (200 × 103 Pa)(300 × 10−6 m3 ) =
Wout =
30 J
2
2
Because Wout
= QH − QC , the heat exhausted is
QC =QH − Wout =(225 J + 90 J) − 30 J =315 J − 30 J =285 J
19.16. Model: The heat engine follows a closed cycle.
Solve: (a) The work done by the gas per cycle is the area inside the closed p-versus-V curve. We get
1 (300 kPa − 100 kPa)(600 cm3 − 200 cm3 ) =
1 (200 × 103 Pa)(400 × 10−6 m3 ) =
Wout =
40 J
2
2
The heat exhausted is QC = 180 J + 100 J = 280 J. The thermal efficiency of the engine is
η=
Wout
Wout
40 J
=
=
= 0.13
QH QC + Wout 280 J + 40 J
(b) The heat extracted from the hot reservoir is QH =QC + Wout =320 J.
19.17. Model: The Brayton cycle involves two adiabatic processes and two isobaric processes. The adiabatic processes involve compression and expansion through the turbine.
Solve: The thermal efficiency for the Brayton cycle is ηB = 1 − rp(1−γ )/γ , where γ = CP /CV and rp is the pressure
ratio. For a diatomic gas γ = 1.4. For an adiabatic process,
p1V1γ = p2V2γ
⇒
p2 /p1 = (V1/V2 )γ
Because the volume is halved, V2 = 12 V1 so
rp = p2 /p1 = (2)γ = 21.4 = 2.639
The efficiency is
ηB =
1 − (2.639) −0.4/1.4 =
0.24
19.18. Model: The refrigerator follows a closed cycle.
Solve: (a) The net work done on the refrigerator in one cycle is
Win =
−Ws =
−78 J + 119 J =
41 J
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19-6
Chapter 19
This is the work needed to push the heat from the cold reservoir up to the hot reservoir. The heat exhausted to the hot
reservoir is QH = 105 J. From the first law of thermodynamics, the heat extracted from the cold reservoir is
Win + QC = QH
⇒ QC = QH − Win = 105 J − 41 J = 64 J
(b) The coefficient of performance is
=
K
QC 64 J
=
= 1.6
Win 41 J
Assess: This is a reasonable value for the coefficient of performance for a refrigerator.
Section 19.5 The Limits of Efficiency
Section 19.6 The Carnot Cycle
19.19. Model: The efficiency of a Carnot engine (ηCarnot ) depends only on the temperatures of the hot and cold
reservoirs. On the other hand, the thermal efficiency (η ) of a heat engine depends on the heats QH and QC .
Solve: (a) According to the first law of thermodynamics, =
QH Wout + QC . For engine (a), QH = 500 J, QC = 200 J
and Wout = 300 J, so the first law of thermodynamics is obeyed. For engine =
(b), QH 500
=
J, QC 200 J and
(c) QH 300
Wout = 200 J, so the first law is violated. For engine=
=
J, QC 200 J and Wout = 100 J, so the first law of
thermodynamics is obeyed.
(b) For the three heat engines, the maximum or Carnot efficiency is
1−
ηCarnot =
TC
300 K
1−
0.50
=
=
TH
600 K
Engine (a) has
η=
1−
QC Wout 300 J
= =
=
0.60
QH QH 500 J
This is larger than ηCarnot , thus violating the second law of thermodynamics. For engine (b),
W
QH
200 J
500 J
ηη
= out =
=0.40 < Carnot
so the second law is obeyed. Engine (c) has a thermal efficiency of
100 J
300 J
ηη
=
=0.33 < Carnot
so the second law of thermodynamics is obeyed.
19.20. Model: For a refrigerator Q=
H QC + Win , and the coefficient of performance and the Carnot coefficient of
performance are
=
K
QC
TC
=
, K Carnot
Win
TH − TC
Solve: (a) For refrigerator (a) QH =
QC + Win (60 J =
40 J + 20 J), so the first law of thermodynamics is obeyed. For
refrigerator (b) 50
=
J 40 J + 10 J, so the first law of thermodynamics is obeyed. For the refrigerator (c)
40 J ≠ 30 J + 20 J, so the first law of thermodynamics is violated.
(b) For the three refrigerators, the maximum coefficient of performance is
K
=
Carnot
TC
300 K
=
= 3
TH − TC 400 K − 300 K
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Heat Engines and Refrigerators
19-7
For refrigerator (a),
K=
QC 40 J
=
= 2 < K Carnot
Win 20 J
so the second law of thermodynamics is obeyed. For refrigerator (b),
K=
QC 40 J
=
= 4 > K Carnot
Win 10 J
so the second law of thermodynamics is violated. For refrigerator (c),
K=
30 J
= 1.5 < K Carnot
20 J
so the second law is obeyed.
19.21. Model: The efficiency of a Carnot engine depends only on the absolute temperatures of the hot and cold
reservoirs.
Solve: The efficiency of a Carnot engine is
ηCarnot =1 −
TC
TH
⇒ 0.60 =1 −
TC
(427 + 273) K
⇒ TC =280 K =7°C
Assess: A “real” engine would need a lower temperature than 7°C to provide 60% efficiency because no real
engine can match the Carnot efficiency.
19.22. Model: The efficiency of an ideal engine (or Carnot engine) depends only on the temperatures of the hot and
cold reservoirs.
Solve: (a) The engine’s thermal efficiency is
Wout
Wout
10 J
=
=
=0.40 =40%
QH QC + Wout 15 J + 10 J
η=
(b) The efficiency of a Carnot engine is ηCarnot = 1 − TC /TH . The minimum temperature in the hot reservoir is found
as follows:
293 K
0.40 =1 −
⇒ TH =488 K =215°C
TH
This is the minimum possible temperature. In a real engine, the hot-reservoir temperature would be higher than
215°C because no real engine can match the Carnot efficiency.
19.23. Model: Assume that the heat engine follows a closed cycle.
Solve: (a) The engine’s thermal efficiency is
Wout
Wout
200 J
=
=
=0.25 =25%
QH QC + Wout 600 J + 200 J
η=
(b) The thermal efficiency of a Carnot engine is ηCarnot = 1 − TC /TH . For this to be 25%,
0.25 =1 −
TC
(400 + 273) K
⇒ TC =504.8 K =232°C
19.24. Solve: (a) The efficiency of the Carnot engine is
ηCarnot =1 −
TC
300 K
=1 −
=0.40 =40%
TH
500 K
(b) An engine with power output of 1000 W does Wout = 1000 J of work during each ∆t =1 s. A Carnot engine has a
heat input that is
Wout
1000 J
Qin
=
=
= 2500 J
0.40
ηCarnot
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19-8
Chapter 19
during each ∆t =1 s. The rate of heat input is 2500 J/s = 2500 W.
(c) Wout
= Qin − |Qout |, so the heat output during ∆t =1 s is |Qout | =Qin − Wout =1500 J . The rate of heat output is
thus 1500 J/s = 1500 W.
19.25. Model: We will use Equation 19.27 for the efficiency of a Carnot engine.
ηCarnot = 1 −
TC
TH
We are given TH = 673 K and the original efficiency ηCarnot = 0.40.
Solve: First solve for TC .
T=
) (673 K)(1 − 0.40)
= 404 K
C TH (1 − ηCarnot=
Solve for TC′ again with η ′Carnot =0.60.
′
′ =
= 269 K
T=
) (673 K)(1 − 0.60)
C TH (1 − ηCarnot
The difference of these TC values is 135 K, so the temperature of the cold reservoir should be decreased by 135°C to
raise the efficiency from 40% to 60%.
Assess: We expected to have to lower TC by quite a bit to get the better efficiency.
19.26. Model: The maximum possible efficiency for a heat engine is provided by the Carnot engine.
Solve: The maximum efficiency is
1−
ηη
max =Carnot =
TC
(273 + 20) K
1−
0.6644
=
=
TH
(273 + 600) K
Because the heat engine is running at only 30% of the maximum efficiency, ηη
=
(0.30) max =
0.1993. The amount of
heat that must be extracted is
QH =
Wout
η
=
1000 J
= 5.0 kJ
0.1993
19.27. Model: We are given TH = 773 K and TC = 273 K, therefore (by Equation 19.27) the Carnot efficiency is
273 K
nCarnot =
1 − 773
=
0.647 We are also given ηη
= 0.60( Carnot ).
K
Solve: Rearrange Equation 19.5:=
QH Wout (1 − η ). Wout is the same for both engines, so it cancels.
QH
Wout (1 − ηη
)
1 − 0.60( Carnot ) 1 − 0.60(0.647)
=
=
=
= 1.7
(QH )Carnot Wout (1 − ηη
)
1 − Carnot
1 − 0.647
Carnot
Assess: This engine requires 1.7 times as much heat energy during each cycle as a Carnot engine to do the same
amount of work.
19.28. Model: The coefficient of performance of a Carnot refrigerator depends only on the temperatures of the cold
and hot reservoirs.
Solve: (a) The Carnot performance coefficient of a refrigerator is
K Carnot =
TC
(−20 + 273) K
=
= 6.325 ≈ 6.3
TH − TC (20 + 273) K − (−20 + 273) K
(b) The rate at which work is done on the refrigerator is found as follows:
K=
QC
Q
200 J/s
⇒ Win = C =
= 32 J/s = 32 W
Win
K
6.325
(c) The heat exhausted to the hot side per second is
QH= QC + Win= 200 J/s + 32 J/s= 232 J/s ≈ 0.23 kW
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Heat Engines and Refrigerators
19-9
19.29. Model: The minimum possible value of TC occurs with a Carnot refrigerator.
Solve: (a) For the refrigerator, the coefficient of performance is
Q
K =C
Win
⇒ QC =
KWin =
(5.0)(10 J) =
50 J
The heat energy exhausted per cycle is
QH =QC + Win =50 J + 10 J = 60 J
(b) If the hot-reservoir temperature is 27°C = 300 K, the lowest possible temperature of the cold reservoir can be
obtained as follows:
T
K Carnot = C
TH − TC
TC
⇒ 5.0 =
300 K − TC
19.30. Model: Equation 19.27 gives ηCarnot = 1 −
⇒ TC =250 K =−23°C
TC
. We are given ηCarnot
= 1/ 3.
TH
Solve:
1−
ηCarnot =
TC 1
TC 2
3
= ⇒
= ⇒ TH =TC
TH 3
TH 3
2
Equation 19.28 gives the coefficient of performance for the Carnot refrigerator.
TC
=
TH − TC
K Carnot
=
TC
=
− TC
3T
2 C
1
= 2
−1
3
2
Assess: This result is in the ballpark for coefficients of performance.
19.31. Solve: The work done by the engine is equal to the change in the gravitational potential energy. Thus,
Wout =
∆U grav ==
mgh (2000 kg)(9.8 m/s 2 )(30 m) =
588,000 J
The efficiency of this engine is

=
0.40 Carnot =
0.40 1 −
ηη


TC 
(273 + 20) K 
0.40 1 −
0.3484
=
=
TH 
(273
+ 2000) K 

The amount of heat energy transferred is calculated as follows:
η=
Wout
QH
⇒ QH =
Wout
η
=
588,000 J
= 1.7 × 106 J
0.3484
19.32. Solve: The mass of the water is
 10−3 m3   1000 kg 
(100 × 10−3 L) 
0.100 kg
=
 1 L   m3 


The heat energy is removed from the water in three steps: (1) cooling from +15°C to 0°C, (2) freezing at 0°C, and (3)
cooling from 0°C to −15°C. The three heat energies are
Q1 = mc∆T = (0.100 kg)(4186 J/kg K)(15 K) = 6279 J
Q2 = mLf = (0.100 kg)(3.33 × 105 J/kg) = 33,300 J
Q3 = mc∆T = (0.100 kg)(2090 J/kg K)(15 K) = 3135 J
QC = Q1 + Q2 + Q3 = 42,714 J
Using the performance coefficient,
=
K
QC
Win
⇒ 4=
.0
42,714 J
Win
⇒ W
=
in
42,714 J
= 10,679 J
4.0
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19-10
Chapter 19
The heat exhausted into the room is thus
QH = QC + Win = 42,714 J + 10,679 J = 5.3 × 104 J
19.33. Solve: An adiabatic process has Q = 0 and thus, from the first law, Ws = −∆Eth . For any ideal-gas process,
∆Eth = nCV ∆T , so Ws =
− nCV ∆T . We can use the ideal-gas law to find
T
=
pV
nR
T
⇒ ∆=
∆( pV ) ( pV )f − ( pV )i pf Vf − piVi
=
=
nR
nR
nR
Consequently, the work is
CV
 p V − piVi 
Ws =−nCV ∆T =−nCV  f f
( pf Vf − piVi )
 =−
nR
R


Because C=
P CV + R, we can use the specific heat ratio γ to find
C
CV
C + R CV /R + 1
=
⇒
CV
C V /R
γ = P =V
CV
1
=
R γ −1
With this, the work done in an adiabatic process is
C
pf Vf − piVi
1
Ws =
( pf Vf − piVi ) =
− V ( pf Vf − piVi ) =
−
R
1− γ
γ −1
19.34. Model: We are given TH = 323 K and TC = 253 K. See Figure 19.11.
Solve: Every second, the refrigerator must draw enough heat from the cold reservoir to compensate for the heat lost
through the stainless-steel panel. Therefore, the heat transferred from the cold reservoir to the system is
QC= k
A
(0.40 m)(0.40 m)
∆T ∆=
t (14 W/m K)
[25°C − (−20°C)](1.0 s)= 10,080 J
L
0.010 m
Using the coefficient of performance for a Carnot refrigerator, we can find the energy required to operate for one
second:
TC
Q
Q (T − T ) (10,080 J)(70 K)
= C ⇒ Win = C H C =
= 2.8 kJ
K Carnot =
TH − TC Win
TC
253 K
The power required is therefore
=
P = Win ∆t (2.8
=
kJ)(1.0 s) 2.8 kW.
Assess: This is much more power than is required for the Brayton-cycle refrigerator of Example 19, which shows
why refrigerators are insulated with more than simple steel doors.
19.35. Solve: For any heat engine, η = 1 − QC /QH . For a Carnot heat engine, ηCarnot= 1 − TC /TH . Thus a property
of the Carnot cycle is that QC /QH = TC /TH . Consequently, the coefficient of performance of a Carnot refrigerator is
K Carnot
=
QC
QC
QC /QH
TC /TH
TC
=
=
=
=
Win QH − QC 1 − QC /QH 1 − TC /TH TH − TC
19.36. Model: We are given TH = 298 K and TC = 273 K. See Figure 19.11.
Solve: QC =mLf =(10 kg)(3.33 × 105 J/kg) =3.33 × 106 J.
(a) For a Carnot cycle ηCarnot = 1 −
Q
TC
Q
T
but that must also equal η = 1 − C , so C = C .
QH
TH
QH TH
QH = QC
TH
 298 K 
6
= (3.33 × 106 J) 
 = 3.6 × 10 J
TC
 273 K 
(b)
Win =QH − QC =3.63 × 106 J − 3.33 × 106 J =0.30 × 106 J =3.0 × 105 J
Assess: This is a reasonable amount of work to freeze 10 kg of water.
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Heat Engines and Refrigerators
19-11
19.37. Model: We will use the Carnot engine to find the maximum possible efficiency of a floating power plant.
Solve: The efficiency of a Carnot engine is
ηη
max = Carnot = 1 −
TC
(273 + 5) K
= 1−
= 0.0825 ≈ 8.3%
TH
(273 + 30) K
19.38. Model: The ideal gas in the Carnot engine follows a closed cycle in four steps. During the isothermal expansion at temperature TH , heat QH is transferred from the hot reservoir into the gas. During the isothermal compression
at TC , heat QC is removed from the gas. No heat is transferred during the remaining two adiabatic steps.
Solve: The thermal efficiency of the Carnot engine is
1
ηCarnot =−
TC Wout
=
TH QH
⇒ 1−
W
323 K
= out
573 K 1000 J
436 J
⇒ Wout =
Using Q=
H QC + Wout , we obtain
Qisothermal= QC= QH − Wout= 1000 J − 436 J ≈ 0.56 kJ
19.39. Solve: Substituting into the formula for the efficiency of a Carnot engine,
ηCarnot =1 −
TC
TH
⇒ 0.25 =1 −
TC
TC + 80 K
⇒ TC =240 K =−33°C
The hot-reservoir temperature is TH =
TC + 80 K =
320 K =°
47 C.
19.40. Solve: From the thermal efficiency of the Carnot engine, we can find the work done each cycle:
1−
ηCarnot =
TC Wout
=
TH QH
 273 K 
⇒ Wout =
10 J
1 −
 (25 J) =
 455 K 
The work required to lift a 10 kg mass 10 m is W
= Fd
= (10 kg)(9.8 m/s 2 )(10 m) = 980 J. At 10 J/cycle, the Carnot
engine will have to cycle 98 times to do this work.
19.41. Model: Assume the soda is essentially made of water. We are given TH = 328 K, TC = 253 K, and
QH = 250 J. See Figure 19.11.
Solve: The total amount of heat to transfer from the soda is
(QC )Total = Mc∆T = V ρ c∆T = (5.00 × 10−4 m3 )(1000 kg/m3 )(4190 J/kg K)(20 K) = 41,900 J
For a Carnot cycle ηCarnot = 1 −
Q
T
TC
Q
but also η = 1 − C , so C = C . Therefore, the heat extracted from the soda
QH TH
TH
QH
each cycle is
TC
 253 K 
=
QC Q=
(250 J)  =
H
 192.8 J
TH
 328 K 
To remove 41,900 J will therefore take
(QC )Total 41,900 J
= = 218 cycles
QC
192.8 J
19.42. Solve: (a) Q1 is given as 1000 J. Using the energy transfer equation for the heat engine,
QH =
QC + Wout
⇒ Q1 =+
Q2 Wout
⇒ Q2 =−
Q1 Wout
The thermal efficiency of a Carnot engine is
η =1 −
TC
W
300 K
=1 −
=0.50 = out
TH
600 K
Q1
Q2 = Q1 − ηη
Q1 = Q1 (1 − ) = (1000 J)(1 − 0.50) = 500 J
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19-12
Chapter 19
To determine Q3 and Q4 , we turn our attention to the Carnot refrigerator, which is driven by the output of the heat
engine with Win = Wout . The coefficient of performance is
K=
400 K
TC
Q
Q
Q
=
= 4.0 = C = 4 = 4
TH − TC 500 K − 400 K
Win Wout ηQ1
Q4 =KηQ1 =(4.0)(0.50)(1000 J) =2000 J
Using now the energy transfer equation Win + Q4 =
Q3 , we have
Q3 =Wout + Q4 =ηQ1 + Q4 =(0.50)(1000 J) + 2000 J =2500 J
(b) From part (a) Q3 = 2500 J and Q1 = 1000 J, so Q3 > Q1.
(c) Although Q1 = 1000 J and Q3 = 2500 J, the two devices together do not violate the second law of thermodynamics. This is because the hot and cold reservoirs are different for the heat engine and the refrigerator.
19.43. Solve: The work done by the Carnot engine powers the refrigerator, so (Wout )Carnot Eng = (Win )Refrigerator . We
are given that TH = 350 K and TC = 250 K for both the Carnot engine and the refrigerator and (QH )Carnot Eng =
10.0 J for the Carnot engine. The work done by the Carnot engine is
1−
η=
TC (Wout )Carnot Eng
=
TH (QH )Carnot Eng
 250 K 
2.857 J
⇒ (Wout )Carnot Eng =
1 −
 (10.0 J) =
 350 K 
The heat extracted from the cold reservoir by the refrigerator may be found from it coefficient of performance:
K
=
QC
QC
=
(Win ) Refrigerator (Wout )Carnot Eng
J) 5.713 J
=
⇒ QC (2.00)(2.857
=
The heat exhausted by the refrigerator to the hot reservoir may be found from the first law of thermodynamics:
(Win ) Refrigerator + QC = QH
⇒ QH = 2.857 J + 5.713 J= 8.57 J
Assess: The work done on the refrigerator is less than the heat exhausted to the hot reservoir, as expected.
19.44. Model: A heat pump is a refrigerator that is cooling the already cold outdoors and warming the indoors with
its exhaust heat.
Solve: (a) The coefficient of performance for this heat pump is K = 5.0 = QC /Win , where QC is the amount of heat
removed from the cold reservoir. QH is the amount of heat exhausted into the hot reservoir. Q=
H QC + Win , where
Win is the amount of work done on the heat pump. We have
QC =
5.0Win
⇒ QH =
5.0Win + Win =
6.0Win
If the heat pump is to deliver 15 kJ of heat per second to the house, then
QH =
15 kJ =
6.0Win
15 kJ
⇒ Win = =
2.5 kJ
6.0
In other words, 2.5 kW of electric power is used by the heat pump to deliver 15 kJ/s of heat energy to the house.
(b) The monthly heating cost in the house using an electric heater is
 15 kJ 
 3600 s  1$ 

 (200 h) 

 = $270
 s 
 1 h  40 MJ 
The monthly heating cost in the house using a heat pump is
 2.5 kJ 
 3600 s  1$ 

 (200 h) 

 = $45
s


 1 h  40 MJ 
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Heat Engines and Refrigerators
19-13
19.45. Visualize: We are given TC = 275 K,=
TH 295 K. We are also given that in one second Win = 100 J and
=
QC (1 s)(100 kJ/min)(1 min/60
=
s) 1667 J.
Solve: The coefficient of performance of a refrigerator is given in Equation 19.8.
K=
QC 1667 J
=
= 16.67
Win 100 J
However the coefficient of performance of a Carnot refrigerator is given in Equation 19.28.
K Carnot=
TC
275 K
=
= 13.75
TH − TC 20 K
However, informal statement #8 of the second law says that the coefficient of performance cannot exceed the Carnot
coefficient of performance, so the salesman is making false claims. You should not buy the DreamFridge.
Assess: The second law imposes real-world restrictions.
19.46. Solve: The maximum possible efficiency of the heat engine is
1−
ηmax =
TC
300 K
1−
0.40
=
=
TH
500 K
The efficiency of the engine designed by the first student is
η1 =
Wout 110 J
=
= 0.44
QH
250 J
Because ηη
1 > max , the first student has proposed an engine that would violate the second law of thermodynamics. His or
her design will not work. The efficiency of the engine designed by the second student is ηη
2 =90 J / 250 J =0.36 < max
in agreement with the second law of thermodynamics. Applying the first law of thermodynamics,
QH =
QC + Wout
⇒ 250 J =
170 J + 90 J
we see the first law is violated. This design will not work as claimed. The design by the third student satisfies the first
law of thermodynamics because QH =QC + Wout =250 J. The thermal efficiency of this engine is ηη
3 =0.36 < max ,
which satisfies the second law of thermodynamics. The data presented by students 1 and 2 are faulty. Only student 3
has an acceptable design.
19.47. Model: The power plant is to be treated as a heat engine.
Solve: (a) Every hour 300 metric tons or 3 × 105 kg of coal is burnt. The volume of coal is
 3 × 105 kg  m3 
(24 h) = 4800 m3


 1500 kg 
1
h



The height of the room will be 48 m.
(b) The thermal efficiency of the power plant is
Wout
7.50 × 108 J/s
7.50 × 108 J
=
=
=0.32 =32%
QH  3 × 105 kg  28 × 106 J   1 h  2.333 × 109 J


 


kg
 1h

  3600 s 
h=
Assess: An efficiency of 32% is typical of power plants.
19.48. Model: The power plant is treated as a heat engine.
Solve: We are given that
=
TH 300°C
= 573 K and=
TC 25°C
= 298 K.
(a) The maximum possible thermal efficiency of the power plant is
ηmax =1 −
TC
298 K
=1 −
=0.48 =48%
TH
573 K
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19-14
Chapter 19
6
(b) In one second, the plant generates W=
QH 3000 × 106 J of heat energy to replace
out 1000 × 10 J of work and =
the energy taken from the hot reservoir to heat the water. The plant’s actual efficiency is
=
η
Wout 1000 × 106 J
=
0.=
33 33%
QH 3000 × 106 J
(c) Because Q=
H QC + Wout ,
QC = QH − Wout
⇒ QC = 3.0 × 109 J/s − 1.0 × 109 J/s = 2.0 × 109 J/s
The mass of water that flows per second through the condenser is
L  1 hr   10−3 m3   1000 kg 

4
m = 1.2 × 108 


 (1 s) = 3.333 × 10 kg
h  3600 s   1 L   m3 

The change in the temperature as QC = 1.0 × 109 J of heat is transferred to m =.
3 333 × 104 kg of water is
QC =mc∆T
⇒ 2.0 × 109 J =(3.333 × 104 kg)(4186 J/kg K)∆T
⇒ ∆T =14°C
The exit temperature is 18°C + 14°C =
32°C.
19.49. Model: The power plant is treated as a heat engine.
Solve: The mass of water per second that flows through the plant every second is
L  1 hr   10−3 m3   1000 kg 

4
m = 1.0 × 108 


 = 2.778 × 10 kg/s
h  3600 s   1 L   m3 

The amount of heat transferred per second to the cooling water is thus
QC =mc∆T =(2.778 × 104 kg/s)(4186 J/kg K)(27°C − 16°C) =1.279 × 109 J/s
The amount of heat per second input into the power plant is
QH =
Wout + QC =.
0 750 × 109 J/s + 1.279 × 109 J/s =.
2 029 × 109 J/s
Finally, the power plant’s thermal efficiency is
Wout 0.750 × 109 J/s
=
=0.37 =37%
QH 2.029 × 109 J/s
η=
19.50. Solve: (a) The energy supplied in one day is

Wout = 1.0 × 109

J  3600 s  24 h 
13


 = 8.6 × 10 J
s  1 h  1 d 
(b) The volume of water =
is V 1=
km3 109 m3. The amount of energy is
 1000 kg 
15
QH = mc∆T = (109 m3 ) 
 (4190 J/kg K)(1 K) = 4 × 10 J
 m3 
(c) While it’s true that the ocean contains vast amounts of thermal energy, that energy can be extracted to do useful
work only if there is a cold reservoir at a lower temperature. That is, the ocean has to be the hot reservoir of a heat
engine. But there’s no readily available cold reservoir, so the ocean’s energy cannot readily be tapped. There have
been proposals for using the colder water near the bottom of the ocean as a cold reservoir, pumping it up to the surface where the heat engine is. Although possible, the very small temperature difference between the surface and the
ocean depths implies that the maximum possible efficiency (the Carnot efficiency) is only a few percent, and the
efficiency of any real ocean-driven heat engine would likely be less than 1%—perhaps much less. Thus the second
law of thermodynamics prevents us from using the thermal energy of the ocean. Save your money. Don’t invest.
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Heat Engines and Refrigerators
19-15
19.51. Visualize: If we do this problem on a “per-second” basis then in one second =
QC (1 s)(5.0 × 105 J/min)
(1 min/60 s) =8.33 × 103 J. QH = (1 s)(8.0 × 105 J/min)(1 min/ 60 s) = 13.33 × 103 J.
Solve: (a) Again, in one second
Win =QH − QC =13.33 × 103 J − 8.33 × 103 J =5.0 × 103 J
Since this is per second, the power required by the compressor is P = 5.0 kW.
(b) The coefficient of performance is
QC 8.33 × 103 J
=
= 1.7
Win 5.0 × 103 J
K=
Assess: The result is typical for air conditioners.
19.52. Model: The heat engine follows a closed cycle with process 1 → 2 and process 3 → 4 being isochoric and
process 2 → 3 and process 4 → 1 being isobaric. For a monatomic gas, CV =
3
2
R and CP =
5
2
R.
Visualize: Please refer to Figure P19.52.
Solve: (a) The first law of thermodynamics is Q =
∆Eth + WS. For the isochoric process 1 → 2, WS 1→2= 0 J. Thus,
Q1→2 =
3750 J =
∆Eth =
nCV ∆T
=
∆T
3750 J
3750 J
3750 J
=
=
= 301 K
nCV
(1.0 mol) 32 R
(1.0 mol) 32 (8.31 J/mol K)
( )
( )
T2 − T1 = 300.8 K ⇒ T2 = 300.8 K + 300 K = 601 K
To find volume V2 ,
nRT (1.0 mol)(8.31 J/mol K)(300 K)
=8.31 × 10−3 m3
V2 =V1 = 1 =
p1
3.0 × 105 Pa
The pressure p2 can be obtained from the isochoric condition as follows:
p2 p1
T
 601 K 
5
5
= ⇒ p2 = 2 p1 =
 (3.00 × 10 Pa) =6.01 × 10 Pa
T2 T1
T1
 300 K 
With the above values of p2 , V2 , and T2 , we can now obtain p3 , V3 , and T3. We have
V3 =
2V2 =
1.662 × 10−2 m3
p3 =p2 =6.01 × 105 Pa
T3 T2
=
V3 V2
⇒ T3 =
V3
T2 = 2T2 =1202 K
V2
For the isobaric process 2 → 3,
Q2→3 = nCP ∆T = (1.0 mol)
WS
2 →3
( 52 R ) (T3 − T2 ) = (1.0 mol) ( 52 ) (8.31 J/mol K)(601 K) = 12,480 J
= p3 (V3 − V2 ) = (6.01 × 105 Pa)(8.31 × 10−3 m3 ) = 4990 J
∆Eth= Q2→3 − WS
2→3=
12,480 J − 4990 J= 7490 J
We are now able to obtain p4 , V4 , and T4 . We have
V4 =
V3 =
1.662 × 10−2 m3
p4 =p1 =3.00 × 105 Pa
T4 T3
=
p4 p3
⇒ T4 =
p4
T3 =
p3
 3.00 × 105 Pa 

 (1202 K) = 600 K
5
 6.01 × 10 Pa 
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19-16
Chapter 19
For isochoric process 3 → 4,
Q3→4 = nCV ∆T = (1.0 mol)
WS
3→ 4
( 32 R ) (T4 − T3 ) = (1.0 mol) ( 32 ) (8.31 J/mol K)(−602) = −7500 J
=
0 J ⇒ ∆Eth =
Q3→4 − WS
3→ 4
=
−7500 J
For isobaric process 4 → 1,
Q4→1 =nCP ∆T =(1.0 mol) 52 (8.31 J/mol K)(300 K − 600 K) =−6230 J
WS
4→1
=p4 (V1 − V4 ) =(3.00 × 105 Pa) × (8.31 × 10−3 m3 − 1.662 × 10−2 m3 ) =−2490 J
∆Eth =Q4→1 − WS
4→1
=−6230 J − (−2490 J) =−3740 J
Q (kJ)
WS (J)
1→ 2
2→3
3→4
4 →1
Net
0
4990
0
–2490
2500
3750
12,480
–7500
–6230
2500
∆Eth (kJ)
3750
7490
–7500
–3740
0
(b) The thermal efficiency of this heat engine is
Wout
Wout
2500 J
=
=
=0.15 =15%
QH Q1→2 + Q2→3 3750 J + 12,480 J
η=
Assess: Note that more than two significant figures are retained in part (a) because the results are intermediate. For a
closed cycle, as expected, (Ws ) net = Qnet and (∆Eth ) net =
0J
19.53. Model: The heat engine follows a closed cycle. For a diatomic gas, CV = 52 R and C=
P
7
R.
2
Visualize: Please refer to Figure P19.53.
Solve: (a) Since T1 = 293 K, the number of moles of the gas is
n=
p1V1 (0.5 × 1.013 × 105 Pa)(10 × 10−6 m3 )
=
=2.08 × 10−4 mol
RT1
(8.31 J/mol K)(293 K)
At point
2, V2 4=
=
V1 and p2 3 p1. The temperature is calculated as follows:
p1V1 p2V2
=
T1
T2
⇒ T2=
p2 V2
T=
= 3516 K
1 (3)(4)(293 K)
p1 V1
At point 3, V=
p1. The temperature is calculated as before:
3 V=
2 4V1 and p=
3
=
T3
p3 V3
=
T1 (1)(4)(293
=
K) 1172 K
p1 V1
For process 1 → 2, the work done is the area under the p-versus-V curve. That is,
Ws =
(0.5 atm)(40 cm3 − 10 cm3 ) + 12 (1.5 atm − 0.5 atm)(40 cm3 − 10 cm3 )
 1.013 × 105 Pa 
(30 × 10−6 m3 )(1 atm) 
3.04 J
=
 =

1 atm


The change in the thermal energy is
∆Eth = nCV ∆T = (2.08 × 10−4 mol) 52 (8.31 J/mol K)(3516 K − 293 K) = 13.93 J
The heat is Q= Ws + ∆Eth= 16.97 J. For process 2 → 3, the work done is Ws = 0 J and
Q =∆Eth =nCV ∆T =n
( 52 R ) (T3 − T2 )
=
(2.08 × 10−4 mol) 52 (8.31 J/mol K)(1172 K − 3516 K) =
−10.13 J
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Heat Engines and Refrigerators
19-17
For process 3 → 1,
Ws = (0.5 atm)(10 cm3 − 40 cm3 ) = (0.5 × 1.013 × 105 Pa)( −30 × 10−6 m3 ) = −1.52 J
∆Eth =nCV ∆T =(2.08 × 10−4 mol) 52 (8.31 J/mol K)(293 K − 1172 K) =−3.80 J
The heat is Q =∆Eth + Ws =−5.32 J.
Ws (J)
Q (J)
∆Eth
3.04
0
−1.52
1.52
16.97
−10.13
−5.32
1.52
13.93
−10.13
−3.80
0
1→ 2
2→3
3 →1
Net
(b) The efficiency of the engine is
1.52 J
0.090 =
9.0%
=
16.97 J
W
QH
η =net =
(c) The power output of the engine is
 revolutions  1 min  Wnet   500 
500 


= 
 (1.52 J/s) = 13 W
min

 60 s  revolution   60 
Assess: Note that more than two significant figures are retained in part (a) because the results are intermediate. For a
closed cycle, as expected, (Ws ) net = Qnet and (∆Eth ) net =
0 J.
19.54. Model: For the closed cycle, process 1 → 2 is isothermal, process 2 → 3 is isobaric, and process 3 → 1 is
isochoric.
Visualize: Please refer to Figure P19.54.
Solve: (a) We first need to find the conditions at points 1, 2, and 3. We can then use that information to find WS and
Q for each of the three processes that make up this cycle. Using the ideal-gas equation the number of moles of the gas
is
n=
p1V1 (1.013 × 105 Pa)(600 × 10−6 m3 )
=
= 0.0244 mol
RT1
(8.31 J/mol K)(300 K)
We are given that γ = 1.25, which means this is not a monatomic or a diatomic gas. The specific heats are
CV =
R
= 4R
1− γ
CP = CV + R = 5 R
At point 2, process 1 → 2 is isothermal, so we can find the pressure p2 as follows:
p1V1 =
p2V2
⇒
V
6.00 × 10−4 m3
p2 =1 p1 =
p =
3 p1 =
3 atm =
3.039 × 105 Pa
3 1
−4
V2
2.00 × 10 m
At point 3, process 2 → 3 is isobaric, so we can find the temperature T3 as follows:
V2 V3
=
T2 T3
V
6.00 × 10−4 m3
⇒ T3 = 3 T2 =
T2 =3T2 =900 K
V2
2.00 × 10−4 m3
V (m3 )
Point
P (Pa)
1
1.0 =
atm 1.013 × 10
2
3.0=
atm 3.039 × 10
3
3.0=
atm 3.039 × 10
5
5
5
T (K)
6.00 × 10
−4
300
2.00 × 10
−4
300
6.00 × 10
−4
900
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19-18
Chapter 19
Process 1 → 2 is isothermal:
(WS )12 =p1V1 ln(V2 /V1 ) =
−66.8 J
Q12 =
(WS )12 =
−66.8 J
Process 2 → 3 is isobaric:
(WS ) 23 = p2∆V = p2 (V3 − V2 ) = 121.6 J
Q23= nCP ∆T = nCP (T3 − T2 )= 608.3 J
Process 3 → 1 is isochoric:
(WS )31 = 0 J
Q31 =nCV ∆T =nCV (T1 − T3 ) =−486.7 J
We find that
(WS )cycle =
−66.8 J + 121.6 J + 0 J =
54.8 J
Qcycle =
−66.8 J + 608.3 J − 486.7 J =
54.8 J
These are equal, as they should be. Knowing that the work done=
is Wout (W
=
S )cycle 54.8 J/cycle, an engine operating at 20 cycles/s has a power output of
 54.8 J   20 cycle 
J
Pout
= 
=

 1096 = 1096 W ≈ 1.10 kW
s
s

 cycle  
(b) Only Q23 is positive, so Q=
in Q=
23 608 J. Thus, the thermal efficiency is
W
Qin
54.8 J
0.0901 =
9.01%
=
608.3 J
η =out =
19.55. Model: For the closed cycle of the heat engine, process 1 → 2 is isobaric, process 2 → 3 is isochoric, and
process 3 → 1 is adiabatic. CV = 32 R and CP = 52 R for a monatomic gas, so γ = 5/3.
Visualize: Please refer to Figure P19.55.
Solve: (a) We can use the adiabat 3 → 1 to calculate p1 as follows:
p1V1γ= p3V3γ
γ
⇒
5/3
 600 cm3 
V 
p=
p3  3  = (100 kPa) 
= 1981 kPa
1
3


 V1 
 100 cm 
T1 can be determined by taking the ratio of the ideal-gas equation applied to points 1 and 2. This gives
p1V1 T1
=
p2V2 T2
V1
100 cm3
=
T1 T=
(600 K)
= 100 K
2
V2
600 cm3
where we have used the fact that p1 = p2 . Applying the same strategy at point 3 gives
p2V2 T2
=
p3V3 T3
p
 100 kPa 
T3= T2 3= (600 K) 
=
 30.3 K
p2
 1981 kPa 
where we have used the fact that V2 = V3. Before we calculate the work and heat exchanged for each cycle, we need
to know the number of moles. This may be calculated by applying the ideal gas law at any point on the cycle:
n
=
p1V1 (1981 kPa)(100 × 10−6 m3 )
=
= 0.238 mol
RT1
(8.31 J/mol K)(3600 K)
Now we can calculate Ws , Q, and ∆Eth for the three processes involved in the cycle. For process 1 → 2,
( )
Q1→2 =
nCP (T2 − T1 ) =
n ( 52 R ) (T2 − T1 ) =.
2 476 kJ
∆Eth,1→2 =
nCV (T2 − T1 ) =
n 32 R (T2 − T1 ) =
1.486 kJ
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Heat Engines and Refrigerators
19-19
The work done Ws 1→2 is the area under the p-versus-V graph. We have
Ws 1→2 =
(1981 kPa)(600 × 10−6 m3 − 100 × 10−6 m3 ) =
0.991 kJ
For process 2 → 3, Ws 2→3 = 0 J and
( )
Q2→3 =
nCV (T3 − T2 ) =
n 32 R (T3 − T2 ) =
∆Eth, 2→3 =
−1.693 kJ
For process 3 → 1, Q3→1 = 0 J and
( )
∆Eth, 3→1 =
nCV (T1 − T3 ) =
n 32 R (T1 − T3 ) =.
0 2072 kJ
Because ∆Eth =W + Q and W = −Ws , Ws
3→1
= −∆Eth,3→1 = −0.2072 kJ for process 3 → 1.
Ws (kJ)
1→ 2
2→3
3 →1
Net
∆Eth (kJ)
Q (kJ)
2.476
−1.693
0
0.783
0.991
0
−0.207
0.783
1.486
−1.693
0.207
0
(b) The thermal efficiency of the engine is
Wout
783 J
=
=0.32 =32%
QH 2476 J
η=
Assess: Note that more than two significant figures are retained in part (a) because the results are intermediate. As
expected for a closed cycle, (Ws ) net = Qnet and (∆Eth ) net =
0 J.
19.56. Model: For the closed cycle of the heat engine, process 1 → 2 is isochoric, process 2 → 3 is adiabatic, and
process 3 → 1 is isothermal. For a diatomic gas CV = 52 R and γ = 75 .
Solve: (a) From the graph V2 = 1000 cm3.
The pressure p2 lies on the adiabat from 2 → 3. We can find the pressure as follows:
γ
γ
p2V2 =p3V3
γ
⇒
 4000 cm3 
V 
p2 =p3  3  =(1.00 × 105 Pa) 
 1000 cm3 
 V2 


7/5
=6.964 × 105 Pa ≈ 696 kPa
The temperature T2 can be obtained from the ideal-gas equation relating points 1 and 2:
p1V1 p2V2
=
T1
T2
⇒ T2 = T1
 6.964 × 105 Pa 
p2 V2
= (300 K) 
(1) = 522.3 K ≈ 522 K
 4.00 × 105 Pa 
p1 V1


(b) The number of moles of the gas is
n=
p1V1 (4.00 × 105 Pa)(1.00 × 10−3 m3 )
=
= 0.1604 mol
RT1
(8.31 J/mol K)(300 K)
For isochoric process 1 → 2, Ws =
0 J and
Q =∆Eth =nCV ∆T =n
( 52 R ) ∆T =741.1 J
For adiabatic process 2 → 3, Q =
0 J and
∆Eth =nCV ∆T =n
( 52 R ) (T3 − T2 ) =−741.1 J
Using the first law of thermodynamics, ∆Eth = Ws + Q, which means Ws = −∆Eth = +741.1 J. Ws can also be determined from
W
=
s
p3V3 − p2V2 nR (T3 − T2 )
=
=
1− γ
1− γ
( 43 J/K ) (300 K − 522.3 K)
=
( − 52 )
741.1 J
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19-20
Chapter 19
For isothermal process 3 → 1, ∆Eth =
0 J and
Ws =
nRT1 ln
V1
=
−554.5 J
V3
Using the first law of thermodynamics, ∆Eth =
−Ws + Q, Q =
Ws =
−554.5 J.
∆Eth (J)
1→ 2
2→3
3 →1
Net
Ws (J)
741.1
−741.1
0
0
0
741.1
−554.5
186.6
Q (J)
741.1
0
−554.5
186.6
(c) The work per cycle is 187 J and the thermal efficiency is
W 186.6 J
=0.25 =25%
η= s =
QH 741.1 J
19.57. Model: For this heat engine, process 1 → 2 is adiabatic, process 2 → 3 is isothermal, and process 3 → 1 is
isobaric. For a diatomic gas, CV = 52 R, CP = 72 R, and γ = 1.40.
Solve: (a) From the graph, p1 = 100 kPa. Point 1 is connected by an adiabatic process to point 2, where p2 = 400 kPa
and V2 = 1000 cm3. So p1V1γ = p2V2γ and thus
1/γ
1/1.40
 p2 
 400 kPa 
V1 V2=
=
=
(1000 cm3 ) 
2692 cm3 ≈ 2690 cm3
 

 100 kPa 
 p1 
Point 1 is connected by an isochoric process to point 3, where V3 = 4000 cm3 and T3 = 400 K. Thus
 2692 cm3 
 V1 
=
=
269 K
T1 T=
 (400 K) 
2
3

 V2 
 4000 cm 
Altogether, p1 = 100 kPa, V1 = 2690 cm3 , and
=
T1 269 K.
(b) For the adiabatic process 1 → 2, Q = 0 and Ws = −∆Eth . For any process, ∆Eth = nCV ∆T . The number of moles
can be found from point 2:
n=
pV (4.0 × 105 Pa)(0.0010 m3 )
=
= 0.120 mol
RT
(8.31 J/mol K)(400 K)
Thus
∆Eth =
(0.120 mol)
327 J =
− Ws
( 52 R ) (400 K − 269 K) =
For the isothermal process 2 → 3, ∆Eth =
0 and Q = Ws . The work done is
 V3 
=
=
Ws nRT2 ln =
J Q
 553
 V2 
For the isobaric process 3 → 1,
Ws =p∆V =(1.0 × 105 Pa)(0.00269 m3 − 0.00400 m3 ) =−131 J
The heat exhausted to the cold reservoir is
Q =nCP ∆T =(0.120 mol)
( 72 R ) (269 K − 400 K) =−458 J
Finally, from the first law, ∆Eth =−
Q Ws =
−327 J.
1→2
2→3
3→1
Net
∆Eth
Ws
327
0
−327
0
−327
553
−131
95
Q
0
553
−458
95
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Heat Engines and Refrigerators
19-21
(c) The work per cycle is Wout = 95 J. Heat is input only during process 2 → 3, so QH = 553 J and the engine’s
thermal efficiency is
W
95 J
η = out =
=0.17 =17,
QH 553 J
Assess: As expected for a closed cycle, (∆Eth ) net =
0 and (Ws ) net = Qnet .
19.58. Model: For the closed cycle of the refrigerator, process 1 → 2 is isochoric, process 2 → 3 is adiabatic, process 3 → 4 is isochoric, and process 4 → 1 is adiabatic. For a monatomic gas CV = 32 R and γ = 53 .
Visualize: Please refer to the figure below.
Solve: (a) The number of moles of gas may be found by applying the ideal-gas equation to point 2. The result is
n
=
p2V2 (150 kPa )(1.00 × 10−4 m3 )
=
= 7.22 × 10−3 mol
RT2
(8.31 J/mol K)(250 K)
The temperatures at points 3 and 4 may be found using Table 19.1:
γ −1
2/3
γ −1
2/3
 100 cm3 
 V2 
T3 T=
(250 K)  =
461 K
=

2
 40 cm3 
 V3 


 100 cm3 
 V1 
T4 T=
(200 K)  =
368 K
=

1
 40 cm3 
 V4 


Only the adiabatic segments do work, so the total work done by the system is
WS = WS,2→3 + WS,4→1 = − nCv (∆T2→3 + ∆T4→1 ) = − n
=
−(7.22 × 10−3 mol)
( 23 R ) (T3 − T2 + T1 − T4 )
−3.97 J
( 32 ) (8.31 J/molK)(461 K − 250 K + 200 K − 369 K) =
Thus, the work done on the system is Win =
−WS =
−3.79 J. During the adiabatic segments, no heat is exchanged with
the heat reservoirs, so heat is exchanged only during the isochoric segments. For a refrigerator, the heat exchanged
with the cold reservoir is the heat that is put into the system (i.e., > 0), which occurs in segment 1 → 2. With the help
of Table 19.1, this is
QC = nCV ∆T1→ 2 = n
( 32 R ) (T2 − T1) =
(7.22 × 10−3 mol)
( 32 ) (8.31 J/molK)(250 K − 200 K) =
4.50 J
Thus, the coefficient of performance is
=
K
QC 4.50 J
=
= 1.19
Win 3.71 J
(b) The power needed to run the refrigerator is
cycles 
−1 
=
=
P (W
in cycle )  60
 227 W
s 

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19-22
Chapter 19
19.59. Model: Process 1 → 2 of the cycle is isochoric, process 2 → 3 is isothermal, and process 3 → 1 is isobaric.
For a monatomic gas, CV = 32 R and C
=
P
5
2
R.
Visualize: Please refer to Figure P19.59.
Solve: (a) At point 1, the pressure=
1000 × 10−6 m3 =
1 × 10−3 m3.
p1 1 =
atm 1.013 × 105 Pa and the volume V1 =
The number of moles is
n=
0.120 g
= 0.03 mol
4 g/mol
Using the ideal-gas law,
T
=
1
p1V1 (1.013 × 105 Pa)(1.0 × 10−3 m3 )
=
= 406 K ≈ 0.4 kK
nR
(0.030 mol)(8.31 J/mol K)
At point 2, the pressure p2 =5 atm =5.06 × 105 Pa and V2 = 1 × 10−3 m3. The temperature is
T2
=
p2V2 (5.06 × 105 Pa)(1.0 × 10−3 m3 )
=
= 2030 K ≈ 2 kK
nR
(0.030 mol)(8.31 J/mol K)
At point 3, the pressure is p3 =
1 atm =
1.013 × 105 Pa and the temperature is T=
3 T=
2 2030 K. The volume is
p
 5 atm 
(1 × 10−3 m3 ) 
5 × 10−3 m3
V3 =
V2 2 =
=
1
atm
p3


(b) For the isochoric process 1 → 2, W1→2 = 0 J and
Q1→2= nCV ∆T= (0.030 mol)
( 32 R ) (2030 K − 406 K)=
607 J
For the isothermal process 2 → 3, ∆Eth 2→3 =
0 J and
 5.0 × 10−3 m3 
V
Q2→3 =
W2→3 =
nRT2 ln 3 =
(0.030 mol)(8.31 J/mol K)(2030 K)ln 
=
815 J
 1.0 × 10−3 m3 
V2


For the isobaric process 3 → 1,
W3→1 =p3∆V =(1.013 × 105 Pa)(1.0 × 10−3 m3 − 5.0 × 10−3 m3 ) =−405 J
Q3→1 =nCP ∆T =(0.030 mol)
( 52 ) (8.31 J/mol K)(406 K − 2030 K) =−1012 J
The total work done is Wnet = W1→2 + W2→3 + W3→1 = 410 J. The total heat input is QH = Q1→2 + Q2→3 =1422 J. The
thermal efficiency of the engine is
=
η
Wnet 410 J
=
= 29,
QH 1422 J
(c) The maximum possible efficiency of a heat engine that operates between Tmax and Tmin is
406 K
Tmin
1−
80,
=
=
2030 K
Tmax
1−
ηmax =
Assess: The actual efficiency of an engine is less than the maximum possible efficiency.
19.60. Model: The process 2 → 3 of the heat engine cycle is isochoric and the process 3 → 1 is isobaric. For a
monatomic gas CV = 32 R and CP = 52 R.
Solve: (a) The three temperatures are
=
T1
p1V1 (4.0 × 105 Pa)(0.025 m3 )
=
= 601.7 K ≈ 0.60 kK
nR
(2.0 mol)(8.31 J/mol K)
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Heat Engines and Refrigerators
=
T2
p2V2 (6.0 × 105 Pa)(0.050 m3 )
=
= 1805.1 K ≈ 1.8 kK
nR
(2.0 mol)(8.31 J/mol K)
=
T3
p3V3 (4.0 × 105 Pa)(0.050 m3 )
=
= 1203.4 K ≈ 1.2 kK
nR
(2.0 mol)(8.31 J/mol K)
19-23
(b) For process 1 → 2, the work done is the area under the p-versus-V graph. The work and the change in internal
energy are
1 (6.0 × 105 Pa − 4.0 × 105 Pa)(0.050 m3 − 0.025 m3 ) + (4.0 × 105 Pa)(0.050 m3 − 0.025 m3 )
Ws =
2
= 1.25 × 104 J
∆Eth= nCV ∆T= (2.0 mol)
=
(2.0 mol)
( 32 R ) (T2 − T1)
3.00 104 J
( 32 ) (8.31 J/mol K)(1805.1 K − 601.7 K) =×
The heat input is Q= Ws + ∆Eth= 4.25 × 104 J. For isochoric process 2 → 3, Ws = 0 J and
Q =∆Eth =nCV ∆T =(2.0 mol) 32 (8.31 J/mol K)(1203.4 K − 1805.1 K) =−1.50 × 104 J
For isobaric process 3 → 1, the work done is the area under the p-versus-V curve. Hence,
Ws =
(4.0 × 105 Pa)(0.025 m3 − 0.050 m3 ) =
−1.0 × 104 J
∆Eth =nCV ∆T =n
( 32 R ) (T1 − T3 ) =(2.0 mol) 32 (8.31 J/mol K)(601.7 K − 1203.4 K) =−1.5 ×104 J
The heat input is Q = WS + ∆Eth = −2.50 × 104 J.
∆ Eth (J)
1→2
3.0 × 10
2→3
−1.5× 104
3→1
−1.5× 104
0
Net
WS (J)
1.25 × 10
0
4
Q (J)
4.25 × 104
4
−1.50 × 104
−1.0 × 104
−2.50 × 104
2.5 × 103
2.5 × 103
(c) The thermal efficiency is
=
η
Wnet
2.5 × 103 J
=
= 5.9,
QH 4.25 × 104 J
19.61. Model: The closed cycle in this heat engine includes adiabatic process 1 → 2, isobaric process 2 → 3, and
isochoric process 3 → 1. For a diatomic gas, CV=
5
R, CP=
2
7
R, and γ=
2
7
=
5
1.4.
Visualize: Please refer to Figure P19.61.
Solve: (a) We can find the temperature T2 from the ideal-gas equation as follows:
T2
=
p2V2 (4.0 × 105 Pa)(1.0 × 10−3 m3 )
=
= 2407 K ≈ 2.4 kK
nR
(0.020 mol)(8.31 J/mol K)
We can use the equation p2V2γ = p1V1γ to find V1,
1/γ
1/1.4
 4.0 × 105 Pa 
p 
V1 =
V2  2  =
(1.0 × 10−3 m3 ) 
 1.0 × 105 Pa 
 p1 


2.692 × 10−3 m3
=
The ideal-gas equation can now be used to find T1,
T1
=
p1V1 (1.0 × 105 Pa)(2.692 × 10−3 m3 )
=
= 1620 K ≈ 1.6 kK
nR
(0.020 mol)(8.31 J/mol K)
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19-24
Chapter 19
At point 3, V3 = V1 so we have
T3
=
p3V3 (4 × 105 Pa)(2.692 × 10−3 m3 )
=
= 6479 K ≈ 6.5 kK
nR
(0.020 mol)(8.31 J/mol K)
(b) For adiabatic process 1 → 2, Q = 0 J, ∆Eth =
−Ws , and
p2V2 − p1V1 nR (T2 − T1 ) (0.020 mol)(8.31 J/mol K)(2407 K − 1620 K)
=
=
= −327.0 J
1− γ
1− γ
(1 − 1.4)
For isobaric process 2 → 3,
Q= nCP ∆T= n ( 72 R ) (∆T )= (0.020 mol) 72 (8.31 J/mol K)(6479 K − 2407 K)= 2369 J
Ws =
∆Eth= nCV ∆T= n
( 52 R ) ∆T=
1692 J
The work done is the area under the p-versus-V graph. Hence,
Ws = (4.0 × 105 Pa)(2.692 × 10−3 m3 − 1.0 × 10−3 m3 ) = 677 J
For isochoric process 3 → 1, Ws = 0 J and
∆Eth =Q =nCV ∆T =(0.020 mol)
∆ Eth (J)
1→2
2→3
3→1
Net
327
1692
−2019
0
( 52 ) (8.31 J/mol K)(1620 K − 6479 K) =−2019 J
WS (J)
−327
677
0
350
Q (J)
0
2369
−2019
350
(c) The engine’s thermal efficiency is
=
η
Wnet
350 J
=
= 0.15
= 15,
QH 2369 J
19.62. Model: For the closed cycle of the heat engine, process 1 → 2 is isothermal, process 2 → 3 is isobaric, and
process 3 → 1 is isochoric. For a diatomic gas CV = 52 R and γ = 75 .
Visualize: Please refer to the figure below.
Solve: (a) Begin by expressing the pressure, volume, and temperature in terms of the pressure, volume, and temperature at
point 1. In the isothermal expansion 1 → 2, the volume is halved so the pressure must double (ideal gas equation).
Therefore p2 = 2p1. Because 2 → 3 is isobaric, p=
p=
3
2 2 p1. We are given that V2 = V1/2 and that V3 = V1. Finally,
we know that T2 = T1 because they are on the same isotherm, and the ideal gas equation gives
=
T3
p3V3 2 p1V1
= = 2T1
nR
nR
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Heat Engines and Refrigerators
19-25
The table below summarizes:
p1
V1
T1
p2 = 2p1
V2 = V1/2
T2 = T1
p3 = 2p1
V3 = V1
T3 = 2T1
With the help of Table 19.1, we can find expressions for the work and heat for each segment in the cycle. The results
are given in the table below.
WS
Q
1→2
nRT1 ln(V2 /V1 ) = −nRT1 ln 2
−nRT1 ln 2
2→3
p2 (V1 − V2 ) = (2 p1 )(V1/2) = p1V1 = nRT1
7
nCP (T3 − T2 ) =
nRT1
2
3→1
0
nCV (T1 − T3 ) =
− 52 nRT1
The heat transferred from the hot reservoir into the heat engine (> 0) is done in the isochoric segment 2 → 3. The
total work done by the system is
Wout =
− nRT1 ln 2 + nRT1 =
nRT1 (1 − ln 2)
The thermal efficiency is therefore
Wout nRT1 (1 − ln 2)
=
= 0.088
= 8.8,
η =
7 nRT
QH
1
2
(b) The thermal efficiency of a Carnot engine operating between T1 and T3 is
ηCarnot =−
1
TC
T
=−
1 1 =0.5 =
50,
TH
T3
Assess: The efficiency is much less than the Carnot efficiency.
19.63. Model: The closed cycle of the heat engine involves the following four processes: isothermal expansion,
isochoric cooling, isothermal compression, and isochoric heating. For a monatomic gas CV = 32 R.
Visualize:
Solve: Using the ideal-gas law,
p1
=
nRT1 (0.20 mol)(8.31 J/mol K)(600 K)
=
= 4.986 × 105 Pa
V1
2.0 × 10−3 m3
At point 2, because of the isothermal conditions, T2= T=
1 600 K and
 2.0 × 10−3 m3 
V
p2 =
p1 1 =
(4.986 × 105 Pa) 
2.493 × 105 Pa
=
 4.0 × 10−3 m3 
V2


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19-26
Chapter 19
3
At point 3, because it is an isochoric process, V=
3 V=
2 4000 cm and
 300 K 
T
p3 =
p2 3 =
(2.493 × 105 Pa) 
1.247 × 105 Pa
=
T2
 600 K 
Likewise at point 4, T=
4 T=
3 300 K and
 4.0 × 10−3 m3 
V
p4 =
p3 3 =
(1.247 × 105 Pa) 
2.493 × 105 Pa
=
 2.0 × 10−3 m3 
V4


Let us now calculate Wnet = W1→2 + W2→3 + W3→4 + W4→1. For the isothermal processes,
V2
=
W1→2 nRT
=
(0.20 mol)(8.31 J/mol K)(600=
K)ln(2) 691.2 J
1 ln
V1
W3→4 = nRT3 ln
V4
= (0.20 mol)(8.31 J/mol K)(300 K)ln
V3
( 12 ) = −345.6 J
For the isochoric processes, W=
=
Wnet 345.6 J ≈ 350 J. Because
2 →3 W
4→1 0 J. Thus, the work done per cycle is=
Q
= WS + ∆Eth ,
Q1=
0 J 691.2 J
→ 2 W1→ 2 + ( ∆Eth )1=
→ 2 691.2 J + =
For the first isochoric process,
Q2→3= nCV ∆T= (0.20 mol)
( 32 R ) (T3 − T2 )
=
(0.20 mol) 32 (8.31 J/mol K)(300 K − 600 K) =
−747.9 K
For the second isothermal process
Q3→4 =W3→4 + (∆Eth )3→4 =−345.6 J + 0 J =−345.6 J
For the second isochoric process,
( 32 R ) (T1 − T4 )
(0.20 mol) ( 23 ) (8.31=
J/mol K)(600 K − 300 K)
Q4→1= nCV ∆T= n
=
747.9 K
Thus, QH = Q1→2 + Q4→1 = 1439.1 J. The thermal efficiency of the engine is
=
η
Wnet 345.6 J
=
= 0.24
= 24,
QH 1439.1 J
19.64. Model: Processes 2 → 1 and 4 → 3 are isobaric. Processes 3 → 2 and 1 → 4 are isochoric.
Visualize:
Solve: (a) Except in an adiabatic process, heat must be transferred into the gas to raise its temperature. Thus heat is
transferred in during processes 4 → 3 and 3 → 2. This is the reverse of the heat engine in Example 19.2.
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Heat Engines and Refrigerators
19-27
(b) Heat flows from hot to cold. Since heat energy is transferred into the gas during processes 4 → 3 and 3 → 2,
which end with the gas at temperature 2700 K, the reservoir temperature must be T > 2700 K. This is the hot
reservoir, so the heat transferred is QH . Similarly, heat energy is transferred out of the gas during processes 2 → 1
and 1 → 4. This requires that the reservoir temperature be T < 300 K. This is the cold reservoir, and the energy transferred during these two processes is QC .
(c) The heat energies were calculated in Example 19.2, but now they have the opposite signs.
QH = Q43 + Q32 = 7.09 × 105 J + 15.19 × 105 J = 22.28 × 105 J
QC = Q21 + Q14 = 21.27 × 105 J + 5.06 × 105 J = 26.33 × 105 J
(d) For a counterclockwise cycle in the pV diagram, the work is Win . Its value is the area inside the curve, which is Win =
(∆p )(∆V ) = ( 2 × 101,300 Pa )(2 m3 ) = 4.05 × 105 J. Note that W=
in QC − QΗ , as expected from energy conservation.
(e) No. A refrigerator uses work input to transfer heat energy from the cold reservoir to the hot reservoir. This device
uses work input to transfer heat energy from the hot reservoir to the cold reservoir.
19.65. Solve: (a) If you wish to build a Carnot engine that is 80% efficient and exhausts heat into a cold reservoir at
0°C, what temperature (in °C) must the hot reservoir be?
(b)
(0°C + 273)
273
0.80 =−
1
⇒
=0.20 ⇒ TH =
1.1 × 103°C
(TH + 273)
TH + 273
19.66. Solve: (a) A refrigerator with a coefficient of performance of 4.0 exhausts 100 J of heat in each cycle. What
work is required each cycle and how much heat is removed each cycle from the cold reservoir?
(b) We have 4.0
= QC /Win ⇒ Q=
C 4Win . This means
QH =QC + Win = 4Win + Win =5Win
⇒ Win =
QH 100 J
=
= 20 J
5
5
Hence, QC = QH − Win = 100 J − 20 J = 80 J.
19.67. Solve: (a) A heat engine operates at 20% efficiency and produces 20 J of work in each cycle. What is the net
heat extracted from the hot reservoir and the net heat exhausted in each cycle?
(b) We have 0.20 = 1 − QC /QH . Using the first law of thermodynamics,
Wout = QH − QC = 20 J
⇒
QC = QH − 20 J
Substituting into the definition of efficiency,
0.20 = 1 −
QH − 20 J
20 J 20 J
= 1−1+
=
QH
QH
QH
⇒ QH =
20 J
= 100 J
0.20
The heat exhausted is QC =QH − 20 J =100 J − 20 J =80 J.
19.68. Solve: (a)
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19-28
Chapter 19
In this heat engine, 400 kJ of work is done each cycle. What is the maximum pressure?
(b)
1
( pmax − 1.0 × 105 Pa)(2.0 m3 ) = 4.0 × 105 J ⇒ pmax = 5.0 × 105 Pa = 500 kPa
2
19.69. Model: The heat engine follows a closed cycle, starting and ending in the original state.
Visualize: The figure indicates the following seven steps. First, the pin is inserted when the heat engine has the initial conditions. Second, heat is turned on and the pressure increases at constant volume from 1 to 3 atm. Third, the
pin is removed. The flame continues to heat the gas and the volume increases at constant pressure from 50 cm3 to
100 cm3. Fourth, the pin is inserted and some of the weights are removed. Fifth, the container is placed on ice and
the gas cools at constant volume to a pressure of 1 atm. Sixth, with the container still on ice, the pin is removed. The
gas continues to cool at constant pressure to a volume of 50 cm3. Seventh, with no ice or flame, the pin is inserted
back in and the weights returned bringing the engine back to the initial conditions and ready to start over.
Solve: (a)
(b) The work done per cycle is the area inside the curve:
Wout =∆
( p )(∆V ) =(2 × 101,300 Pa)(50 × 10−6 m3 ) =
10 J
(c) Heat energy is input during processes 1 → 2 and 2 → 3, so Q=
H Q12 + Q23 . This is a diatomic gas, with
CV = 32 R and CP = 52 R. The number of moles of gas is
n
=
Process
1
→
2
is
p1V1 (101,300 Pa)(50 × 10−6 m3 )
=
= 0.00208 mol
RT1
(8.31 J/mol K)(293 K)
isochoric,
=
T2 (p2 /p1 )=
T1 3=
T1 879 K.
so
Process
2
→
3
is
isobaric,
so
=
T3 (V3/V2 )=
T2 2=
T2 1758 K. Thus
Q12= nCV ∆T=
5
nR (T2
2
− T1 )=
5
(0.00208 mol)(8.31 J/mol K)(586 K)=
2
25.32 J
Q23= nCP ∆T=
7
nR (T3
2
− T2 )=
7
(0.00208 mol)(8.31 J/mol K)(879 K)=
2
53.18 J
Similarly,
Thus QH = 25.32 J + 53.18 J = 78.50 J and the engine’s efficiency is
=
η
Wout 10.13 J
=
= 0.13
= 13,
QH 78.50 J
19.70. Model: System 1 undergoes an isochoric process and system 2 undergoes an isobaric process.
Solve: (a) Heat will flow from system 1 to system 2 because system 1 is hotter. Because there is no heat input from
(or loss to) the outside world, we have Q1 + Q2 =
0 J. Heat Q1, which is negative, will change the temperature of
system 1. Heat Q2 will both change the temperature of system 2 and do work by lifting the piston. But these consequences of heat flow don’t change the fact that Q1 + Q2 =
0 J. System 1 undergoes constant volume cooling from
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Heat Engines and Refrigerators
19-29
T1i = 600 K to Tf . System 2, whose pressure is controlled by the weight of the piston, undergoes constant pressure
heating from T2i = 300 K to Tf . Thus,
Q1 + Q2 = 0 J = n1CV (Tf − T1i ) + n2CP (Tf − T2i ) = n1
( 32 R ) (Tf − T1i ) + n2 ( 52 R ) (Tf − T2i )
Solving this equation for Tf gives
=
Tf
3n1T1i + 5n2T2i 3(0.060 mol)(600 K) + 5(0.030 mol)(300 K)
=
= 464 K
3n1 + 5n2
3(0.060 mol) + 5(0.030 mol)
(b) Knowing Tf , we can compute the heat transferred from system 1 to system 2:
Q
=
=
) n2
2 n2CP (Tf − T2i
(c) The change of thermal energy in system 2 is
∆Eth= n2CV ∆T= n2
( 52 R ) (Tf − T2i=)
( 32 R ) (Tf − T2i=)
3
Q=
5 2
102 J
61.2 J
According to the first law of thermodynamics, Q=
2 WS + ∆Eth . Thus, the work done by system 2 is
WS= Q − ∆Eth= 102.0 J − 61.2 J= 40.8 J. The work is done to lift the weight of the cylinder and the air above it by a
height ∆y. The weight of the air is wair =
pA =
pp r 2 =
(101.3 × 10 N/m 2 )p (0.050 m) 2 =
795.6 N. Therefore,
W
=
=
y
s ( wcyl + wair ) ∆y ⇒ ∆
Ws
40.8 J
=
= 0.050 m
( wcyl + wair ) (2.0 kg)(9.8 m/s 2 ) + 795.6 N
(d) The fraction of heat converted to work is
Ws 40.8 J
=
= 0.40
= 40,
Q2 102.0 J
19.71. Model: Process 1 → 2 and process 3 → 4 are adiabatic, and process 2 → 3 and process 4 → 1 are isochoric.
Visualize: Please refer to Figure CP19.71.
Solve: (a) For adiabatic process 1 → 2, Q12 = 0 J and
=
W12
p2V2 − p1V1 nR (T2 − T1 )
=
1− γ
1− γ
For isochoric process 2 → 3,
=
W23 0 J and
=
Q23 nCV (T3 − T2 ). For adiabatic process 3 → 4, Q34 = 0 J and
=
W34
p4V4 − p3V3 nR(T4 − T3 )
=
1− γ
1− γ
For isochoric process 4 → 1,
=
W41 0 J and
=
Q41 nCV (T1 − T4 ). The work done per cycle is
Wnet = W12 + W23 + W34 + W41 =
nR (T2 − T1 )
nR (T4 − T3 )
nR
+0J+
+0J =
(T2 − T1 + T4 − T3 )
1− γ
1− γ
1− γ
(b) The thermal efficiency of the heat engine is
W
QH
1−
η =out =
QC
nC (T − T )
|Q |
=
1 − 41 =
1− V 4 1
QH
|Q23|
nCV (T3 − T2 )
The last step follows from the fact that T3 > T2 and T4 > T1. We will now simplify this expression further as follows:
γ
γ −1
pV= pVV = nRTV
γ −1
γ −1
γ −1
⇒ nRT1V1 = nRT2V2
γ −1
 V1 
γ −1
⇒ T=
 = T1r
2 T1 
 V2 
Similarly, T3 = T4r γ −1. The equation for thermal efficiency now becomes
T −T
4
1
1−
η=
γ −1
T4r
− T1r γ −1
1
=
1 − γ −1
r
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19-30
Chapter 19
(c)
19.72. Model: For the Diesel cycle, process 1 → 2 is an adiabatic compression, process 2 → 3 is an isobaric expansion, process 3 → 4 is an adiabatic expansion, and process 4 → 1 is isochoric.
Visualize: Please refer to CP19.72.
Solve: (a) It will be useful to do some calculations using the compression ratio, which is
=
r
Vmax V1 1050 cm3
= =
= 21
Vmin V2
50 cm3
The number of moles of gas is
p1V1 (1.013 × 105 Pa)(1050 × 10−6 m3 )
=
= 0.0430 mol
RT1
(8.31 J/mol K)[(25 + 273) K]
n
=
For an adiabatic process,
p1V1γ =p2V2γ
γ
⇒
V 
p2 = 1  p1 =r γ p1 =211.40 × 1 atm =71.0 atm =7.19 × 106 Pa
 V2 
γ −1
V 
⇒ T2 =  1  T1 = r γ −1T1 = 210.41 × 298 K = 1007 K
T1V1 = T2V2
 V2 
Process 2 → 3 is an isobaric heating with Q = 1000 J. Constant pressure heating obeys
Q
Q= nCP ∆T ⇒ ∆T=
nCP
γ −1
γ −1
The gas has a specific heat ratio=
γ 1.40
= 7/5, thus CV = 52 R and CP = 72 R. Knowing Cp , we can calculate first
∆T = 800 K and then T3= T2 + ∆T= 1807 K. Finally, for an isobaric process we have
V2 V3
=
T2 T3
⇒ V3 =
T3
1807 K
V2 =
(50 × 10−6 m3 ) = 89.7 × 10−6 m3
T2
1007 K
Process 3 → 4 is an adiabatic expansion to V4 = V1. Thus,
p3V3γ = p4V4γ
1.4
γ
⇒
 89.7 × 10−6 m3 
V 
p4 = 3  p3 =
 1050 × 10−6 m3 
 V4 


(7.19 × 106 Pa) =2.30 × 105 Pa =2.27 atm
γ −1
 89.7 × 10−6 m3 
T3 
=
 1050 × 10−6 m3 


V 
−1
T3V3γ=
T4V4γ −1 ⇒ =
T4  3 
 V4 
Point
P
1
1.00=
atm 1.013 ×10 Pa
2
71.0 atm
= 7.19 ×10 Pa
50.0 × 10
3
4
0.4
(1807 K)
= 675 K
V
3
T
298 K= 25°C
m3
1007 =
K 734°C
71.0 atm
= 7.19 ×106 Pa
89.7 × 10−6 m3
1807=
K 1534°C
2.27 atm
= 2.30 ×105 Pa
1050 × 10−6 m3
675=
K 402°C
5
6
1050 × 10
−6
−6
m
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Heat Engines and Refrigerators
19-31
(b) For adiabatic process 1 → 2,
(Ws )12 =
p2V2 − p1V1
= −633 J
1− γ
For isobaric process 2 → 3,
(Ws ) 23 = p2∆V = p2 (V3 − V2 ) = 285 J
For adiabatic process 3 → 4,
=
(Ws )34
p4V4 − p3V3
= 1009 J
1− γ
For isochoric process 4 → 1, (Ws ) 41 = 0 J. Thus,
(Ws )cycle =
Wout =
(Ws )12 + (Ws ) 23 + (Ws )34 + (Ws ) 41 =
661 J
(c) The efficiency is
=
η
Wout
661 J
=
= 66.1% ≈ 66%
QH 1000 J
(d) The power output of one cylinder is
 661 J   2400 cycle   1 min 
J
=
= 26.4 kW
 26,440



cycle
min
60
sec
s






For an 8-cylinder engine the power will be 211 kW or 283 horsepower.
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