Physics 41 Chapter 22 HW Serway 7th Edition
Conceptual Questions: 1, 3, 8, 12 Problems: 9, 13, 20, 23, 27, 39, 48, 54, 55
Conceptual Questions: 1, 3, 8, 12
Q22.1
First, the efficiency of the automobile engine cannot exceed the Carnot efficiency: it is limited by the
temperature of burning fuel and the temperature of the environment into which the exhaust is dumped. Second,
the engine block cannot be allowed to go over a certain temperature. Third, any practical engine has friction,
incomplete burning of fuel, and limits set by timing and energy transfer by heat.
Q22.3
A higher steam temperature means that more energy can be extracted from the steam. For a constant temperature
T −T
T
heat sink at Tc , and steam at Th , the efficiency of the power plant goes as h c = 1 − c and is maximized for a high
Th
Th
Th .
Q22.8
Q22.12
(a)
When the two sides of the semiconductor are at different temperatures, an electric potential
(voltage) is generated across the material, which can drive electric current through an
external circuit. The two cups at 50°C contain the same amount of internal energy as the pair
of hot and cold cups. But no energy flows by heat through the converter bridging between
them and no voltage is generated across the semiconductors.
(b)
A heat engine must put out exhaust energy by heat. The cold cup provides a sink to absorb
output or wasted energy by heat, which has nowhere to go between two cups of equally
warm water.
For an expanding ideal gas at constant temperature, the internal energy stays constant. The
gas must absorb by heat the same amount of energy that it puts out by work. Then its
⎛V ⎞
ΔQ
= nR ln ⎜ 2 ⎟
entropy change is ΔS =
T
⎝ V1 ⎠
For a reversible adiabatic expansion ΔQ = 0 , and ΔS = 0 . An ideal gas undergoing an
irreversible adiabatic expansion can have any positive value for ΔS up to the value given in
part (a).
(a)
(b)
Problems: 9, 13, 20, 23, 24, 27, 39, 48, 54, 55
P22.9
Tc = 703 K
Th = 2 143 K
ΔT 1 440
=
= 67.2%
Th 2 143
(a)
ec =
(b)
Qh = 1.40 × 105 J , Weng = 0.420 Qh
P
=
Weng
Δt
=
5.88 × 10 4 J
= 58.8 kW
1s
13. An ideal gas is taken through a Carnot cycle. The isothermal expansion occurs at 250°C, and the isothermal compression takes place at 50.0°C. The gas takes in 1 200 J of energy from the hot reservoir during the isothermal expansion. Find (a) the energy expelled to the cold reservoir in each cycle and (b) the net work done by the gas in each cycle. P22.13 Isothermal expansion at Th = 523 K
Isothermal compression at Tc = 323 K
Gas absorbs 1 200 J during expansion.
P22.23
(a)
⎛T ⎞
⎛ 323⎞
Q c = Q h ⎜ c ⎟ = 1200 J⎜
= 741 J
⎝ 523⎟⎠
⎝ Th ⎠
(b)
W eng = Q h − Q c = (1200 − 741) J= 459 J
( COP)Carnot refrig =
Q
Tc 4.00
=
= 0.013 8 = c
ΔT 289
W
∴ W = 72.2 J per 1 J energy removed by heat.
Section 22.5
Gasoline and Diesel Engines
P22.27 In a cylinder of an automobile engine, just after combustion, the gas is confined to a volume of 50.0 cm3 and has an initial pressure of 3.00 × 106 Pa. The piston moves outward to a final volume of 300 cm3 and the gas expands without energy loss by heat. (a) If γ = 1.40 for the gas, what is the final pressure? (b) How much work is done by the gas in expanding? γ
γ
(a)
PV
i i = Pf V f
⎛V
Pf = Pi ⎜ i
⎜ Vf
⎝
γ
1.40
⎞
⎛ 50.0 cm 3 ⎞
= 244 kPa
⎟ = ( 3.00 × 106 Pa ) ⎜
⎟
3
⎟
⎝ 300 cm ⎠
⎠
Vi
(b)
∫
W = PdV
Vi
⎛V ⎞
P = Pi ⎜ i ⎟
⎝V ⎠
γ
Integrating,
⎛ 1 ⎞
W =⎜
i i
⎟ PV
⎝ γ − 1⎠
= 192 J
⎡ ⎛V
⎢1 − ⎜ i
⎢ ⎜⎝ V f
⎣
⎞
⎟
⎟
⎠
γ −1
⎤
3 0.400 ⎤
⎡ ⎛
⎞
⎥ = ( 2.50) ( 3.00 × 106 Pa )( 5.00 × 10−5 m 3 ) ⎢1 − ⎜ 50.0 cm3 ⎟
⎥
3
00
cm
⎥
⎢
⎝
⎠
⎥⎦
⎣
⎦
⎛ Vf ⎞
P22.39 ΔS = nR ln ⎜ ⎟ = R ln 2 = 5.76 J K
⎝ Vi ⎠
There is no change in temperature for an ideal gas.
FIG. P22.39
P22.20
(a)
First, consider the adiabatic process D → A :
γ
γ
γ
PD VD = PA VA
so
53
⎛V ⎞
⎛ 10.0 L ⎞
PD = PA ⎜ A ⎟ = 1 400 kPa ⎜
⎟ = 712 kPa
⎝ 15.0 L ⎠
⎝ VD ⎠
Also
⎛ nRTD
⎜
⎝ VD
⎞ γ ⎛ nRTA
⎟ VD = ⎜
⎠
⎝ VA
or
⎛V ⎞
TD = TA ⎜ A ⎟
⎝ VD ⎠
γ −1
⎞ γ
⎟ VA
⎠
⎛ 10.0 ⎞
= 720 K ⎜
⎟
⎝ 15.0 ⎠
23
= 549 K
Now, consider the isothermal process C → D : TC = TD = 549 K
⎛V
PC = PD ⎜ D
⎝ VC
PC =
γ
⎞ ⎡ ⎛ VA ⎞ ⎤ ⎛ VD ⎞ PA VAγ
P
=
⎢
⎟
⎟=
⎟ ⎥⎜
A ⎜
γ −1
⎠ ⎢⎣ ⎝ VD ⎠ ⎥⎦ ⎝ VC ⎠ VCVD
1 400 kPa ( 10.0 L )
24.0 L ( 15.0 L )
53
23
= 445 kPa
Next, consider the adiabatic process B → C : PBVBγ = PCVCγ
But, PC =
⎛V ⎞
PA VAγ
from above. Also considering the isothermal process, PB = PA ⎜ A ⎟
VCVDγ −1
⎝ VB ⎠
⎛V
Hence, PA ⎜ A
⎝ VB
⎞ γ ⎛ PA VAγ ⎞ γ
10.0 L ( 24.0 L )
V V
= 16.0 L
V which reduces to VB = A C =
⎟ VB = ⎜
γ −1 ⎟ C
V
V
V
15.0 L
D
⎠
⎝ C D ⎠
⎛V ⎞
⎛ 10.0 L ⎞
Finally, PB = PA ⎜ A ⎟ = 1 400 kPa ⎜
⎟ = 875 kPa
⎝ 16.0 L ⎠
⎝ VB ⎠
State
A
B
C
D
(b)
P(kPa)
1 400
875
445
712
V(L)
10.0
16.0
24.0
15.0
For the isothermal process A → B :
T(K)
720
720
549
549
ΔEint = nCV ΔT = 0
⎛V ⎞
⎛ 16.0 ⎞
so Q = −W = nRT ln ⎜ B ⎟ = 2.34 mol ( 8.314 J mol ⋅ K ) ( 720 K ) ln ⎜
⎟ = +6.58 kJ
⎝ 10.0 ⎠
⎝ VA ⎠
For the adiabatic process B → C :
Q= 0
⎡3
⎤
ΔEint = nCV ( TC − TB ) = 2.34 mol ⎢ ( 8.314 J mol ⋅ K ) ⎥ ( 549 − 720) K = −4.98 kJ
⎣2
⎦
and W = −Q + ΔEint = 0 + ( −4.98 kJ) = −4.98 kJ
ΔEint = nCV ΔT = 0
For the isothermal process C → D :
⎛V ⎞
⎛ 15.0 ⎞
and Q = −W = nRT ln ⎜ D ⎟ = 2.34 mol ( 8.314 J mol ⋅ K ) ( 549 K ) ln ⎜
⎟ = −5.02 kJ
⎝ 24.0 ⎠
⎝ VC ⎠
Finally, for the adiabatic process D → A :
Q= 0
⎡3
⎤
ΔEint = nCV ( TA − TD ) = 2.34 mol ⎢ ( 8.314 J mol ⋅ K ) ⎥ ( 720 − 549) K = +4.98 kJ
⎣2
⎦
and W = −Q + ΔEint = 0 + 4.98 kJ = +4.98 kJ
Process
Q(kJ)
W(kJ)
A→B
B→C
C→D
D→A
ABCDA
+6.58
0
–5.02
0
+1.56
–6.58
–4.98
+5.02
+4.98
–1.56
ΔEint (kJ)
0
–4.98
0
+4.98
0
The work done by the engine is the negative of the work input. The output work Weng is
given by the work column in the table with all signs reversed.
(c)
e=
Weng
Qh
ec = 1 −
=
−WABCD 1.56 kJ
=
= 0.237 or 23.7%
QA →B
6.58 kJ
Tc
549
= 1−
= 0.237 or 23.7%
720
Th
P22.24 COP = 0.100COPCarnot cycle
or
⎛Q ⎞
⎛
⎞
1
= 0.100 ⎜ h ⎟
= 0.100 ⎜
⎟
W
⎝ W ⎠Carnot cycle
⎝ Carnot efficiency ⎠
Qh
⎛ Th ⎞
293 K
⎛
⎞
= 0.100 ⎜
⎟ = 0.100 ⎜
⎟ = 1.17
W
⎝ 293 K − 268 K ⎠
⎝ Th − Tc ⎠
Qh
FIG. P22.24
Thus, 1.17 joules of energy enter the room by heat for each joule of w ork done.
25. An ideal refrigerator or ideal heat pump is equivalent to a Carnot engine running in reverse. That is, energy Qc is taken in from a cold reservoir and energy Qh is rejected to a hot reservoir. (a) Show that the work that must be supplied to run the refrigerator or heat pump is W
(b) Show that the coefficient of performance of the ideal refrigerator is P22.25
(a)
COP =
=
Tc
Th – Tc
(b)
⎡T − T ⎤
W = Qc ⎢ h c⎥ .
⎣ Tc ⎦
We have the definition of the coefficient of performance for a refrigerator,
Qc
CO P =
.
W
Tc
.
Using the result from part (a), this becomes
CO P =
Th − Tc
ΔScold =
−1 000 J
600 K
+750 J
350 K
(a)
ΔSU = ΔShot + ΔScold = 0.476 J K
(b)
ec = 1 −
T1
= 0.417
T2
Weng = ec Qh = 0.417 ( 1 000 J) = 417 J
(c)
⎡ (Q h )
⎤
W = Qh − Qc = Qc ⎢
− 1⎥ .
⎢ Qc
⎥
⎣
⎦
Q h Th
We have already shown that for a Carnot cycle (and only for a Carnot cycle)
=
.
Tc
Qc
For a complete cycle, ΔEint = 0 and
Therefore,
P22.48 ΔShot =
Th – Tc
Qc Tc
Wnet = 417 J− 250 J = 167 J
T1ΔSU = 350 K ( 0.476 J K ) = 167 J
*P22.54
(a)
For the isothermal process AB, the work on the gas is
⎛V ⎞
WAB = − PA VA ln ⎜ B ⎟
⎝ VA ⎠
⎛ 50.0 ⎞
WAB = −5 ( 1.013 × 105 Pa )( 10.0 × 10−3 m 3 ) ln ⎜
⎟
⎝ 10.0 ⎠
WAB = −8.15 × 103 J
where we have used 1.00 atm = 1.013 × 105 Pa
FIG. P22.54
−3
1.00 L = 1.00 × 10 m
and
3
WBC = − PB ΔV = − ( 1.013 × 105 Pa ) ⎡⎣( 10.0 − 50.0) × 10−3 ⎤⎦ m 3 = +4.05 × 103 J
WCA = 0 and Weng = −WAB − WBC = 4.10 × 103 J = 4.10 kJ
(b)
ΔEint, AB = 0
Since AB is an isothermal process,
and
QAB = −WAB = 8.15 × 103 J
For an ideal monatomic gas,
CV =
3R
5R
and CP =
2
2
TB = TA =
TC =
Also,
5
−3
PBVB ( 1.013 × 10 )( 50.0 × 10 ) 5.06 × 10 3
=
=
nR
R
R
5
−3
PCVC ( 1.013 × 10 )( 10.0 × 10 ) 1.01× 103
=
=
nR
R
R
⎛ 3 ⎞⎛ 5.06×103 −1.01×103 ⎞
QCA = nCV ΔT = 1.00⎜ R ⎟⎜
⎟ = 6.08 kJ
R
⎝ 2 ⎠⎝
⎠
so the total energy absorbed by heat is QAB + QCA = 8.15 kJ+ 6.08 kJ = 14.2 kJ
(c)
QBC = nCP ΔT =
QBC =
5
5
( nRΔT ) = PBΔVBC
2
2
5
(1.013 × 105 ) ⎡⎣(10.0 − 50.0) × 10−3 ⎤⎦ = −1.01× 104 J = −10.1 kJ
2
Weng
(e)
A Carnot engine operating between Thot = TA = 5060/R and Tcold = TC = 1010/R has
efficiency 1 − Tc/Th = 1 − 1/5 = 80.0%.
The three-process engine considered in this problem has much lower efficiency.
QAB + QCA
=
4.10 × 103 J
= 0.288 or 28.9%
1.42 × 104 J
e=
Qh
=
Weng
(d)
*P22.55
At point A,
n = 1.00 mol
PV
i i = nRTi
At point B,
3PV
i i = nRTB so
TB = 3Ti
At point C,
( 3Pi )( 2Vi ) = nRTC
and
Pi ( 2Vi ) = nRTD
so
and
TC = 6Ti
At point D,
TD = 2Ti
FIG. P22.55
The heat for each step in the cycle is
found using CV =
3R
5R
and CP =
:
2
2
QAB = nCV ( 3Ti − Ti ) = 3nRTi
QBC = nCP ( 6Ti − 3Ti ) = 7.50nRTi
QCD = nCV ( 2Ti − 6Ti ) = −6nRTi
QDA = nCP ( Ti − 2Ti ) = −2.50nRTi
Qentering = Qh = QAB + QBC = 10.5nRTi
(a)
Therefore,
(b)
Qleaving = Qc = QCD + QDA = 8.50nRTi
(c)
Actual efficiency,
e=
(d)
Carnot efficiency,
ec = 1 −
Qh − Qc
Qh
= 0.190
Tc
T
= 1 − i = 0.833
6Ti
Th
The Carnot efficiency is much higher.