Chapter 22
Questions: 2, 4, 10, 16, 18
Problems: 1, 5, 7, 8, 11, 15, 23, 25, 39, 45, 61
Questions
2. Only temporarily. The heat dissipated by the coils in the back of the
refrigator results in a net positive heat transfer to the kitchen.
4. The energy required to cool the cooling water would require more
energy than you would gain from an increased e¢ ciency.
10. The COP is given by
COP =
Qc
Tc
Qc
=
=
:
W
Qh Qc
Th Tc
So it is clear the smaller the temperature di¤erence between Th and Tc the
higher the COP. Basically it requires much less work for a small temperature
change versus a large temperature change.
16. Material that could withstand higher temperatures would increase
the maximum possible e¢ ciency of any thermal cycle which is given by
e=1
Tc
;
Th
thus saving energy.
18. It is the quality of energy that enables us to do work with the energy.
No process is perfectly reversible, hence eventually the quality of the energy
makes it useless as an energy source.
Problems
1. (a) The number of ways to arrange 6 distinguishable eggs in a carton
for 12 eggs is
N = 12
11
10
9
8
7 = 12!=6! = 924:
(b) The number of ways to arrange 6 distinguishable eggs in 6 speci…c
locations is
M = 6 5 4 3 2 1 = 6! = 72:
5. For reversible cycles the e¤eciency is e = 1 Tc =Th : Thus (a) e =
1 273=373 = :268: (b) e = 1 277=298 = :070: (c) e = 1 300=1273 = :764:
7. For a Carnot engine e = 1 Tc =Th so that
Tc
Tc
= :777 = 1
;
Th
4:25
= 4:25 :223 = :948K
e = 1
Tc
1
8. (a) From the …rst law, we know for a cycle that
W = Qh
Qc = (890
470) J = 420J.
(b) The e¢ ciency is
e=
420
W
=
= :472:
Qh
890
(c) The temperature of the cool reservoir is found from the e¤eciency of
a Carnot engine,
Tc
Tc
=1
;
Th
550
:528 = 290K = 17C.
e = :472 = 1
Tc = 550
(d) The power produced is
P =
420J
W
=
= 9:24kW.
t
1=22cyc/s
11. (a) The rate of heat extraction is
Qc
m
=
C T = 2:8 104 kg/s (4184J/kgK) 8:5K = 996MW
t
t
(b) Since the rate of work is
W
Qh Qc
Qh
=
= 750MW =
t
t
t
the rate of energy extraction from the fuel is
996MW,
Qh
= (750 + 996) MW = 1746MW.
t
(c) The e¤eciency is
e=
750
W= t
=
= :430:
Qh = t
1746
Thus the minimum highest temperature can be found from the e¤eciency of
a Carnot cycle
288
= :430;
Th
288
=
= 505K = 232C.
1 e
e = 1
TH
2
15. During a cycle W = Qh Qc and the e¤eciency is e = W=Qh : Given
that W = 350J and Qh = 900J the e¤eciency is given by (a)
e = 350=900 = :39:
(b) The rejected heat Qc is
Qc = Qh
W = 550J.
(c) The maximum temperature is found from the e¤eciency of a Carnot
cycle,
Tc
;
Th
e = 1
Th =
Tc
1
e
=
283
= 464K = 191C
:61
23. (a) The de…nition for COP is
COP =
Qc
Tc
Qc
=
=
;
W
Qh Qc CC Th Tc
where CC denotes a Carnot cycle. Assuming a Carnot cycle for the heat
pump then in the summer the COP is
COP =
278
= 13:2:
21
COP =
275
= 3:5:
78
In the winter the COP is
(b) The work is given by W = Qh Qc and the ratio of work to heat
removed from the cooler temperature bath is W=Qc so that
W =
1
W
Qc =
Qc :
Qc
COP
So the work required for each Joule of heat removed from the house in the
summer is
W = :075J.
3
(c) In the winter the work required for each Joule of heat removed from
the environment is
W = :286J.
25. (a) Consider the cycle in …gure 22-29. During the isothermal expansion the heat transfer to the gas is
Qh = nRT ln
VF
= P V ln 2:
Vi
For the conditions in …gure 22-29 we have P V = 8atm 1L= 808kPaL=
808J. Thus the heat transfer to the gas is
Qh = 808J ln 2 = 560J.
(b) The heat rejected is given by
Qc =
nRT ln
VF
= P V ln 2:
Vi
From the conditions in …gure 22-29 we have P V = 4:1atm 1:612L= 4:1
101kPa 1:612L= 667:5J. The heat transfer to the cold bath is
Qc = 667:5J ln 2 = 462:7J.
(c) The work done is
W = Qh
Qc = (560
462:7) J = 97:3J.
(d) The e¤eciency is
e=
W
97:3
=
= :174
Qc
560
(e) From the ideal gas law the high temperature is found from
P V = nRTh = 808kPaL = :2 8:314
Th = 808= (:2 8:314) = 486K.
Th ;
The low temperature is found from
P V = nRTc = 667:5kPaL = :2 8:314
Tc = 667:5=(:2 8:314) = 401:4K.
4
Tc ;
The Carnot e¤eciency is
401:4
= :174;
486
Tc
=1
Th
e=1
which agrees with part (d).
39. Given 5 moles of an ideal diatomic gas with CV = 5R=2 which is
initially at 1atm and 300K,
(a) what is the change in entropy if the gas is heated to 500K at constant
volume? The change in entropy is given by the integral
Z
Z 500
nCV dT
5
500
25
5
dQ
=
=5
R ln
= 8:314 ln = 53J=K
S=
T
T
2
300
2
3
300
(b) Under constant pressure CP = CV + R = 7R=2: Thus the change in
entropy is
7
S = 53J=K = 74:3J=K.
5
(c) Since Q = 0 for an adiabatic process S = 0:
45. The work that could have been done from the adabatic expansion is
found from the …rst law,
W =
Z
P dV = Pi Vi
Z
Vf
Vi
W = Pi Vi
Vi
+1
Vf
dV
= Pi Vi
V
+1
=(
1) =
V
+1
Vf
+1
Vi
Pi Vi
;
Pi Vi Vf
1
+1
:
Solving for Pi yields
Pi =
(
Vi
1) W
(
1) W=Vi
:
+1 =
Vi Vf
1 (Vi =Vf ) 1
For this expansion W = 6:5J, Vi = 10 6 m3 , and Vf = 1m3 : Substituting in
these values for = 1:4; we …nd
Pi =
:4
6:5 106
= 2:61
1 10 2:4
106 Pa ' 2610kPa ' 26atm
61. (a) There is no work done during the processes with constant volume.
From …gure 22-33, during the adiabatic compression from state 1 to state 2,
5
the work done on the gas is
W1
W1
2
2
=
Z
=
P dV = P1 V1
"
P1 V1
V1
1
5
Z
V1 =5
V1
+1
P1 V1
dV
V
=
V
+1
#
V1
+1
=
P1 V1
+1 V1 =5
V1
(5
1
;
1)
1
:
From …gure 22-33, during the adabatic expansion from state 3 to state 4 the
work done by the gas is
Z
Z V1
3P2 (V1 =5)
dV
V
=
V +1 V11 =5 :
W3 4 = P dV = P3 V3
+1
V1 =5 V
Since the pressure in state 2 was obtained via adiabatic expansion from state
1, we have
P2
V1
5
= P1 V1 ;
P2 = 5 P1 :
Substituting this for P2 into the relation for the work done from state 3 to
state 4 yields
W3
4
W3
4
3P1 V1
V +1 V1
+1 1
P1 V1
5 1 1 :
= 3
1
=
+1
=5
+1
=
3P1 V1
1
+1
5
1
;
The total work done during the cycle is
W = W3
4
+ W1
2
=2
P1 V1
5
1
1
1 :
From …gure 22-33, the total heat added during the combustion phase is
Qh = nCV
Qh = 2
T = nCV
5
1 CV
R
PV
2 CV
2 CV
=
P2 V1 = 5
P1 V1 ;
nR
5 R
5 R
P1 V1 :
6
Now CV = R= (
1) : Thus the …nal expression for the heat added during
the compression cycle is
Qh =
2
1
5
P1 V1
1
The e¢ ciency is
eOtto =
(5
W
=2
Qh
1
1)
1
1
2
5
1
=
1
5
5
1
1
=1
51 :
Note that everything cancels except for the compression factor (5) and .
Clearly as the compression increases the e¢ ciency increases.
(b) The maximum temperature occurs just after the combustion, state
3, while the lowest temperture just after the exhaust, state 1. From the
ideal gas law the maximum temperature, Tmax ; in terms of the minimum
temperature, Tmin ; is
P3 V3
3P2 V1 =5
3 P2
3
Tmax
=
=
=
= 5 ;
Tmin
P1 V1
P1 V1
5 P1
5
Tmax
= 3 5 1:
Tmin
(c) The Carnot e¤eciency is
eCarnot = 1
51
3
Tmin
=1
Tmax
=
5
1
5
1=3
1
The Carnot e¤eciency is greater than the Otto e¢ ciency by a factor of
eCarnot
5
=
eOtto
5
1
7
1
1=3
> 1:
1