Chapter 22
Questions: 2, 4, 9, 10, 15, 16, 18, 19
Problems: 1, 2, 8, 11, 13, 15, 17, 23, 25, 39, 45, 54
Questions
2. Only temporarily. The heat dissipated by the coils in the back of the
refrigerator 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.
9. It is not operating in a closed cycle. This enables it to take advantage
of the potential energy of the water.
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.
17. A free expansion is an irreversible process where the gas is NOT
passing through states in quasi-equilibrium. However, the temperature of the
gas remains constant during the process. Even though it is also an adiabatic
process, we can evaluate the change in entropy via a reversible path, i.e. an
isothermal process in which dQ = dW:. Then the integral representing the
change in entropy is:
Z f
Z f
Z f
Z f
dW
dV
dV
Vf
dQ
=
=
P
= nR
= nR ln
> 0:
S=
T
T
T
V
Vi
i
i
i
i
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.
19. Assume that you have a perfect power plant that has an energy
source. Since it is 100% e¢ cient, it performs work equal to the energy extracted from its energy source. It is this work that drives a heat pump. The
1
heat output of the heat pump satis…es the 1st law of thermodynamics for
a cyclic process, Qh Qc = W: For a heatpump, all of these quantities are
negative. Hence we can write jQh j jQc j = jW j : From the 2nd law of thermodynamics jQh j < jW j : So no, the heat output of the heat pump can never
be greater (or even equal to) than the work provided by the power plant.
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! = 665; 280:
(b) The number of ways to arrange 6 distinguishable eggs in 6 speci…c
locations is
M = 6 5 4 3 2 1 = 6! = 720
Hence the fraction is
f=
1
720
=
= 1:1
665; 280
924
10
3
2. (a) The number of states is N = 24 = 16:
(b) The number of states corresponding to all four particles traveling in
the same direction, + or is N = 2.
(c) The probability of …nding all the moving in one direction (+ or ) is
P = 2=16 = 1=8 = 12:5%
(d) For 10 distinguishable molecules these results are
N = 210 = 1024
N = 2
P = 2=1024 = 1=512 = :195%
(e) With an Avogadro’s type of number, most of the molecules will be
indistinguishable from those of the same type. Hence this exercise is not
relevant!
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=
W
420
=
= :472 = 47:2%:
Qh
890
2
(c) The temperature of the cool reservoir is found from the e¢ ciency of
a Carnot engine,
Tc
Tc
=1
;
Th
550
:528 = 290K = 17 C.
e = :472 = 1
Tc = 550
(d) The power produced is
P =
W
420J
=
= 9:24kW:
t
1=22cyc=s
11. (a) The rate of heat output to the cooling water, Qc = t; is
Qc
m
=
C T = 2:8
t
t
104 kg=s (4184J=kgK) 8:5K = 996M W
Since the rate of work is
W
Qh Qc
Qh
=
= 750M W =
t
t
t
996M W;
the rate of energy extraction from the fuel is
Qh
= (750 + 996) M W = 1746M W:
t
(b) The e¢ ciency is
e=
W= t
750
=
= :43 = 43%:
Qh = t
1746
(c) Thus the minimum highest temperature can be found from the e¢ ciency of a Carnot cycle
288
= :43 = 43%;
Th
288
=
= 505K = 232 C.
1 e
e = 1
TH
13. From the 1st law of thermodynamics for a cyclic process, the rate of
work and input/output heat is
W = Qh
Qc = 2
3
1011 W:
Since the e¢ ciency is 1=3; the heat ejected to the low temperature reservoir
is:
e =
Qc
W
Qh
=
Qc
=
Qh
Qh
2
2
=
Qh = 3W = 2
3
3
1
3
2
1011 W = 4
1011 M W
The cooling water experiences a rise in temperature of 5 C; hence the ‡ow
rate is found from
Qc = mCwat T ! m =
4 1011
'2
4184 5
107 kg=s
This is approximately 10% greater than the ‡ow at the mouth of the Mississippi.
15. (a) During a cycle W = Qh Qc and the e¢ ciency is e = W=Qh :
Given that W = 350J and Qh = 900J the e¢ ciency is given by
e = 350=900 = :39 = 39%:
(b) The rejected heat Qc is
Qc = Qh
W = 550J:
(c) The maximum temperature is found from the e¢ ciency of a Carnot
cycle,
Tc
;
Th
e = 1
Th =
Tc
1
e
=
283
= 464K = 191 C
:61
17. For a refrigerator
COP =
Qc
Qc
:67Lf
:67 334kJ
= 4:2 ! W =
=
=
= 53:3kJ
W
4:2
4:2
4:2
23. (a) The de…nition for COP is
COP =
Qc
Qc
Tc
=
=
;
W
Qh Qc CC Th Tc
4
where CC denotes a Carnot cycle. Assuming a Carnot cycle for the heat
pump then in the summer the COP is
COPsum =
278
= 13:2:
21
COPwin =
275
= 3:53:
78
In the winter the COP is
(b) The ratio of work to heat removed from the cooler temperature bath
is W=Qc and
W
1
= :0755
=
Qc
COP
So the work required for each Joule of heat removed from the house in the
summer is
W = :0755J:
(c) In the winter the ratio of work to of heat supplied to the house is
W=Qh = 1
Tc
=1
Th
275
= :221:
353
So the work required for each Joule of heat supplied to the house is
W = :221J:
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
810J: Thus the heat transfer to the gas is
1L = 810kP aL =
Qh = 810J ln 2 = 562J:
(b) The heat rejected during the isothermal compression is given by
Qc =
nRT ln
5
Vf
= P V ln 2:
Vi
From the conditions in …gure 22-29 we have
P V = 4:1atm
1:612L = 4:1
101:3kP a
1:612L = 670J:
The heat transfer to the cold bath is
Qc = 670J ln 2 = 464J:
(c) The work done is
W = Qh
Qc = (562
464) J = 98J:
(d) The e¢ ciency is
e=
W
98
=
= :174 = 17:4%
Qh
562
(e) From the ideal gas law the high temperature is found from
P V = nRTh = 810kP aL = :2 8:314
Th = 810= (:2 8:314) = 487K:
Th ;
The low temperature is found from
P V = nRTc = 673kP aL = :2 8:314
Tc = 670=(:2 8:314) = 403K:
Tc ;
The Carnot e¢ ciency is
e=1
Tc
=1
Th
403
= :172 = 17:2%;
487
which agrees (to within roundo¤ errors) 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
dQ
nCv dT
5
500
25
5
S=
=
=5
R ln
= 8:314 ln = 53J=K
T
T
2
300
2
3
300
6
(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 was lost due to an adiabatic free expansion is
W = nRT ln (Vf =Vi ) = Pi Vi ln (Vf =Vi ) :
Solving for Pi yields
Pi =
W
6:5
=
= 470:5kP a
6
Vi ln (Vf =Vi )
10 ln 106
53. (a) The net heat added during the isothermal expansion is equal to
the work done during this process. Hence the net heat to the gas during this
cycle (a Sterling engine works on this cycle) is
Qin = nCv T + nRTh ln (Vf =Vi ) = nR
Qin = :2
8:314
3
300 + 600 ln 2
2
3
T + T ln (Vf =Vi )
2
= 1440J
The net work done is
W = nR (Th
Tc ) ln (Vf =Vi ) = :2
8:314
(b) The e¢ ciency of this cycle is
e = 346=1440 = :24 = 24%
7
300 ln 2 = 346J