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19.1
THE SECOND LAW OF THERMODYNAMICS
EXERCISES
Sections 19.2 and 19.3 The Second Law of Thermodynamics and Its Applications
14.
The efficiency of a reversible engine, operating between two absolute temperatures,
TT
hc
>
, is given by Equation
19.3.
(a)
eT
T
ch
=−
=
1
1
273 373
26 8
//
.
%
.
(b)
eTTT T
T
h
h
=
=
=
()
/
.
%
.
∆
21298
7 05
(c)
With room
temperature at
Te
c
=° =
=
20
980 1273
77 0
C,
/
. %.
15.
INTERPRET
This problem is about the thermal efficiency of a heat engine.
DEVELOP
If it were a reversible engine, its efficiency would be that of a Carnot engine, given by Equation 19.3:
e
T
T
Carnot
c
h
1
EVALUATE
Substituting the values given in the problem, we obtain
e
T
T
Carnot
c
h
K
K
=− =−
×
≈
−
11
27
5600
1
4 82
10
99 9
4
.
..
5
%
ASSESS
The engine efficiency is almost 100%. This is too good to be true.
16.
We can solve Equation 19.3 for the low temperature to find
T
=
(.
)
(
.
1
1
0 777 4 25 K)
0.948 K.
17.
INTERPRET
This problem is about a Carnot engine that operates via the Carnot cycle.
DEVELOP
The efficiency of an engine, by definition is,
e
W
Q
h
=
where
WQ
h
and
are the work done and heat absorbed per cycle.
EVALUATE
(a)
From the equation above, the efficiency of the engine is
e
W
Q
h
==
=
350
900
38 9
J
J
.%
(b)
h
and
are related to the heat rejected per cycle by the first law of thermodynamics (since
∆
U
per cycle is
zero). The relation is
QQW
=
−
=−=
900
350
550
JJJ
(c)
For a Carnot engine operating between two temperatures,
QQ
h c
,
=
so
Q
Q
h
c
=
=
=
283
900
550
463
K
J
J
K
=
°
190 C
ASSESS
The maximum temperature
T
h
is greater than
T
c
, as our calculation confirms. Note that Carnot’s theorem
applies to the ratio of absolute temperatures.
18.
From Equation 19.4, COP
rev
=
=
TT T
ch c
/(
)
/
.
.
273 30
9 10
19.
INTERPRET
This problem is about the work done by a refrigerator to freeze water. Heat of transformation is
involved in the phase change.
19
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View Full Document 19.2
Chapter 19
DEVELOP
The amount of heat that must be extracted in order to freeze the water is
Qm
L
cf
==
=
(.
)
(
)
0 67
334
224
kg
kJ/kg
kJ
The work consumed by the refrigerator while extracting this heat is given by Equation 19.4,
WQ
c
=
/COP.
EVALUATE
Substituting the values given, we obtain
W
Q
c
=
COP
kJ
kJ
224
42
53 3
.
.
ASSESS
A COP of 4.2 means that each unit of work can transfer 4.2 units of heat from inside the refrigerator.
A smaller COP would mean that more work is required to freeze the water.
Section 19.4 Entropy and Energy Quality
20.
Since the temperature is constant during a change of phase, Equation 19.6 gives
∆∆
SQ
T
m
L
T
f
=
//
()
(
)
/
1
334
273
kg
kJ/kg
K
1.22 kJ/K.
×=
21.
INTERPRET
This problem asks for the entropy increase after heating up water.
DEVELOP
For a substance with constant specific heat (in this case at constant pressure),
dQ
mcdT
=
, and the
change in entropy is
∆
S
dQ
T
mc
dT
T
mc
T
T
T
T
=
∫∫
1
2
2
1
1
2
ln
EVALUATE
Substituting the values given, we have
∆
Sm
c
T
T
=
=⋅
( .
)( .
)ln
2
1
0 25
4 184
3
kg
kJ/kg K
68
275
K
283 K
J/K
=
ASSESS
The final entropy of the system has increased.
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This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at University of California, Berkeley.
 Fall '10
 Pheong

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