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The Second Law of Thermodynamics
2021
20.60.
IDENTIFY:
For a reversible isothermal process,
.
Q
S
T
Δ=
For a reversible adiabatic process,
0
Q
=
and
0.
S
Δ
=
The Carnot cycle consists of two reversible isothermal processes and two reversible adiabatic processes.
SET UP:
Use the results for the Stirling cycle from Problem 20.50.
EXECUTE:
(a)
The graph is given in Figure 20.60.
(b)
For a reversible process,
, and so
, and
dQ
dS
dQ T dS
T
==
,
Qd
QT
d
S
∫
∫
which is the area under the curve
in the
TS
plane.
(c)
H
Q
is the area under the rectangle bounded by the horizontal part of the rectangle at
H
T
and the verticals.
C

Q
is the area bounded by the horizontal part of the rectangle at
C
T
and the verticals. The net work is then
HC
,
QQ
−
the area bounded by the rectangle that represents the process. The ratio of the areas is the ratio of the lengths of the
vertical sides of the respective rectangles, and the efficiency is
HH
.
WTT
e
−
(d)
As explained in Problem 20.50, the substance that mediates the heat exchange during the isochoric expansion
and compression does not leave the system, and the diagram is the same as in part (a). As found in that problem,
the ideal efficiency is the same as for a Carnotcycle engine.
EVALUATE:
The derivation of
e
Carnot
using the concept of entropy is much simpler than the derivation in
Section 20.6, but yields the same result.
Figure 20.60
20.61.
IDENTIFY:
The temperatures of the icewater mixture and of the boiling water are constant, so
.
Q
S
T
The heat
flow for the melting phase transition of the ice is
f
.
Qm
L
= +
SET UP:
For water,
5
f
3.34 10 J/kg.
L
=×
EXECUTE:
(a)
The heat that goes into the icewater mixture is
54
f
(0.160 kg)(3.34 10 J/kg)
5.34 10 J.
L
×
This is same amount of heat leaves the boiling water, so
4
5.34 10 J
143 J/K.
373 K
Q
S
T
−×
Δ= =
=
−
(b)
4
196 J/K
273 K
Q
S
T
×
=
+
(c)
For any segment of the rod, the net heat flow is zero, so
0.
S
Δ
=
(d)
tot
143 J/K 196 J/K
53 J/K.
S
−
+
=
+
EVALUATE:
The heat flow is irreversible, since the system is isolated and the total entropy change is positive.
20.62.
IDENTIFY:
Use the expression derived in Example 20.6 for the entropy change in a temperature change.
SET UP:
For water,
4190 J/kg K.
c
=⋅
20 C
293.15 K,
=
°
65 C
338.15 K
=
°
and 120 C
393.15 K.
=
°
EXECUTE:
(a)
3
21
ln(
)
(250 10 kg)(4190 J kg K)ln(338.15 K 293.15 K) 150 J K.
Sm
cTT
−
=
×
⋅
=
(b)
3
element
(250 10 kg)(4190 J kg K)(338.15 K
293.15 K)
120 J/K.
393.15 K
mc T
S
T
−
−Δ − ×
⋅
−
=
=
−
(c)
The sum of the result of parts (a) and (b) is
system
30 J/K.
S
Δ
=
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Chapter 20
EVALUATE:
(d)
Heating a liquid is not reversible. Whatever the energy source for the heating element, heat is
being delivered at a higher temperature than that of the water, and the entropy loss of the source will be less in
magnitude than the entropy gain of the water. The net entropy change is positive.
20.63.
IDENTIFY:
Use the expression derived in Example 20.6 for the entropy change in a temperature change. For the
value of
T
for which
S
Δ
is a maximum,
()
/
0
.
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 Spring '06
 Buchler
 Physics, Second Law Of Thermodynamics

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