206
Chapter 20
E
XECUTE
:
(
)
(
)
3
4
C
f
0.0400 kg
334
10
J/kg
1.336
10
J.
Q
mL
= −
= −
×
= −
×
C
C
H
H
Q
T
Q
T
= −
gives
(
)
(
)
(
) (
)
4
4
H
H
C
C
1.336
10
J
373.15 K
273.15 K
1.825
10
J.
Q
T
T
Q
= −
= − −
×
= +
×
⎡
⎤
⎣
⎦
3
C
H
4.89
10
J.
W
Q
Q
=
+
=
×
E
VALUATE
:
For a heat engine,
C
Q
is negative and
H
Q
is positive. The heat that comes out of the engine
(
0)
Q
<
goes into the ice (
0
Q
>
).
20.23.
I
DENTIFY
:
The power output is
.
W
P
t
=
The theoretical maximum efficiency is
C
Carnot
H
1
.
T
e
T
=
−
H
.
W
e
Q
=
S
ET
U
P
:
4
H
1.50
10
J.
Q
=
×
C
350 K.
T
=
H
650 K.
T
=
1 hp
746 W.
=
E
XECUTE
:
C
Carnot
H
350 K
1
1
0.4615.
650 K
T
e
T
=
−
=
−
=
4
3
H
(0.4615)(1.50
10
J)
6.923
10
J;
W
eQ
=
=
×
=
×
this is the
work output in one cycle.
3
4
(240)(6.923
10
J)
2.77
10
W
37.1 hp.
60.0 s
W
P
t
×
=
=
=
×
=
E
VALUATE
:
We could also use
C
C
H
H
Q
T
Q
T
= −
to calculate
4
3
C
C
H
H
350 K
(1.50
10
J)
8.08
10
J.
650 K
T
Q
Q
T
⎛
⎞
⎛
⎞
= −
= −
×
= −
×
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
Then
3
C
H
6.92
10
J,
W
Q
Q
=
+
=
×
the same as previously calculated.
20.24.
I
DENTIFY
and
S
ET
U
P
:
C
Carnot
H
1
.
T
e
T
=
−
C
Carnot
H
C
.
T
K
T
T
=
−
E
XECUTE
:
(a)
C
H
(1
).
T
T
e
=
−
H
H
H
(1
)
1
.
(1
)
T
e
e
K
T
T
e
e
−
−
=
=
−
−
E
VALUATE
:
(b)
When
1,
e
→
0.
K
→
When
0,
e
→
.
K
→ ∞
1
e
→
when
C
H
.
Q
Q
<<
C
Q
is small in this limit. That is good for an engine since
C
Q
is wasted. But it is bad
for a refrigerator since
C
Q
is what is useful.
0
e
→
when
C
H
Q
Q
→
and
W
is very small. That is bad for an
engine but good for a refrigerator.
20.25.
I
DENTIFY
:
Q
S
T
Δ
=
for each object, where
T
must be in kelvins. The temperature of each object remains constant.
S
ET
U
P
:
For water,
5
f
3.34
10
J/kg.
L
=
×
E
XECUTE
:
(a)
The heat flow into the ice is
5
5
f
(0.350 kg)(3.34
10
J/kg)
1.17
10
J.
Q
mL
=
=
×
=
×
The heat flow
occurs at
273 K,
T
=
so
5
1.17
10
J
429 J/K.
273 K
Q
S
T
×
Δ
=
=
=
Q
is positive and
S
Δ
is positive.
(b)
5
1.17
10
J
Q
= −
×
flows out of the heat source, at
298 K.
T
=
5
1.17
10
J
393 J/K.
298 K
Q
S
T
−
×
Δ
=
=
= −
Q
is
negative and
S
Δ
is negative.
(c)
tot
429 J/K
(
393 J/K)
36 J/K.
S
Δ
=
+ −
= +
E
VALUATE
:
For the total isolated system,
0
S
Δ
>
and the process is irreversible.
20.26.
I
DENTIFY
:
Apply
system
0
Q
=
to calculate the final temperature.
.
Q
mc T
=
Δ
Example 20.6 shows that
2
1
ln(
/
)
S
mc
T
T
Δ
=
when an object undergoes a temperature change.
S
ET
U
P
:
For water
4190 J/kg
K.
c
=
⋅
Boiling water has
100.0
C
373 K.
T
=
=
°
E
XECUTE
:
(a)
The heat transfer between 100
C
°
water and 30 C
°
water occurs over a finite temperature
difference and the process is irreversible.
(b)
2
2
(270 kg)
(
30.0
C)
(5.00 kg)
(
100 C)
0.
c T
c T
−
+
−
=
°
°
2
31.27 C
304.42 K.
T
=
=
°
(c)
304.42 K
304.42 K
(270 kg)(4190 J/kg
K)ln
(5.00 kg)(4190 J/kg
K)ln
.
303.15 K
373.15 K
S
⎛
⎞
⎛
⎞
Δ
=
⋅
+
⋅
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
4730 J/K
(
4265 J/K)
470 J/K.
S
Δ
=
+ −
= +
E
VALUATE
:
system
0,
S
Δ
>
as it should for an irreversible process.
20.27.
I
DENTIFY
:
Both the ice and the room are at a constant temperature, so
.
Q
S
T
Δ
=
For the melting phase transition,
f
.
Q
mL
=
Conservation of energy requires that the quantity of heat that goes into the ice is the amount of heat that
comes out of the room.
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The Second Law of Thermodynamics
207
S
ET
U
P
:
For ice,
3
f
334
10
J/kg.
L
=
×
When heat flows into an object,
0,
Q
>
and when heat flows out of an
object,
0.
Q
<
E
XECUTE
:
(a)
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 Spring '06
 Buchler
 Physics, Thermodynamics, Entropy, Heat, ΔS, ΔU

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