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206
Chapter 20
EXECUTE:
()
( )
34
Cf
0.0400 kg 334 10 J/kg
1.336 10 J.
Qm
L
=−
×
×
CC
HH
QT
= −
gives
( )
44
C
C
1.336 10 J
373.15 K
273.15 K
1.825 10 J.
T
Q
=− −
×
=+
×
⎡⎤
⎣⎦
3
CH
4.89 10 J.
WQ Q
=+= ×
EVALUATE:
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.
IDENTIFY:
The power output is
.
W
P
t
=
The theoretical maximum efficiency is
C
Carnot
H
1.
T
e
T
H
.
W
e
Q
=
SET UP:
4
H
1.50 10 J.
Q
=×
C
350 K.
T
=
H
650 K.
T
=
1 hp
746 W.
=
EXECUTE:
C
Carnot
H
350 K
1
1
0.4615.
650 K
T
e
T
=
43
H
(0.4615)(1.50 10 J)
6.923 10 J;
We
Q
==
×
=
×
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
×
= ×
=
EVALUATE:
We could also use
= −
to calculate
C
H
350 K
(1.50 10 J)
8.08 10 J.
650 K
T
QQ
T
⎛⎞
×
×
⎜⎟
⎝⎠
Then
3
6.92 10 J,
the same as previously calculated.
20.24.
IDENTIFY
and
SET UP:
C
Carnot
H
T
e
T
C
Carnot
HC
.
T
K
TT
=
−
EXECUTE:
(a)
(1
).
TT e
H
)
1
.
)
Te
e
K
TT e e
−
−
−−
EVALUATE:
(b)
When
1,
e
→
0.
K
→
When
0,
e
→
.
K
→∞
1
e
→
when
.
<<
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
→
and
W
is very small. That is bad for an
engine but good for a refrigerator.
20.25.
IDENTIFY:
Q
S
T
Δ=
for each object, where
T
must be in kelvins. The temperature of each object remains constant.
SET UP:
For water,
5
f
3.34 10 J/kg.
L
EXECUTE:
(a)
The heat flow into the ice is
55
f
(0.350 kg)(3.34 10 J/kg)
1.17 10 J.
L
×
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
Q
×
flows out of the heat source, at
298 K.
T
=
5
393 J/K.
298 K
Q
S
T
−×
=
−
Q
is
negative and
S
Δ
is negative.
(c)
tot
429 J/K
( 393 J/K)
36 J/K.
S
+
−
=
+
EVALUATE:
For the total isolated system,
0
S
Δ
>
and the process is irreversible.
20.26.
IDENTIFY:
Apply
system
0
Q
=
to calculate the final temperature.
.
cT
=
Δ
Example 20.6 shows that
21
ln(
/ )
Sm
cTT
when an object undergoes a temperature change.
SET UP:
For water
4190 J/kg K.
c
=⋅
Boiling water has
100.0 C
373 K.
T
°
EXECUTE:
(a)
The heat transfer between 100 C
°
water and 30 C
°
water occurs over a finite temperature
difference and the process is irreversible.
(b)
22
(270 kg) (
30.0 C)
(5.00 kg) (
100 C)
0.
−+
−
=
°°
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
⋅
+
⋅
4730 J/K
( 4265 J/K)
470 J/K.
S
+−
=
+
EVALUATE:
system
0,
S
Δ>
as it should for an irreversible process.
20.27.
IDENTIFY:
Both the ice and the room are at a constant temperature, so
.
Q
S
T
For the melting phase transition,
f
.
L
=
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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View Full DocumentThe Second Law of Thermodynamics
207
SET UP:
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
<
EXECUTE:
(a)
Irreversible because heat will not spontaneously flow out of 15 kg of water into a warm room to
freeze the water.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics, Heat

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