Version PREVIEW – Homework 10 – Savrasov – (39825)
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Carnot Efficiency
001
10.0 points
A heat engine operates between two reservoirs
at 18
.
9
◦
C and 168
◦
C.
What is the maximum efficiency possible
for this engine?
Correct answer: 0
.
338095.
Explanation:
Given :
T
h
= 168
◦
C = 441 K
and
T
c
= 18
.
9
◦
C = 291
.
9 K
.
Efficiency
e
= 1

T
c
T
h
= 1

291
.
9 K
441 K
=
0
.
338095
.
Entropy in Expansion
002
10.0 points
58
.
1 mol of an ideal gas is allowed to undergo
a free expansion.
If the initial volume is 20 cm
3
and the final
volume is 100 cm
3
, find the change in entropy.
Correct answer: 777
.
054 J
/
K.
Explanation:
Given :
n
= 58
.
1 mol
,
V
1
= 20 cm
3
,
and
V
2
= 100 cm
3
.
The free expansion is not quasistatic;
we
have to use a quasistatic process connecting
the initial and the final equilibrium states to
calculate the change of entropy. Since the two
states have the same temperature, we connect
them with an isothermal process which gives
Δ
S
=
n R
ln
parenleftbigg
V
2
V
1
parenrightbigg
= (58
.
1 mol)(8
.
31 J
/
mol
·
K) ln
parenleftbigg
100 cm
3
20 cm
3
parenrightbigg
=
777
.
054 J
/
K
.
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 Fall '09
 RichardScalettar
 Energy, Work, Heat, Correct Answer, Heat engine, Carnot cycle

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