baum (lmb2768) – HW11 – distler – (56295)
1
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001 (part 1 of 2) 10.0 points
A particular engine has a power output of
6 kW and an e
ffi
ciency of 14%.
If the engine expels 11470 J of thermal
energy in each cycle, find the heat absorbed
in each cycle.
Correct answer: 13337
.
2 J.
Explanation:
Let :
P
= 6 kW
,
= 14% = 0
.
14
,
and
Q
c
= 11470 J
.
=
W
Q
h
=
Q
h

Q
c
Q
h
= 1

Q
c
Q
h
Q
c
Q
h
= 1

Q
h
=
Q
c
1

=
11470 J
1

0
.
14
=
13337
.
2 J
.
002 (part 2 of 2) 10.0 points
Find the time for each cycle.
Correct answer: 0
.
311202 s.
Explanation:
The net work done by a heat engine through
a cyclic process (
Δ
U
= 0) is
W
=
Q
h

Q
c
,
and from
P
=
W
t
t
=
Q
h

Q
c
P
=
13337
.
2 J

11470 J
6 kW
1 kW
1000 W
=
0
.
311202 s
.
003
10.0 points
What is the entropy decrease in 1
.
31 mol of
helium gas which is cooled at 1 atm from room
temperature 290 K to a final temperature
of 5
.
16 K?
The specific heat of helium is
21 J
/
mol
·
K.
Correct answer: 110
.
836 J
/
K.
Explanation:
Let :
n
= 1
.
31 mol
,
T
1
= 290 K
,
and
T
2
= 5
.
16 K
.
Since in this cooling process pressure is con
stant,
d Q
=
n c
p
dT ,
where
n
is the number of moles of helium.
The change of entropy is
Δ
S
=
d Q
T
=
T
2
T
1
n c
p
d T
T
=
n c
p
log
T
2
T
1
= (1
.
31 mol)(21 J
/
mol
·
K)
×
log
5
.
16 K
290 K
=

110
.
836 J
/
K
,
a decrease of
110
.
836 J
/
K
.
004
10.0 points
An ice cube melts on the warm sidewalk on
a hot summer day.
Let the entropies of the
ice cube, of the pavement and of the ice cube
sidewalk system be
S
ice
,
S
sw
and
S
sys
.
What happens to these entropies?
1.
S
ice
increases,
S
sw
increases,
S
sys
in
creases
2.
S
ice
decreases,
S
sw
increases,
S
sys
de
creases
3.
S
ice
increases,
S
sw
decreases,
S
sys
does
not change.
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baum (lmb2768) – HW11 – distler – (56295)
2
4.
S
ice
increases,
S
sw
decreases,
S
sys
in
creases
correct
Explanation:
Heat is added to the ice cube, increasing its
entropy. Heat is removed from the sidewalk,
decreasing its entropy.
The total entropy
change, however, is positive.
005 (part 1 of 2) 10.0 points
Use
Stefan’s
law
to
calculate
the
to
tal
power
radiated
per
unit
area
by
a
tungsten
filament
at
a
temperature
of
1830 K.
The StephanBoltzmann constant
is 5
.
6696
×
10

8
W
/
m
2
/
K
4
. Assume the fila
ment is an ideal radiator.
Correct answer: 6
.
35853
×
10
5
W
/
m
2
.
Explanation:
Let :
T
= 1830 K
.
From Stefan’s law
p
≡
P
A
=
σ
T
4
= (5
.
6696
×
10

8
W
/
m
2
/
K
4
)(1830 K)
4
=
6
.
35853
×
10
5
W
/
m
2
.
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 Spring '09
 IRENEPOLYCARPOU
 Physics, Thermodynamics, Power, Correct Answer

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