le (txl65) – Heat – mohanty – (81440)
1
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printout
should
have
14
questions.
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before answering.
001
10.0points
The specific heat of a metal (similar to cop
per) is 0
.
092 cal
/
g
·
◦
C.
The latent heat of
vaporization of a liquid (similar to liquid ni
trogen) is 48 cal
/
g. A 3 kg block of the metal
at 24
◦
C is dropped into a large vessel of the
liquid at 77 K, which is the boiling point of
the liquid.
How many kilograms of the liquid boil away
by the time the metal reaches 77 K?
Correct answer: 1
.
265 kg.
Explanation:
Let :
m
Cu
= 3 kg
,
c
Cu
= 0
.
092 cal
/
g
·
◦
C
,
T
c
= 24
◦
C = 293 K
,
T
o
= 273 K
,
T
= 77 K
,
and
L
N
2
= 48 cal
/
g
.
The change in heat is
Δ
Q
=
m
Cu
c
Cu
(
T
Cu

T
) =
m
N
2
L
N
2
m
N
2
=
m
Cu
c
Cu
[
T
Cu

T
]
L
N
2
=
(3 kg) (0
.
092 cal
/
g
·
◦
C)
48 cal
/
g
×
(297 K

77 K)
=
1
.
265 kg
.
002
10.0points
What is the final equilibrium temperature
when 24 g of milk at 24
◦
C is added to 102 g of
coffee at 74
◦
C? Assume the specific heats of
milk and coffee are the same as that of water,
and neglect the specific heat of the container.
Correct answer: 64
.
4762
◦
C.
Explanation:
Let :
m
m
= 24 g
,
T
m
= 24
◦
C
,
m
c
= 102 g
,
and
T
c
= 74
◦
C
.
The amount of heat given out by coffee
should be equal to the amount of heat ab
sorbed by milk, so
c
m
m
m
(
T
f

T
m
) =
c
c
m
c
(
T
c

T
f
)
m
m
T
f

m
m
T
m
=
m
c
T
c

m
c
T
f
(
m
m
+
m
c
)
T
f
=
m
c
T
c
+
m
m
T
m
T
f
=
m
c
T
c
+
m
m
T
m
m
m
+
m
c
=
(102 g) (74
◦
C) + (24 g) (24
◦
C)
24 g + 102 g
=
64
.
4762
◦
C
.
003
10.0points
Two grams of water are sealed in a rigid con
tainer; then the water is vaporized by heating.
976 cal of heat is needed for vaporization.
What is the change in internal energy of the
water?
Correct answer: 4085
.
54 J.
Explanation:
Let :
Q
= 976 cal
.
The water is sealed in the rigid container,
so
W
= 0 and
Δ
U
=
Q
= (976 cal)
·
4
.
186 J
1 cal
=
4085
.
54 J
.
004
10.0points
A 20 kg
piece
of metal (specific
heat
of
0
.
06 cal
/
g
·
◦
C) having a temperature of 81
◦
C
is added to 0
.
7 kg of water having a tempera
ture of 27
◦
C.
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le (txl65) – Heat – mohanty – (81440)
2
What is the final equilibrium temperature
of the system?
Assume there is no loss of
heat to the surroundings. The specific heat of
water is 1 cal
/
g
·
◦
C.
Correct answer: 61
.
1053
◦
C.
Explanation:
Let :
m
1
= 20 kg
,
c
1
= 0
.
06 cal
/
g
·
◦
C
,
T
1
= 81
◦
C
,
m
2
= 0
.
7 kg
,
c
2
= 1 cal
/
g
·
◦
C
,
and
T
2
= 27
◦
C
.
There is no loss of heat to the surroundings,
so
m
1
c
1
(
T
1

T
f
) =
m
2
c
2
(
T
f

T
2
)
m
1
c
1
T
1

m
1
c
1
T
f
=
m
2
c
2
T
f

m
2
c
2
T
2
m
1
c
1
T
1
+
m
2
c
2
T
2
= (
m
2
c
2
+
m
1
c
1
)
T
f
T
f
=
m
1
c
1
T
1
+
m
2
c
2
T
2
m
1
c
1
+
m
2
c
2
Since
m
1
c
1
T
1
+
m
2
c
2
T
2
= (20 kg) (0
.
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 Spring '11
 sdsd
 Thermodynamics, Energy, Correct Answer, Qlost, Qcool

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