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JEE Main Physics Class 11 Properties of Solids and Liquids Questions Part 1 Questions
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© examsnet.com
Question : 72
Total: 100
The specific heat capacity of a metal at low temperature(T) is given as
C
p
(
k
J
K
−
1
k
g
−
1
)
=
32
(
T
400
)
3
A 100 gram vessel of this metal is to be cooled from 20ºK to 4ºK by a special refrigerator operating at room temperature (27°C). The amount of work required to cool the vessel is
[2011 RS]
greater than 0.148 kJ
between 0.148 kJ and 0.028 kJ
less than 0.028 kJ
equal to 0.002 kJ
Validate
Solution:
Required work = energy released
Here,
Q
=
∫
m
c
d
T
=
4
∫
20
0.1
×
32
×
(
T
3
400
3
)
d
T
=
4
∫
20
3.2
64
×
10
6
T
3
d
T
=
5
×
10
−
8
4
∫
20
T
3
d
T
=
0.002
k
J
Therefore, required work
=
0.002
k
J
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