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Test Index
Class 11 NEET Physics Thermodynamics and Kinetic Theory
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© examsnet.com
Question : 57
Total: 105
When 1 kg of ice at 0°C melts to water at 0°C, the resulting change in its entropy, taking latent heat of ice to be 80 cal/°C, is
(2011)
273 cal/K
8
×
10
4
cal/K
80 cal/K
293 cal/K
Validate
Solution:
Heat required to melt 1 kg ice at 0°C to water at 0°C is
Q
=
m
i
c
e
L
i
c
e
=
(
1
k
g
)
(
80
c
a
l
/
g
)
=
(
1000
g
)
(
80
c
a
l
/
g
)
=
8
×
10
4
cal
Change in entropy,
Δ
S
=
Q
T
=
8
×
10
4
c
a
l
(
273
K
)
=
293
cal/K
Note
: In the question paper unit of latent heat of ice is
given to be cal/°C. It is wrong. The unit of latent heat of ice is cal/g.
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