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Question

A vessel containing 100 g water at 0C is left in the middle of a room. In 15 minutes, the temperature of the water rises by 2C. When an equal amount of ice is placed in the vessel, it melts in 10 hours. The specific heat of fusion of ice is
(Take specific heat capacity of water as 1 cal/g.deg)

A
40 cal/g
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B
50 cal/g
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C
70 cal/g
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D
80 cal/g
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Solution

The correct option is D 80 cal/g
Given that,
Mass of water = Mass of ice, m=100 g
Specific heat capacity of water, s=1 cal/g.deg
Change in temperature of water in 15 min, ΔT=2C

The amount of heat required to rise the temperature of 100 g of water from 0C to 2C

=msΔT=100×1×2=200 cal

200 cal heat is supplied in 15 minutes.
The rate of heat supply per second = 20015×60=29 cal/s

So,total heat supplied in 10 hours,
Q=10×60×60×29=8000 cal

8000 cal melts same amount of ice (=100 g).
The heat required to melt 1 g of ice = 8000 cal100 g=80 cal/g.
So, latent heat of fusion of ice is 80 cal/g

Hence, option (d) is correct.

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