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Question

10 gm of ice at 20C is dropped into a calorimeter containing 10 gm of water at 10C. The specific heat of water is twice that of ice. Neglect heat capacity of the calorimeter. When equilibrium is reached, the calorimeter will contain :-

A
10 gm ice and 10 gm of water
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B
20 gm of water
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C
5 gm ice and 15 gm of water
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D
20 gm ice
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Solution

The correct option is A 10 gm ice and 10 gm of water
As we know Q=mcΔθ
Q1=10×1×10=100cal
Q2=10×0.5(0(20))+10×80=(100+800)cal=900cal
As Q1<Q2,
so ice will not completely melt and final temeprature = 0C
As heat given by water in cooling up to 0C is only just sufficient to increase the temperature of ice from 20C to 0C
, hence mixture in equilibrium will consist of 10gm of ice and 10gm of water, both at 0∘C.

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