Question

# 10 gm of ice at −20∘C is dropped into a calorimeter containing 10 gm of water at 10∘C. 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 waterAs we know Q=mcΔθQ1=10×1×10=100calQ2=10×0.5(0−(−20))+10×80=(100+800)cal=900calAs Q1<Q2, so ice will not completely melt and final temeprature = 0∘CAs heat given by water in cooling up to 0∘C is only just sufficient to increase the temperature of ice from −20∘C to 0∘C , hence mixture in equilibrium will consist of 10gm of ice and 10gm of water, both at 0∘C.

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