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Question

Water of mass m2=1 kg is contained in a copper calorimeter of mass m1=1 kg. Their common temperature is t=10oC. Now a piece of ice of mass m3=2 kg and temperature −11oC is dropped into the calorimeter. Neglecting any heat loss, the final temperature of the system is [specific heat of copper 0.1 Kcal/kgoC, specific heat of water =1 Kcal/kgoC, specific heat of ice =0.5 Kcal/kgoC, latent heat of fusion of ice =78.7 Kcal/kg]

A
0oC
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B
4oC
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C
4oC
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D
2oC
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Solution

The correct option is B 0oC
Heat is lost by calorimeter plus water while it is gained by ice.
Heat lost = heat gained
Qlost=Qcalorimeter+Qwater
=mcalorimeter×scalorimeter×(100)+mwater×swater×1
=1×1×10+1×0.1×10
=11kcal
Qgained=mice×sice×(0(11))=2×0.5×11=11kcal
Thus, all the heat lost by calorie meter and water is used for melting the ice from 110C to 00C.
Hence, the final temp of the system will be 00C, ice and water will co-exist at 00C.

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