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Water of mass `m_(2) = 1 kg` is contained in a copper calorimeter of mass `m_(1) = 1 kg`. Their common temperature `t = 10^(@)C`. Now a piece of ice of mass `m_(2) = 2 kg` and temperature is `-11^(@)C` dropped into the calorimeter. Neglecting any heat loss, the final temperature of system is. [specific heat of copper `=0.1 Kcal//kg^(@)C`, specific heat of water `= 1 Kcal//kg^(@)C`, specific heat of ice `= 0.5 Kcal//kg^(@)C`, latent heat of fusion of ice `= 78.7 Kcal//kg`]
A. `0^(@)C`
B. `4^(@)C`
C. `-4^(@)C`
D. `-2^(@)C`

1 Answer

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Correct Answer - A
Loss in heat from calorimeter+water as temperature change form `10^(@)C` to `0^(@)C`
`=m_(1)C_(1)10+m_(2)C_(2)10`
`=1xx1xx10+1xx0.1xx10=11kcal`
Gain in heat of ice as its temperature changes from `-11^(@)C` to `0^(@)C`
`=m_(3)C_(3)xx2xx0.5xx11 = 11kcal`
Hence ice and water will coexist at `0^(@)C` without any phase change.

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