1 g ice at `0^@C` is placed in a calorimeter having 1 g water at `40^@C`. Find equilibrium temperature and final contents. Assuming heat capacity of calorimeter is negligible small.

1 g ice at `0^@C` is placed in a calorimeter having 1 g water at `40^@

1.	1 g ice at `0^@C` is placed in a calorimeter having 1 g water at `40^@C`. Find equilibrium temperature and final contents. Assuming heat capacity of calorimeter is negligible small.
Answer» The heat available on water `=mcDeltat=1xx80=80cal` The heat available on water `=mcDeltat=1=1xx1xx(40-0)` `=40cal` Threfore, entire heat of water is utilized to melt the ice and its temperature falls to `0^@C` ice is still at `0^@C`. So equilibrium temperature of contents remains `0^@C`. Let m is the amount of ice that melts by absorbing 40 cal heat. Then `mxx80=40` or `m=(1)/(2)g` Final contents: ice `=1-(1)/(2)=(1)/(2)g` Water`=1+(1)/(2)=(3)/(2)g`

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1 g ice at `0^@C` is placed in a calorimeter having 1 g water at `40^@C`. Find equilibrium temperature and final contents. Assuming heat capacity of calorimeter is negligible small.

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