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.

Explanation

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`