1 g ice at `-40^@C`. Is placed in a container having 1 g water at `1^@C`. Find equilibrium temperature. Assume heat capacity of container is negligibly small.

Explanation

The heat available when water cools from `10^@C` to `0^@C`
`=mcDeltaT=1xx1xx(10-0)=10cal`
Let temperature of ice be T after taking this heat
`m_(ice)cDeltaT=10`
or `1xx0.5xx[T-(-40)]=10`
`T+40=20`
`T=-20^@C`
Now the system has 1 g ice at `-20^@C` and 1 g water at `0^@C`. Let m gram water get freezed to bring to ice from `-20^@C` to `0^@C`
Heat gained by ice`=`heat lost by water
`m_(ice)[0-(-20)]=mxx80`
or `1xx0.5xx20=mxx80`
`m=(1)/(8)g`
Thus equilibrium temperature becomes `0^@C`
Final contents: `ice=1g+(1)/(8)G=(9)/(8)g`
water `=1-(1)/(8)=(7)/(8)g`