A 5 g piece of ice at `-20^@C` is put into 10 g of water at `30^@C`. Assuming that heat is exchanged only between the ice and water. The final temperature of the mixture.

Explanation

Maximum heat that can be released by water, when its temperature falls from `30^@C` to `0^@C`.
`DeltaQ_1=m_(omega)s_(omega)(Deltatheta)_(omega)=10xx1xx(30-0)=300cal`
Heat required to convert ice from `-10^@C` to `0^@C`
`DeltaQ_2=m_(ice)s_("ice")(Deltatheta)_("ice")=5xx0.5[0-(-20)]=50cal`
Heat required to melt ice
`DeltaQ_3=m_("ice")L_("ice")=5xx80=400cal`
The heat 300 cal given by water is utilized in converting ice from `-20^@C` to `0^@C` (50cal) and remaining 250 cal is utilized in melting some part of ice. The whole ice is not melted. Water comes to `0^@C` by releasing 300 cal, melted ice becomes water at `0^@C` and remaining ice is at `0^@C` hence equilibrium temperature is `0^@C`.
The amout of ice melted`=(250)/(80)=3.125g`
Contents of mixture:
`ice=5-3.125=1.875g`
water`=10+3.125=13.125g`