Find the result of mixing 0.5 kg ince at `0^@C` with 2 kg water at `30^@C`. Given that latent heat of ice is `L=3.36xx10^5 J//kg` and specific heat of water is `4200 J//kg//K`.

Explanation

In such mixing problems, it is advisable to first convert all the phases of mixing substance into a single phase at a common temperature and keep the excess or required heat for this aside and finally supply or extract that amount of heat to get the final equilibrium temperature of mixture. As in this problem `0.5kg` of ice is given at `0^@C`. To convert it into water at `0^@C` if we require `Q_1` amount of heat, the we have
`Q_1=-mL_f=-0.5xx3.36xx10^5`
`=-1.68xx10^5J`
(`-`ve sign for heat required)
The 2 kg of water is available at `30^@C`. To convert it into `0^@C` , it releases some heat, say `Q_2`. Then we have
`Q_2=msDeltaT`
`=2xx4200xx30=2.52xx10^5J`
Thus we have the final mixture as
Final mixture`=2.5kg` water at `0^@C+2.52xx10^5J-1.68xx10^5J`
`=2.5kg` water at `0^@+8.4xx10^4J`
Thus finally we can supply the availabe heat of the 2.5 kg water at `0^@C` to get find temperature of mixure as
`2.5xx4200xx(T_g-0)=8.4xx10^4`
or `T_f=8^@C`
Thus final result is `8^@C` of 2.5 kg water after mixing.