Find the result of mixing 0.5 kg ice 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 mixign 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 ofheat to get the final equilibrium temperature of the mixture.
As in this problem 0.5 kg of ice is given at `0^(@)C`. To convert it into water at `0^(@)C` if we required `Q_(1)` amount of heat, the we have
`Q_(1)=mL_(f)=-0.5xx3.36xx10^(5)`
`=-1.68xx10^(5)` Joule [-ve sign for heat required]
The water 2 kg is available at `30^(@)C` to convert it into `0^(@)C` it release some heat say `Q_(2)` then we have
`Q_(2)=ms DeltaT`
`=2xx4200xx30=2.52xx10^(5)` Joule
Thus we have the final mixture as
Final mixture `=2.5` kg water at `0^(@)+2.52xx10^(5)J-1.68xx10^(5)`]
`=2.5` kg water at `0^(@)+8.4xx10^(4)`J
Thus finally we can supply the availabel heat to the 2.5 kg wate at `0^(@)C` to get the final temperature of mixture 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.