How should `1 kg` of water at `50^(@)C` be divided in two parts such that if one part is turned into ice at `0^(@)C`. It would release sufficient amount of heat to vapourize the other part. Given that latent heat of fusion of ice is `3.36xx10^(5) J//Kg`. Latent heat of vapurization of water is `22.5xx10^(5) J//kg` and specific heat of water is `4200 J//kg K`.

Explanation

Let x kg of water is frozen then the amount of heat it rrelaeases is
`Q_(1)=x xx 4200xx50+x xx 3.36xx10^(5)` Joule
`x xx 5.46 xx10^(5)` joule
The heat required to vapourize the (1-x) kg of water form `50^(2)C` is
`Q_(2)=(1-x)xx22.5xx10^(5)`
Here we ve taken heat required to vapourize the water as only mass `xx` latent heat of vapourization and not the hear required to first raise the temperature of `(1-x)` kg of water from `50^(@)` to `100^(@)k` plus the mass `xx` latent heat similar as when heat is supplied to water from an external source, it first reaches `100^(@)C` then it vaporizes starts but when heat is taken by water itself if vapourize (evaporation) at `50^(@)C` as in this case. The similar case we see in our general life in cooling of water in a pitcher, drying of cloths hanging in open air etc.
Thus heat `Q_(2)` must be provided by first part of water, we have
`Q_(2)=Q_(1)`
`(1-x)xx22.5xx10^(5)=x xx 5.46xx10^(5)`
or `22.5-22.5=546x`
or `x=22.5/27.96=0.805kg`