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 of water be frozen. Then the amount of heat it releases is
`Q_(1)=x xx4200xx50+xxx3.36xx10^(5)J=x xx5.46xx10^(5) J`
The heat required to vapourize the `(1-x)kg` of water from `50^(@)C` is
`Q_(2)=(1-x)xx22.5xx10^(5)`
Here, we have taken heat required to vapourize the water as only mass x latent heat of vapourization and not the heat required to first raise the tem perature of `(1-x)kg` of water from `50^(@)` to `100^(@)` plus the mass x latent heat similar as when heat is supplied to water from an external source. It first reaches `100^(@)C` and then its vapourization starts, but when heat is taken by water itself, it vapourizes (evaporation) at `50^(@)C` as in this case. Tje simlilar case we see in our general life in coo,ling of water in pitcher, drying of clothes 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 xx5.46xx10^(5)`
or `225.5-22.5x=5.46x` or `x=(22.5)/(25.06)=0.812kg`.