A thermally isolated vessel contains `100g` of water at `0^(@)C`. When air above the water is pumped out, some of the water freezes and some evaporates at `0^(@)C` itself. Calculate the mass of the ice formed such that no water is left in the vessel. Latent heat of vaporization of water at `0^(@)C=2.10xx10^(6)J//kg` and latent heat of fusion of ice `=3.36xx10^(5)J//kg`.

Explanation

Total mass of the water `=M=100g`
Latent heat of vaporization of water at `0^(@)C`
`=L_(1)=21.0xx10^(5)J//Kg`
& Latent heat of fusion of ice `=L_(2)=3.36xx10^(5)J//kg`
Suppose, the mass of the ice formed `=m`
Then, the mass of water evaporated `=M-m`
Heat lost by the water in freezing`=` Heat taken by water in evaporation.
Thus, `mL_(2)=(M-m)L_(1) or m-86 g`