A piece of ice of mass `100 g` and at temperature `0^@ C` is put in `200 g` of water of `25^@ C`. How much ice will melt as the temperature of the water reaches `0^@ C` ? (specific heat capacity of water `=4200 J kg^(-1) K^(-1)` and latent heat of fusion of ice `= 3.4 xx 10^(5) J Kg^(-1)`).

Explanation

The heat released as the water cools down from `25^@ C` to `0^@C` is
`Q = mS Delta theta = (0.2)(4200)(25)=21000 J`
The amount of ice melted by this heat is
`m =(Q)/(L) =(21000)/(3.4 xx 10^(5)) = 62 g`.