19 g water at `30^(@)C` and 5 g of ice at `-20^(@)C` are mixed together in a calorimeter. What is the final temperature of the mixture ? Given specific heat of ice =`0.5 cal g^(-1^@)C^(-1)` and latent heat of fusion of ice = `80 cal g^(-1)`

Explanation

In this case heat is given by water and taken by ice.
Heat available with water to cool from `30^(@)`C to `0^(@)`C
`=mcDeltaT=5xx1xx30`=150cal ltbrlt Heat required by 5g ice to increase its temperature up to `0^(@)`C
`mcDeltaT` = `5xx0.5xx20`=50cal
Out of 150 cal heat available, 50cal is used for increasing tempeature of ice from `-20^(@)`C to `0^(@)`C. The remaining heat 100 cal is used for melting the ice.
If mass of ice melted is mg then
`mxx80 =100 rArr` m=1.25g
Thus, 1.25g ice out of 5g melts and mixture of ice and water is at `0^(@)`C.