5 g of water at `30^@C` and 5 g of ice at `-20^@C` are mixed together in a calorimeter Find the final temperature of the mixure. Assume water equivalent of calorimeter to be negligible, sp. Heats of ice and water are 0.5 and `1 cal//g C^@`, and latent heat of ice is `80 cal//g`

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.