Question

`1kg` ice at `-10^(@)C` is mixed with `1kg` water at `100^(@)C`. Then final equilirium temperature and mixture content.

Answer 1

Heat taken by `1kg` Ice = Heat given by `1kg` water
`1 xx (1)/(2)xx 10 +1 xx 80 +1 xx T = 1 xx (100 -T)`
`85 -100 - 2T rArr 2T = 15`
`theta = (15)/(2) = 7.5^(@)C`, water