Question

When 0.15 kg of ice at `0^(@)C` is mixed with 0.30 kg of water at `50^(@)C` in a container, the resulting temperature is `6.7^(@)C`. Calculate the heat of fusion of ice. `(s_(water) = 4186 J kg^(-1)K^(-1))`

Answer 1

Heat lost by water
`=m_(w)s_(w) (T_(1)-T_(2)) = 0.30 xx 4186 xx (50-6.7)`
`= 54376.14 J`
Heat taken by ice =`m_(i)L + m_(i)s_(w)(T_(2)-T_(0))`
`=0.15 xx L + 0.15 xx 4186 xx (6.7-0)`
`=0.15 L + 4206.93 J`
Heat lost = heat gained
`:. 54376.14 = 0.15 L + 4206.93`
or `L = 3.34 xx 10^(5) J kg^(-1)`.