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))`

Explanation

Heat lost by water `=ms_(w)(theta_(f)-theta_(i))_(w)`
`=(0.30kg)(4186jkg^(-1)K^(-1))(50.0^(@)C-6.7^(@)C)`
= 54376.14 J
Heat required to melt ice `=m_(2)L_(f)=(0.15kg)L_(f)` Heat required to raise temperature of ice water to final temperature `m_(1)s_(w)(theta_(f)-theta_(i)_(1)`
`=(0.15kg)(4186jkg^(-1)K^(-1))(6.7^(@)C-0^(@)C)`ltbrge4206.93 J
Heat lost = heat gained
`54376.14 j =(0.15 kg)L_(f)+4206.93 J`
`l_(f)=3.34xx10^(5)J Kg^(-1)`.