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 ` = m s_(w) (theta_(f) - theta_(1))_(w)`
` = (0.30 kg)(4186 J kg^(-1) K^(-1)) (50.0^(@) C - 6.7 ^(@) C)`
` = 54376.14 J`
Heat required to melt ice ` = m_(2) L_(f) = (0.15 kg) L_(r) `
Heat required to raise temperature of ice water to final temperature ` = m_(1) s_(w) (theta_(f) - theta_(i))_(I)`
` = (0.15 kg) (4186 J kg^(-1) K^(-1)) (6.7 ^(@) C - 0^(@) C)`
` = 4206.93 J`
Heat lost = heat gained
`54376.14 J = (0.15 kg) L_(f) + 4206.93 J`
` L_(f) = 3.34 xx 10^(5) J kg^(-1)`.