When 0.15 kg of ice at 0 ^(@) 0is mixed with 0.30 kg of water at50 ^(@) Cin a container, the resulting temperature is6.7 ^(@) C . Calculate the heat of fusion of ice. (S_("water") = 4186 J kg^(-1) K^(-1))

When 0.15 kg of ice at 0 ^(@) 0is mixed with 0.30 kg of water at50 ^(@

1.	When 0.15 kg of ice at 0 ^(@) 0is mixed with 0.30 kg of water at50 ^(@) Cin a container, the resulting temperature is6.7 ^(@) C . Calculate the heat of fusion of ice. (S_("water") = 4186 J kg^(-1) K^(-1))
Answer» SOLUTION :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)`.