Question

When 0.15 kg of ice of 0 °C 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=4186JKg^-1 K^-1

Answer 1

Explanation:

Heat lost by water = mSw (θf - θi)w = (0.30 kg) (4186 J kg-1°C-1) x (50.0°C - 6.7°C) = 54376.14 J Heat to melt ice = m2Lf = (0.15 kg)Lf Heat to raise temperature of ice water to final temperature = m1Sw (θf - θi) = (0.15 kg) (4188 J kg-1°C-1) (6.7°C - 0°C) = 4206.93 J Heat lost = Heat gained 54.76.14 J = (0.15 kg) Lf + 4206.93 J ∴ Lf = 3.34 x 105 Jkg.