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. Calculating the heat of fusion of ice. (S_water = 4186 J kg^-1°C^-1)

Answer 1

Heat lost by water

= mS_w (θ_f - θ_i)_w = (0.30 kg) (4186 J kg-1°C^-1) x (50.0°C - 6.7°C)

= 54376.14 J

Heat to melt ice = m₂L_f = (0.15 kg)L_f

Heat to raise temperature of ice water to final temperature

= m₁S_w (θ_f - θ_i)

= (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) L_f + 4206.93 J

∴ L_f = 3.34 x 105 Jkg^-1.