Question

A 0.035-kg ice cube at -30.0°C is placed in 0.315 kg of 35.0°C water in a very well-insulated container. The latent heat of fusion for water is Lf = 79.8 kcal/kg.

What is the final temperature of the water, in degrees Celsius?

Answer 1

Mass of the ice cube = m₁ = 0.035 kg

Mass of the water = m₂ = 0.315 kg

Specific heat of ice = C₁ = 0.5 kcal/(kg.^oC)

Specific heat of water = C₂ = 1 kcal/(kg.^oC)

Latent heat of fusion of water = L = 79.8 kcal/kg

Initial temperature of ice = T₁ = -30 ^oC

Initial temperature of water = T₂ = 35 ^oC

Melting point of ice = T₃ = 0 ^oC

Final temperature of the water = T₄

The heat gained by the ice is equal to the heat lost by the water.

m₁C₁(T₃ - T₁) + m₁L + m₁C₂(T₄ - T₃) = m₂C₂(T₂ - T₄)

(0.035)(0.5)(0 - (-30)) + (0.035)(79.8) + (0.035)(1)(T₄ - 0) = (0.315)(1)(35 - T₄)

0.525 + 2.793 + 0.035T₄ = 11.025 - 0.315T₄

0.35T₄ = 7.707

T₄ = 22.02 ^oC

Final temperature of water = 22.02 ^oC