Question

2. (5 points) A 35 g ice cube at 0 °C is placed on a 1.5 kg Al plate initially at 40 °C? What will be the final temperature once the ice melts and the plate and water come to equilibrium? Assume that no heat is exchanged with the environment. Use 333 kJ/kg for the heat of fusion for water, 4.186 kJ/(kg °C) for the specific heat of water, and 0.900 kJ/(kg oC) for the specific heat of Al

Answer 1

Mass of ice = m₁ = 35 g = 0.035 kg

Mass of aluminium plate = m₂ = 1.5 kg

Initial temperature of ice = T₁ = 0 ^oC

Initial temperature of aluminium plate = T₂ = 40 ^oC

Equilibrium temperature = T₃

Latent heat of fusion of water = L = 333 kJ/kg

Specific heat of water = C₁ = 4.186 kJ/(kg.^oC)

Specific heat of aluminium = C₂ = 0.9 kJ/(kg.^oC)

The heat gained by the ice cube is equal to the heat lost by the aluminium plate,

m₁L + m₁C₁(T₃ - T₁) = m₂C₂(T₂ - T₃)

(0.035)(333) + (0.035)(4.186)(T₃ - 0) = (1.5)(0.9)(40 - T₃)

11.655 + 0.14651T₃ = 54 - 1.35T₃

1.49651T₃ = 42.345

T₃ = 28.3 ^oC

Final temperature = 28.3 ^oC