Question

One way to keep the contents of a garage from becoming too cold on a night when a severe subfreezing temperature is forecast is to put a tub of water in the garage. If the mass of the water is 105 kg and its initial temperature is 26.1°C, how much energy must the water transfer to its surroundings in order to freeze completely? The specific heat of water is 4186 J/kg·K, and the latent heat of fusion is 333 kJ/kg.

Answer 1

Mass of the water = m = 105 kg

Initial temperature of the water = T₁ = 26.1 ^oC

Final temperature of the ice = T₂ = 0 ^oC

Specific heat of water = C = 4186 J/kg.K

Latent heat of fusion = L = 333 kJ/kg = 333000 J/kg

Heat transferred by the water = Q

Q = mC(T₁ - T₂) + mL

Q = (105)(4186)(26.1 - 0) + (105)(333000)

Q = 4.644 x 10⁷ J

Amount of heat transferred by the water to the surrounding = 4.644 x 10⁷ J