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 122 kg and its initial temperature is 22.7°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.

E = 5.221 × 10⁷ J

Explanation:

Given:

Mass of water, m = 122 kg

Initial temperature of the water, T₁ = 22.7°C

Specific heat of the water, Cp = 4186 J/kg.K

Latent heat of fusion, = 333 kJ/kg = 333 × 10³ J/kg

Now,

The for energy (E) by the water during freezing is as:

E = Energy released from 22.7°C to get to 0°C + energy released from 0°C to freezing

or

E = mCpΔT +

on substituting the values, we get

E = 122 × 4186 × (22.7° - 0°) + 122 × 333 × 10³

or

E = 5.221 × 10⁷ J

Hence, the total energy released by the water is 5.221 × 10⁷ J

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