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How many kilojoules of heat must be removed to freeze a container of water at 0°C if the water has a mass of 1.50 × 102 g?

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Step-by-step explanation:

The specific heat of fusion of water is the amount of heat that 1 g of ice at 0 °C must absorb to become 1 g of liquid water at 0 °C. It's the same amount of heat that me must remove (with different sign) from 1 g of liquid water at 0°C to convert it to 1 g of ice at 0 °C.

Speficif heat of fusion = 0.334 kJ/g

So to freeze 1 g of liquid water at 0 °C we must remove 0.334 kJ, let's find the amount of heat that we must remove from 1.50 * 10^2 g of water.

Q = - Cf * mass

Q = -0.334 kJ/g * 1.50 * 10^2 g

Q = -50.1 kJ

Answer: -50.1 kJ must be removed.

User Andrew Niefer
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