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A 1.70-kg piece of aluminum that has a temperature of −147 °C is added to 1.00...

A 1.70-kg piece of aluminum that has a temperature of −147 °C is added to 1.00 kg of water that has a temperature of 3.1 °C. At equilibrium the temperature is 0.0°C. Assuming that the heat exchanged with the container and the surroundings is negligible, determine the mass of water that has been frozen into ice.

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Answer #1

Specific heat of aluminum = C1 = 900 J/(kg.oC)

Specific heat of water = C2 = 4186 J/(kg.oC)

Latent heat of fusion of water = L = 334000 J/kg

Mass of the piece of aluminum = m1 = 1.7 kg

Mass of water = m2 = 1 kg

Initial temperature of aluminum = T1 = -147 oC

Initial temperature of water = T2 = 3.1 oC

Final equilibrium temperature = T3 = 0 oC

Mass of water that has been frozen into ice = m3

The heat gained by the piece of aluminum is equal to the heat lost by the water.

m3 = 0.635 kg

Mass of water that has been frozen into ice = 0.635 kg

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