Lead melts at 328 ℃. How much heat is required when 23.0 g of solid lead at 297 K condenses to a liquid at 702 K?

Heat = 1.74 kJ

Given

melts at 328 ℃ + 273 = 601 K

mass = 23 g = 0.023 kg

initial temperature = 297 K

Final tmperature = 702 K

Required

Heat

Solution

1. raise the temperature(297 to 601 K)

c of lead = 0.130 kJ/kg K

Q = 0.023 x 0.13 x (601-297)

Q = 0.909 kJ

2. phase change(solid to liquid)

Q = m.Lf (melting/freezing)

Q = 0.023 x 23 kj/kg = 0.529 kJ

3. raise the temperature(601 to 702 K)

Q = 0.023 x 0.13 x (702-601)

Q = 0.302 kJ

Total heat = 1.74 kJ

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