The heat of fusion for lead at 327 °C is 22.9 kJ/kg and the specific heat of solid lead is 0.130 kJ/kg °C, the specific heat of liquid lead is 0.0900 kJ/kg °C, and the heat of vaporization of lead at 1,620 °C is 871 kJ/kg. What is the energy needed to vaporize 100 grams of lead starting at 0 °C?

Question

Quizplus · Accepted Answer

The energy needed to vaporize 100 grams of lead starting at 0 °C is the sum of the energy needed to raise the temperature of the lead from 0 °C to 327 °C, the energy needed to melt the lead at 327 °C, and the energy needed to vaporize the lead at 1,620 °C. The energy needed to raise the temperature of the lead from 0 °C to 327 °C is:```Q1 = m * c * ΔT = (0.1 kg) * (0.130 kJ/kg °C) * (327 °C - 0 °C) = 4.28 kJ```The energy needed to melt the lead at 327 °C is:```Q2 = m * Lf = (0.1 kg) * (22.9 kJ/kg) = 2.29 kJ```The energy needed to vaporize the lead at 1,620 °C is:```Q3 = m * Lv = (0.1 kg) * (871 kJ/kg) = 87.1 kJ```Therefore, the total energy needed to vaporize 100 grams of lead starting at 0 °C is:```Q = Q1 + Q2 + Q3 = 4.28 kJ + 2.29 kJ + 87.1 kJ = 93.67 kJ```The closest answer choice is A) 105 kJ.

The Heat of Fusion for Lead at 327 °C Is