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