The heat of fusion for lead at 327.0 °C is 22.9 kJ/kg and the specific heat of lead is 0.130 kJ/kg °C. What is the energy needed to melt 100 grams of lead starting at 0 °C?

To solve this problem, we need to follow these steps:

1. First, we need to calculate the energy required to heat the lead from 0 °C to its melting point (327.0 °C). We can use the formula for heat transfer: Q = mcΔT, where m is the mass, c is the specific heat, and ΔT is the change in temperature.

   In this case, m = 100 g = 0.1 kg (since 1 kg = 1000 g), c = 0.130 kJ/kg °C, and ΔT = 327.0 °C - 0 °C = 327.0 °C.

   So, Q = 0.1 kg * 0.130 kJ/kg °C * 327.0 °C = 4.251 kJ.

2. Next, we need to calculate the energy required to melt the lead at its melting point. We can use the formula Q = mL, where m is the mass and L is the heat of fusion.

   In this case, m = 0.1 kg and L = 22.9 kJ/kg.

   So, Q = 0.1 kg * 22.9 kJ/kg = 2.29 kJ.

3. Finally, we add the energy required to heat the lead to its melting point and the energy required to melt it to find the total energy required.

   So, the total energy required is 4.251 kJ + 2.29 kJ = 6.541 kJ.