Problem 5. Some lead having a specific heat capacity of 130 J/ (kg °C) is heated from 27°C to its melting point at 327°C. If the quantity of heat required is 780 kJ, determine the mass of the lead. Quantity of heat, Q = mc ( t 2 – t 1), hence, 780 × 103 J = m × 130 J/ (kg °C) × (327 – 27)°C i.e. 780000 = m × 130 × 300.