Question

A piece of solid lead weighing 45.3 g at a temperature of 308 °C is placed in 453 g of liquid lead at a temperature of 374 °C. After a while, the solid melts and a completely liquid sample remains. Calculate the temperature after thermal equilibrium is reached, assuming no heat loss to the surroundings.

The enthalpy of fusion of solid lead is ΔH_fus = 4.77 kJ/mol at its melting point of 328 °C, and the molar heat capacities for solid and liquid lead are C_solid = 26.9 J/mol K and C_liquid = 28.6 J/mol K.

T_final = ______ °C

Answer 1

Heat lost by liquid water = Heat gained by solid lead

Heat lost by water = Mass of liquid lead * specific heat capacity of liquid * Change in temperature

=> 453g * 28.6 J/gC * (374-Tf)

Heat gained by water = Heat required to convert lead from 308C to 328C + Heat required to convert solid lead at 328C to liquid lead at 328C + Heat required to raise the temperature of liquid from 328 to Tf

=> 45.3g * 26.9 J/gC * (328-308)C + 4770 J/mol * 45.3/207.2 mol + 45.3g * 28.6 J/gC * (Tf - 328)

Equating both these equations to calculate the value of Tf

453 * 28.6 * (374-Tf) = 45.3 * 26.9 * 20 + 4770 * 45.3/207.2 + 45.3 * 28.6 * (Tf-328)

Hence the value of Tf will be 368.03C