A piece of solid bismuth weighing 44.3 g at a temperature of 256 °C is placed in

Question

A piece of solid bismuth weighing 44.3 g at a temperature of 256 °C is placed in 443 g of liquid bismuth at a temperature of 328. °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 bismuth is ?Hfus = 11.0 kJ/mol at its melting point of 271 °C, and the molar heat capacities for solid and liquid bismuth are Csolid = 26.3 J/mol K and Cliquid = 31.6 J/mol K.

Explanation / Answer

Energy balance equation

Heat absorbed by solid to raise the temperature from 256 to 271 + enthalpy of fusion for phase change + heat required to raise the temperature from 271 to T = heat released by liquid from 328 to T

ms x Cps x (271 - 256) + 11 kJ/mol / molecular weight + ms x Cpl x (T - 271) = ml x Cpl x (328 - T)

(44.3 g/209g/mol) x (26.3 J/mol·K x 15 K) + (11*1000 J/mol x 1mol/209g) + (44.3 g/209g/mol) x (31.6 J/mol·K) x (T - 271) = (443 g/209g/mol) x (31.6 J/mol·K) x (328 - T)

83.62 + 52.63 + 6.69T - 1815.16 = - 66.97 T + 21966.16

23645.07 = 73.66 T

T = 321 °C = temperature after thermal equilibrium is reached

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