Question

A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a large tub of ice and water. In 5 minutes of operation of the engine, the heat rejected by the engine melts a mass of ice equal to 4.30×10−2 kg . Throughout this problem use Lf=3.34×105J/kg for the heat of fusion for water. During this time, how much work W is performed by the engine?

Answer 1

The heat delivered to the ice equals mass of malted ice times heat of fusion:
Qc = m∙∆Lf = 4.30×10-^2 kg *3.34×10^5 J /kg = 14362 J

Assuming stationary operation the net energy by work and heat to the engine is zero.
That means the heat absorbed from hot reservoir equals the work done by the engine plus the heat rejected to cold reservoir:
Q_h = Q_c + W
W = Q_h - Q_c

Moreover stationary operation requires that net entropy change of the engine during the process is zero. A Carnot engine operates reversibly, i.e.no entropy is produced in the process. Hence the entropy rise due to absorption of heat from hot reservoir equals entropy loss due to heat rejection:

Q_h/T_h = Q_c/T_c
Q_h = (T_h/T_c)∙Q_c

Therefore
W = (T_h/T_c)∙Q_c - Q_c

W = [(T_h/T_c) - 1]∙Q_c
W = [(373K/273K) - 1]*14362 J
W = 5260.80J