Calculate rate of Q for a Carnot cycle heat engine

Question

A Carnot cycle heat engine has an output power of 50kW. It takes thermal energy from a heat source at 15000°C and rejects heat to a cold reservoir at 1000°C. Calculate the rate of transfer of heat to the cold reservoir in watts.

Answer 1

The efficiency of a Carnot engine is given by:

η=1−T_C/T_H

where:

T_H = 15000 + 273 = 15273 K (absolute temperature of the heat source)

T_c = 1000 + 273 = 1273 K (absolute temperature of the cold reservoir)

Now,

η=1−1273/15273=1−0.0833=0.9167

The efficiency is also given by:

η=W/Q_H

Rearrange for Q_H:

Q_H=W/η ​

Q_H=50,000/0.9167

Q_H≈54,533.5 W

The heat rejected to the cold reservoir is:

Q_C=Q_H−W

Now putting the values Q_H and W:

Q_C=54,533.5−50,000

Q_C≈4,533.5 W

Thus, the rate of heat transfer to the cold reservoir is 4,533.5 W or 4.53 kW.