The ideal heat engine of Carnot performs 75 kJ in one cycle. Heater temperature 100 degrees Celsius,

The ideal heat engine of Carnot performs 75 kJ in one cycle. Heater temperature 100 degrees Celsius, refrigerator 0 degrees. Find the efficiency of the cycle.

The efficiency of the Carnot cycle is determined by the formula:
η = (Tn-To) / Tn * 100% Since in a particular problem Tn = 100C = 373K, To = 0C = 273 K we get:
η = (373-273) / 373 * 100% = 26.8%
It should be noted that the efficiency does not depend on the work being done; this condition in the problem is superfluous.