In an ideal heat engine, the temperature of the heater is 157C, and the temperature of the refrigerator is 28C.

In an ideal heat engine, the temperature of the heater is 157C, and the temperature of the refrigerator is 28C. the heater transferred 10 ^ 5rJ of heat. what work did the engine do?

Initial data: T1 (absolute temperature of the heater) = 430 K (157 ºС); T2 (absolute temperature of the refrigerator) = 301 K (28 ºС); Q (the amount of heat transferred by the heater) = 10 ^ 5 kJ.

1) The efficiency of an ideal heat engine: η = (T1 – T2) / T1 = 430 – 301/430 = 0.3 (30%).

2) The work performed by the heat engine: A = η * Q = 0.3 * 10 ^ 5 = 30 * 10 ^ 3 J (30 kJ).

Answer: The heat engine performed work of 30 kJ.