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.
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