The temperature of the heater of the ideal heat engine is 117 degrees, and the temperature of the refrigerator is 27

| April 4, 2021 | education

The temperature of the heater of the ideal heat engine is 117 degrees, and the temperature of the refrigerator is 27 degrees. The machine receives from the heater in 1 s the amount of heat equal to 200 J. Determine the efficiency of the machine, the amount of heat given to the refrigerator in 1 s, the work done by the machine in 1 s.

Initial data: Тн (machine heater temperature) = 117 ºС (390 К); Тх (refrigerator temperature) = 27 ºС (300 К); Qн (heat from the heater in 1 second) = 200 J.

1) Efficiency of the machine: η = 1 – Тх / Тн = 1 – 300/390 = 0.231 (23.1%).

2) Heat given to the refrigerator of the machine: Qх = Qн * (1 – η) = 200 * (1 – 0.231) = 153.8 J.

3) Machine operation in 1 second: A = (Qн – Qх) / t = (200 – 153.8) / 1 = 46.2 J.

Answer: The efficiency of the machine is 23.1%; the refrigerator will receive 153.8 J of heat; the machine will perform a work of 46.2 J.



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