What is the maximum efficiency of an ideal heat engine if the heater temperature is 455 ° C

What is the maximum efficiency of an ideal heat engine if the heater temperature is 455 ° C, and the temperature of the refrigerator is 273 ° C?

Initial data: Тн (heater temperature) = 455 ºС; Тх (refrigerator temperature) = 273 ºС.

1) Let’s translate the temperatures of the heater and refrigerator into the Kelvin scale: Тн = 455 ºС + 273 К = 728 К; Тх = 273 ºС + 273 К = 546 K.

2) Let’s calculate the efficiency of an ideal heat engine: η = (Тн – Тх) / Тн = (728 – 546) / 728 = 0.25 (25%).

Answer: The efficiency of an ideal heat engine is 25% (0.25).