What is the efficiency of an ideal heat engine if the temperature of the refrigerator is 100 ºС

What is the efficiency of an ideal heat engine if the temperature of the refrigerator is 100 ºС, and the temperature of the heater is three times higher?

The efficiency of an ideal heat engine can be calculated using the formula:
Efficiency = ((T1 – T2) / T1) * 100%, where T1 is the heater temperature (ºC), T2 is the refrigerator temperature (T2 = 100 ºC = 373 K).
Since the temperature of the heater is three times higher than the temperature of the refrigerator, then Т1 = 3 * Т2 = 3 * 100 = 300 ºС = 573 K.
Let’s calculate the efficiency:
Efficiency = ((T1 – T2) / T1) * 100% = (573 – 373) / 573) * 100% = (200/573) * 100% = 34.9%.
Answer. The efficiency of an ideal heat engine is 34.9%.

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