The ideal heat engine operating on the Carnot cycle transfers 7/8 of the amount of heat received

The ideal heat engine operating on the Carnot cycle transfers 7/8 of the amount of heat received from the heater to the refrigerator. Refrigerator temperature 7 C. Determine the heater temperature.

An ideal heat engine operating on the Carnot cycle transfers 7/8 of the amount of heat received from the heater to the refrigerator, that is, Q2 = 7/8 • Q1. Let us use the definition of the efficiency of a heat engine: η = A: Q1, where the work done is A = Q1 – Q2; A = Q1 – 7/8 • Q1; A = 1/8 Q1, we get η = (1/8 Q1): Q1, η = 1/8. On the other hand, η = (T1 – T2): T1 or η = 1 – (T2): T1. The temperature of the refrigerator is 7 ° С or Т2 = (273 + 7) К = 300 K. To determine the temperature of the heater, we express it from the formula Т1 = Т2: (1 – η). T1 = 300: (1 – 1/8). T1 = 342.8 K.
Answer: heater temperature T1 = 342.8 K.

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