The temperatures of the heater and refrigerator of an ideal heat engine are, respectively, 380 and 280 K.

The temperatures of the heater and refrigerator of an ideal heat engine are, respectively, 380 and 280 K. How much will the efficiency of the machine increase if the heater temperature is increased by 200K.

Initial data: T1.1 (initial temperature of the heater of an ideal heat engine) = 380 K; T2 (initial temperature of the refrigerator) = 280 K; Т1.2 (heater temperature after temperature increase) = Т1.1 + 200.

1) Initial efficiency of the machine: η1 = (T1.1 – T2) / T1.1 * 100% = (380 – 280) / 380 * 100% = 26.3%.

2) The efficiency of the machine after increasing the heater temperature: η2 = (T1.2 – T2) / T1.2 * 100% = (580 – 280) / 580 * 100% = 51.7%.

3) Change in the efficiency of the heat engine: η2 / η1 = 51.7 / 26.3 ≈ 2 p.