两热源间的可逆循环满足 qC/qH=TC/THq_C/q_H = T_C/T_HqC/qH=TC/TH,故 ∣qC∣=75|q_C| = 75∣qC∣=75 J,∣w∣=250−75=175|w| = 250-75 = 175∣w∣=250−75=175 J,效率 η=∣w∣/qH=1−TC/TH=70%\eta = |w|/q_H = 1 - T_C/T_H = 70\%η=∣w∣/qH=1−TC/TH=70%。