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t2 = 17*C
t1 - t2 = T1 - T2 = 290K
szukane:
sprawność n
T2 = 273 + t2 = 273 + 17 = 290K
T1 = T2 + 290K = 290K + 290K = 580K
T1 - T2 = 290K
n = [T1-T2]/T1 = 290K/580K = 0,5 = 50%
sprawność silnika wynosi 50%