Thermal Physics - Carnot Cycle

An engine absorbs heat at a temperature of 727^oC and exhaust heat at a temperature of 527^oC. If the engine operates at maximum possible efficiency, for 2000 joules of heat input the mount of work the engine performs is most nearly 

A. 400 J
B. 1450 J
C. 1600 J
D. 2000 J
E. 2760 J

(GR9277 #16)

Solution:

T_H = 727^oC = 727 + 273 = 1000 K
T_C = 527^oC = 527 + 273 = 800 K

Q_H = 2000 J
W = ?

Efficiency of heat engine cycle: η = W/Q_H

Carnot efficiency: η = (T_H − T_C) /T_H = 1 − T_C/T_H

W/Q_H = 1 − T_C/T_H
W = Q_H (1 − T_C/T_H) =  2000  (1 − 800/1000) = 400 J

Answer: A