Physics Question

Question

In the Carno cycle, the gas returned 4 kJ of heat to the cooler. 1 kJ of work was done per cycle. Determine the heater temperature if the cooler temperature is 27 ° C.

Answer 1

If the Carnot cycle is reversible, then

T_h/T_c = Q_h/Q_c ⇒

T_h = (Q_h/Q_c) T_c

W = Q_h - Q_c ⇒ Q_h = W + Q_c

T_h = [(W + Q_c) / Q_c] T_c

W = 1 kJ

Q_c = 4 kJ

T_c = (27 + 273)K = 300 K

T_h = {[(1 + 4) kJ] / (4 kJ)} (300 K) = 375 K = 102°C