Question

A Carnot engine works between 600 K and 300 K. In each cycle of operation, the engine draws 1000 J of heat energy from the source. In order to increase the efficiency to $70\%$, source temperature will have to be

• A. increased by 400 K
• B. decreased by 400 K
• C. increased by 200 K
• D. decreased by 200 K

Answer 1

The correct option is A increased by 400 K

Given $T_L=300K$ $\therefore T_H$ for achieving $\mu =0.7$ is given by $0.7=1-\frac{300}{T_H}$ or $\frac{300}{T_H}=0.3$
or $T_H=1000K$. Since initially the engine was working between 600K and 300K, source temperature has to be increased by 1000 - 600 = 400 K.