Question

A Carnot engine takes $3\times {10}^{6}cal$ of heat from a reservoir at ${627}^{o}C$, and gives it to a sink at ${27}^{o}C$ . The work done by the engine is

• A. $8.4\times {10}^{6}J$
• B. $16.8\times {10}^{6}J$
• C. $0$
• D. $4.2\times {10}^{6}J$

Answer 1

Answer: (A)

$8.4\times {10}^{6}J$

The efficiency of the cycle is given as $\eta =1-\frac{T_2}{T_1}=\frac{W}{Q_1}$. Here $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink. $Q_1$ is the heat taken up from the source and $W$ is the work done in the process. Thus we get $\eta =1-\frac{300}{900}=0.67=\frac{W}{3\times {10}^6}cal$

$⟹W=2.01\times {10}^{6}cal=2.01\times {10}^6\times 4.18J=8.4\times {10}^{6}J$