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

Answer 1

Answer: (B)

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

heat will be transfer from higher temperature body (SOURCE) to lower temperature body(SINK).

Work done is given as, $W=Q_1-Q_2$

So apply the relation $\frac{Q_1}{T_1}=\frac{Q_2}{T_2}$

$\frac{Q_1\times T_2}{T_1}=\frac{12.6\times {10}^6\times 300}{900}=4.2\times {10}^{6}J$

So, $W=Q_1-Q_2={12.6\times 10}^6-{4.2\times 10}^6=8.4\times {10}^6$