Question

A Carnot engine takes in $3000kcal$ of heat from a reservoir at ${627}^{\circ }C$ and gives it to a sink at ${27}^{\circ }C$. 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

The correct option is B $8.4\times {10}^{6}J$

Given : $Q_1=3000\times {10}^{3}cal=3\times {10}^{6}cal$
$T_1={627}^{\circ }C=627+273=900K$
$T_2={27}^{\circ }C=27+273=300K$
Since $\frac{Q_1}{Q_2}=\frac{T_1}{T_2}\Rightarrow Q_2=\frac{T_2}{T_1}\times Q_1$
$Q_2=\frac{300}{900}\times 3\times {10}^6={10}^{6}cal$
work done $W=Q_1-Q_2$
$=3\times {10}^6-{10}^6=2\times {10}^{6}cal$
$=2\times {10}^6\times 4.2J=8.4\times {10}^{6}J$.