Question

A heat engine rejects $600$ cal to the sink at $27^{\circ }C$. Amount of work done by the engine will be (Temperature of source is $227^{\circ }C$ & $J=4.2J/cal$ )

• A. 1680 J
• B. 840 J
• C. 2520 J
• D. None of these

Answer 1

Answer: (A)

$1-\frac{T_2}{T_1}=\frac{W}{Q_1}$
$1-\frac{300}{500}=\frac{W}{Q_1}$
$\frac{W}{Q_1}=\frac{2}{5}$
$Q_1=\frac{5W}{2}$
$\because W=Q_1-Q_2$
Then $Q_2=Q_1-W$
$Q_2=\frac{5W}{2}-W=\frac{3W}{2}$
Then $W=\frac{2Q_2}{3}=\frac{2\times 600}{3}=400cal$
$=400\times 4.2J=1680J$