Question

The efficiency of carnot engine is 0.6. It rejects 20 J of heat to the sink. The work done by the engine is :

• A. 20 J
• B. 30 J
• C. 33.5 J
• D. 50 J

Answer 1

The correct option is B 30 J

As we know, $\eta =1-\frac{Q_r}{Q_{in}}$
Given $Q_r=20$.
$\eta =1-\frac{4}{10}=\frac{3}{5}$
Substituting values back n the equation will give $Q_{in}=50$
Therefore work done will be 30 joule.