Question

For a Carnot engine, the source is at $500\, K$ and the sink at $300\, K$. What is efficiency of this engine?

• A. 0.2
• B. 0.4
• C. 0.6
• D. 0.3

Answer 1

Answer: (B)

0.4

Answer 2

$\eta=\frac{T_{1}-T_{2}}{T_{1}}$
where, $\eta=$ efficiency of engine
$T_{1}=$ temperature of source $=500\, K$
$T_{2}=$temperature of $\sin k=300\, K$
$\eta=\frac{500-300}{500}=\frac{200}{500}=0.4$