Question

Efficiency of a Carnot engine is $50 \%$ when temperature of outlet is $500\, K$. In order to increase efficiency up to $60 \%$ keeping temperature of intake the same what is temperature of outlet?

• A. 200 K
• B. 400 K
• C. 600 K
• D. 800 K

Answer 1

Answer: (B)

$\eta=1-\frac{T_{2}}{T_{1}}$
$\Rightarrow \frac{1}{2}=1-\frac{500}{T_{1}}$
$\Rightarrow \frac{500}{T_{1}}=\frac{1}{2}$ ...(i)
$\frac{60}{100}=1-\frac{T_{2}'}{T_{1}}$
$\Rightarrow \frac{T_{2}'}{T_{1}}=\frac{2}{5}$ ...(ii)
Dividing equation (i) by (ii),
$\frac{500}{T_{2}'}=\frac{5}{4}$
$\Rightarrow T_{2}=400\, K$