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)

400 K

Answer 2

$\eta = 1 - \frac{T_{2}}{T_{1}}$⇒$\frac{1}{2} = 1 - \frac{500}{T_{1}}$⇒ $\frac{500}{T_{1}} = \frac{1}{2}$ …..(i)

$\frac{60}{100} = 1 - \frac{T_{2}'}{T_{1}}$⇒$\frac{T_{2}'}{T_{1}} = \frac{2}{5}$ …..(ii)

Dividing equation (i) by (ii), $\frac{500}{T_{2}'} = \frac{5}{4}$⇒$T_{2} = 400K$