Q. 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) $800\ K$
Explanation
(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)
$\frac{500}{T_{2}'} = \frac{5}{4} \Rightarrow T_{2}^{'} = 400K$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.