Question

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

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

Answer 1

The correct option is B $400K$

Efficiency,$\eta =1-\frac{T_2}{T_1}$

$T_1=$Temperature of Intake

$T_2=$Temperature of Outlet

$\therefore \frac{50}{100}=1-\frac{500}{T_1}\frac{500}{T_1}=\frac{1}{2}\therefore T_1=1000K$

Now,if $\eta =60\frac{60}{100}=1-\frac{T_2}{1000}\frac{T_2}{1000}=1-\frac{60}{100}=\frac{40}{100}\therefore T_2=400K$