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. $600K$
• C. $400K$
• D. $800K$

Answer 1

The correct option is C $400K$

When efficiency is 50$\%$, outlet temperature, $T_2=500K$

We know,

$\eta =1-\frac{T_2}{T_1}⟹\frac{50}{100}=1-\frac{500}{T_1}⟹\frac{500}{T_1}=\frac{100-50}{100}⟹T_1=\frac{500\times 100}{50}=1000K$

To increase the efficiency upto 60$\%$, with $T_1=1000K$, then

$\eta =1-\frac{T_2}{T_1}⟹\frac{60}{100}=1-\frac{T_2}{1000}⟹\frac{T_2}{1000}=\frac{100-60}{100}⟹T_2=\frac{1000\times 40}{100}=400K$

Is the required outlet temperature.