Question

A carnots engine whose sink is at the temperature of $300K$ has an efficiency of $40$. By how much should the temperature of the source be increased so as to increase the efficiency to $50\%$ $?$

• A. $250K$
• B. $275K$
• C. $300K$
• D. $325K$

Answer 1

The correct option is A $250K$

Efficiency of heat engine $n=\frac{(T_2-T_1)}{T_2}$

where $T_2$ is source temperature and $T_1$ is the sink temperature ...

For $n=0.4$

$0.4=(T_2-T_1)/T_2$

$T_1/T_2=0.6$

As $T_1=300K...T_2=\frac{300}{0.6}=500K$

Case 2 : for efficiency n to be increased 50% of its original efficiency, i.e. $n=0.4(1+0.5)=0.6=60\%$

Source temperature can be calculated by the same formula $n=\frac{(T_2-T_1)}{T_2}$

$0.6=\frac{(T_2-300)}{T_2}$ ... (sink temperature is constant)

$T_2=\frac{300}{0.4}=750K$

Increase in source temperature

Del $T=(750-500)=250K...$

That's your answer..

The source temperature should be increased by 250 K so as to increase the efficiency of heat engine by 50% of its original efficiency.