Question

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

• A. 325 $K$
• B. 250 $K$
• C. 380 $K$
• D. 275 $K$

Answer 1

The correct option is B 250 $K$

Efficiency of the carnot engine, $\eta =\frac{T_1-T_2}{T_1}$
where $T_1=$Source temperature
$T_2$ = sink temperature
$0.40=\frac{T_1-300}{T_1}$
$T_1=500$ K
Since $\eta$ is increased by 50% , the new efficiency will be 60%
$0.60=\frac{T_1-300}{T_1}$
$T_1=750$ K
Increase in Temprature$=750-500=250$ K