Question

A Carnot engine whose sink is at 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 $60\%$ ?

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

Answer 1

The correct option is A $250K$

The carnot efficiency is given as $\eta =1-\frac{T_2}{T_1}$. Here $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink.
Thus we get $0.4=1-\frac{300}{T_1}$
or, $T_1=500K$
For 60% efficiency we have the $0.6=1-\frac{300}{T_1}$
or, $T_1=750$

Thus, temperature of the source should be increased by $750-500=250K$.