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

Answer 1

Answer: (D)

250 K

Answer 2

$\eta =1-\frac{ T _{2}}{ T _{1}}$
$\Rightarrow 1-\frac{300}{ T _{1}}=0.4$
$\Rightarrow T _{1}=500 \,K$
now $\eta\, \epsilon=0.4+0.4 \times \frac{50}{100}=0.6$
Therefore $0.6=1-\frac{300}{500+\Delta T }$
$\Rightarrow 500+\Delta T =750$
$\Rightarrow \Delta T =250\, K$