Thermodynamics Ques 41
41. 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?
[2006]
(a) $325 K$
(b) $250 K$
(c) $380 K$
(d) $275 K$
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Answer:
Correct Answer: 41.(b)
Solution:
We know that efficiency of Carnot Engine $\eta=\frac{T_1-T_2}{T_1}$ where, $T_1$ is temperature of source and $T_2$ is temperature of sink
$\therefore \quad 0.40=\frac{T_1-300}{T_1} \Rightarrow T_1-300=0.40 T_1$
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BETA
