Kinetic Theory Of Gases Question 10
Question: A reversible engine converts one-sixth of the heat input into work. When the temperature of the sink is reduced by $ 62{}^\circ C $ , the efficiency of the engine is doubled. The temperatures of the source and sink are
Options:
A) $ 80{}^\circ C $ , $ 37{}^\circ C $
B) $ 95{}^\circ C $ , $ 98{}^\circ C $
C) $ 90{}^\circ C $ , $ 37{}^\circ C $
D) $ 99{}^\circ C $ , $ 37{}^\circ C $
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Answer:
Correct Answer: D
Solution:
[d] Initially $ \eta =( 1-\frac{T _{2}}{T _{1}} )=\frac{W}{Q}=\frac{1}{6} $ …(i)
Finally $ \eta =( 1-\frac{T’_2}{T_1})$
=$( 1-\frac{(T _{2}-62)}{T _{1}} )=1-\frac{T _{2}}{T _{1}}+\frac{62}{T _{1}} $
$ =\eta +\frac{62}{T _{1}} $ …(ii)
It is given that $ \eta ‘=2\eta $ hence solving equation (i) and (ii) $ T _{1}=372K=99{}^\circ C $ and $ T _{2}=310K=37{}^\circ C $
BETA
