Thermodynamics - Result Question 49
51. 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
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(a) $99^{\circ} C, 37^{\circ} C$
(b) $80^{\circ} C, 37^{\circ} C$
(c) $95^{\circ} C, 37^{\circ} C$
(d) $90^{\circ} C, 37^{\circ} C$
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Answer:
Correct Answer: 51. (a)
Solution:
- (a) Initially the efficiency of the engine was $\frac{1}{6}$ which increases to $\frac{1}{3}$ when the sink temperature reduces by $62^{\circ} C$.
$\eta=\frac{1}{6}=1-\frac{T_2}{T_1}$, when $T_2=$ sink temperature
$T_1$ source temperature $\Rightarrow T_2=\frac{5}{6} T_1$
Secondly,
$\frac{1}{3}=1-\frac{T_2-62}{T_1}=1-\frac{T_2}{T_1}+\frac{62}{T_1}=1-\frac{5}{6}+\frac{62}{T_1}$
or, $T_1=62 \times 6=372 K=372-273=99^{\circ} C$
$& T_2=\frac{5}{6} \times 372=310 K=310-273=37^{\circ} C$
52 .
(d) $\eta=1-\frac{T_1}{T_2}$
$T_1=-23^{\circ} C=250 K, T_2=100^{\circ} C=373 K$
$\eta=1-\frac{250}{373}=\frac{373-250}{373}$
