Question

An ideal reversible heat engine converts one sixth of the heat input into work. If the temperature of the sink is reduced by ${62}^{\circ }C$, its efficiency is doubled. Find the temperature of the source and sink respectively (In $\text{Kelvin}$).

• A. $325,375$
• B. $372,310$
• C. $372,248$
• D. $472,310$

Answer 1

The correct option is B $372,310$

An ideal heat engine is a Carnot engine.
Given ,$W=\frac{Q_1}{6}$
$\Rightarrow \eta =\frac{W}{Q_1}=\frac{1}{6}$
Also, $\eta =1-\frac{T_2}{T_1}$
$\therefore 1-\frac{T_2}{T_1}=\frac{1}{6}$
$\Rightarrow \frac{T_2}{T_1}=1-\frac{1}{6}=\frac{5}{6}...\left(i\right)$
when temperature of sink is reduced, $T_2^{\prime }=T_2-62$ and the efficiency $\eta$ becomes twice i.e ${\eta }^{\prime }=\frac{1}{3}$
Thus,
${\eta }^{\prime }=1-\frac{T_2^{\prime }}{T_1}$
$\Rightarrow \frac{1}{3}=1-\frac{(T_2-62)}{T_1}$
$\Rightarrow \frac{T_2-62}{T_1}=\frac{2}{3}$
$\Rightarrow \frac{T_2}{T_1}-\frac{62}{T_1}=\frac{2}{3}...\left(ii\right)$
From Eq.$\left(i\right)\&\left(ii\right)$,
$\frac{62}{T_1}+\frac{2}{3}=\frac{5}{6}$
$\Rightarrow T_1=372\text{K}...\left(iii\right)$
From Eq.$\left(iii\right),\left(i\right)$ we get,
$T_2=\frac{5T_1}{6}$
$\therefore T_2=\frac{5}{6}\times 372=310\text{K}$
Hence temperature of source and sink are respectively $372\text{K},310\text{K}$