Question

A reversible heat engine converts one-fourth of the input into work. When the temperature of the sink is reduced by $52\text{K}$, its efficiency is doubled. The temperature in Kelvin of the source will be

Answer 1

Temperature of the source $\to T_1\text{(in K)}$
Temperature of the sink $\to T_2\text{(in K)}$

So efficiency of the engine,

$\eta =1-\frac{T_2}{T_1}=\frac{W}{Q}$

Here, $W=\frac{Q}{4}$

$\Rightarrow \frac{1}{4}=1-\frac{T_2}{T_1}$

$\Rightarrow \frac{T_2}{T_1}=1-\frac{1}{4}=\frac{3}{4}........\left(1\right)$

Now when temperature of the sink decreased by $52\text{K}$, then the efficiency of the engine

${\eta }^{\prime }=2\times \eta =2\times \frac{1}{4}=\frac{1}{2}$

${\eta }^{\prime }=1-\frac{T_2^{\prime }}{T_1}$

$\Rightarrow \frac{1}{2}=1-\frac{T_2-52}{T_1}$

$\Rightarrow \frac{T_2-52}{T_1}=\frac{1}{2}$

$\Rightarrow T_1=2T_2-104$

Now, from equation $\left(1\right)$, $T_2=\frac{3}{4}{T}_1$

$\Rightarrow T_1=2\times \frac{3}{4}{T}_1-104$

$\Rightarrow 104=\frac{1}{2}{T}_1$

$\Rightarrow T_1=208\text{K}.$

Hence, $208$ is the correct answer.