Question

A reversible engine concerts 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
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 ${}^\circ C$

Answer 1

A reversible heat engine converts heat to work. It gets heat as energy from the source and converts part of it to work and the rest is discharged in the sink. The ratio of Work Done to Heat is called the efficiency of the engine. It is also expressed in heat both the expressions are given below.
Formula Used:
$\eta =\dfrac{\left( {{T}_{2}}-{{T}_{1}} \right)}{{{T}_{2}}}$
$\eta =\dfrac{W}{Q}$

Complete answer:
A reversible heat engine converts heat to work. It gets heat as energy from the source and converts part of it to work and the rest is discharged in the sink. The ratio of Work Done to Heat is called the efficiency of the engine. It is formulated as
$\eta =\dfrac{W}{Q}$
The same relation can be written in terms of temperature
$\eta =\dfrac{\left( {{T}_{2}}-{{T}_{1}} \right)}{{{T}_{2}}}$
Where ${{T}_{2}}$is the source temperature and ${{T}_{1}}$ is the sink temperature. Here the temperature is in Kelvin
In the given question the efficiency is given
${{\eta }_{1}}=\dfrac{W}{Q}$
And

& \dfrac{W}{Q}=\dfrac{1}{6} \\\ & \Rightarrow {{\eta }_{1}}=\dfrac{1}{6} \\\ \end{aligned}$$ Writing in terms of temperature $$\begin{aligned} & {{\eta }_{1}}=\dfrac{1}{6} \\\ & \Rightarrow \dfrac{\left( {{T}_{2}}-{{T}_{1}} \right)}{{{T}_{2}}}=\dfrac{1}{6} \\\ & \Rightarrow 6{{T}_{2}}-6{{T}_{1}}={{T}_{2}} \\\ \end{aligned}$$ $$\Rightarrow {{T}_{2}}=1.2\times {{T}_{2}}$$……….(1) Now When the temperature of the sink is reduced by 62 $${}^\circ C$$, the efficiency of the engine is doubled. So, $${{\eta }_{2}}=2{{\eta }_{1}}=2\times \dfrac{1}{6}=\dfrac{2}{6}$$ In terms of temperature $$\begin{aligned} & {{\eta }_{2}}=\dfrac{\left( {{T}_{2}}-\left( {{T}_{1}}-62 \right) \right)}{{{T}_{2}}}=\dfrac{2}{6} \\\ & \Rightarrow 6{{T}_{2}}-6{{T}_{1}}+372=2{{T}_{2}} \\\ & \Rightarrow 6{{T}_{2}}-2{{T}_{2}}-6{{T}_{1}}+372=0 \\\ & \Rightarrow 4{{T}_{2}}-6{{T}_{1}}+372=0 \\\ \end{aligned}$$ Plugging in Value of $${{T}_{2}}$$ from (1) $$\begin{aligned} & 4(1.2{{T}_{1}})6{{T}_{1}}+372=0 \\\ & \Rightarrow 4.8{{T}_{1}}-6{{T}_{1}}+372=0 \\\ & \Rightarrow 372=1.2{{T}_{1}} \\\ & \Rightarrow {{T}_{1}}=310K \\\ & \Rightarrow {{T}_{1}}=37{}^\circ C \\\ \end{aligned}$$ Putting Value of $$T$$in (1) $$\begin{aligned} & {{T}_{2}}=1.2\times {{T}_{2}} \\\ & \Rightarrow {{T}_{2}}=1.2\times 310=372K \\\ & \Rightarrow {{T}_{2}}=99{}^\circ C \\\ \end{aligned}$$ As $${{T}_{2}}$$is the source temperature and $${{T}_{1}}$$ is the sink temperature. Therefore Source temperature is $$99{}^\circ C$$and Sink Temperature is $$37{}^\circ C$$ **So, Option A is correct.** **Note:** In order to avoid confusion, it is important to note that the engine converts Heat of Source to Work and discharges the excess heat to sink. So, the temperature of the Sink cannot be more than that of the source. At worst they can be equal when $$\eta =0$$.