Question

A Carnot engine has an efficiency of 1/6. When the temperature of the sink is reduced by ${62}^{0}C$, its efficiency is doubled. The temperature of the source and the sink are, respectively:
(a). ${124}^{0}C,\text{ 6}{\text{2}}^{0}C$
(b). ${37}^{0}C,\text{ 9}{\text{9}}^{0}C$
(c). ${62}^{0}C,\text{ 12}{\text{4}}^{0}C$
(d). ${99}^{0}C,\text{ 3}{\text{7}}^{0}C$

Answer 1

- Hint: The efficiency of the Carnot engine increases if the source can be maintained at high temperature and the sink can be maintained at very low temperature.

Complete step-by-step solution -

We are given a Carnot engine which has an efficiency of 1/6. Let ${\text{T}}_{\text{1}}\text{ and }{\text{T}}_2$ be the temperature of the source and the sink respectively. The efficiency of the Carnot engine is given by the formula,
$\eta ={\displaystyle \frac{{\text{T}}_{\text{1}}\text{-}{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}}$
Where,
$\eta$ is the efficiency of the Carnot engine.
It is given in the question that when the temperature of the sink is decreased by ${62}^{0}C$ the efficiency doubles. So the new efficiency after the temperature change be ${\eta }^{\prime }$. So the new efficiency will be ${\eta }^{\prime }=2\times \eta$.
So we are given $\eta ={\displaystyle \frac{1}{6}}$, so ${\eta }^{\prime }={\displaystyle \frac{1}{3}}$
${\displaystyle \frac{1}{6}}={\displaystyle \frac{{\text{T}}_{\text{1}}\text{-}{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}}=1-{\displaystyle \frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}}$ ……equation (1)
${\displaystyle \frac{1}{3}}={\displaystyle \frac{{\text{T}}_{\text{1}}\text{-(}{\text{T}}_{\text{2}}\text{-62)}}{{\text{T}}_{\text{1}}}}=1-{\displaystyle \frac{{\text{T}}_{\text{2}}}{{\text{T}}_{\text{1}}}}+{\displaystyle \frac{62}{{\text{T}}_1}}$……equation (2)

Substituting equation (1) in equation (2), we get
${\displaystyle \frac{1}{3}}={\displaystyle \frac{1}{6}}+{\displaystyle \frac{62}{{\text{T}}_1}}$
${\displaystyle \frac{1}{6}}={\displaystyle \frac{62}{{\text{T}}_1}}$
$\therefore {\text{T}}_1=372K={(372-273)}^{0}C$
$\Rightarrow {\text{T}}_1={99}^{0}C$
${\text{T}}_2={\text{T}}_1\left({\displaystyle \frac{5}{6}}\right)=372\left({\displaystyle \frac{5}{6}}\right)$
${\text{T}}_2=310K={37}^{0}C$
So the temperature of the source is ${99}^{0}C$and the temperature of the sink is $\text{3}{\text{7}}^{0}C$.
The answer to the question is option (D) ${99}^{0}C,\text{ 3}{\text{7}}^{0}C$

Additional Information: Carnot engine was proposed by Leonard Carnot to find out the theoretical efficiency possible for a heat engine. He used basic thermodynamic processes in his theoretical model. It uses the concept of converting heat energy into mechanical energy.

Note: The efficiency of the Carnot engine is defined as the ratio of work done by the engine to the amount of heat drawn from the source.