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,
$\eta =\dfrac{{{\text{T}}_{\text{1}}}\text{-}{{\text{T}}_{\text{2}}}}{{{\text{T}}_{\text{1}}}}$
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 $\eta '$. So the new efficiency will be $\eta '=2\times \eta $.
So we are given $\eta =\dfrac{1}{6}$, so $\eta '=\dfrac{1}{3}$
$\dfrac{1}{6}=\dfrac{{{\text{T}}_{\text{1}}}\text{-}{{\text{T}}_{\text{2}}}}{{{\text{T}}_{\text{1}}}}=1-\dfrac{{{\text{T}}_{\text{2}}}}{{{\text{T}}_{\text{1}}}}$ ……equation (1)
$\dfrac{1}{3}=\dfrac{{{\text{T}}_{\text{1}}}\text{-(}{{\text{T}}_{\text{2}}}\text{-62)}}{{{\text{T}}_{\text{1}}}}=1-\dfrac{{{\text{T}}_{\text{2}}}}{{{\text{T}}_{\text{1}}}}+\dfrac{62}{{{\text{T}}_{1}}}$……equation (2)

Substituting equation (1) in equation (2), we get
$\dfrac{1}{3}=\dfrac{1}{6}+\dfrac{62}{{{\text{T}}_{1}}}$
$\dfrac{1}{6}=\dfrac{62}{{{\text{T}}_{1}}}$
$\therefore \text{ }{{\text{T}}_{1}}=372K={{(372-273)}^{0}}C$
$\Rightarrow \text{ }{{\text{T}}_{1}}={{99}^{0}}C$
${{\text{T}}_{2}}={{\text{T}}_{1}}\left( \dfrac{5}{6} \right)=372\left( \dfrac{5}{6} \right)$
${{\text{T}}_{2}}=310K={{37}^{0}}C$
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.