Question

A Carnot engine having an efficiency of $\dfrac{1}{{10}}$ as a heat engine, is used as a refrigerator. If the work done on the system is $10\,J$, the amount of energy absorbed from the reservoir at lower temperature is:
(A) $100\,J$
(B) $1\,J$
(C) $90\,J$
(D) $99\,J$

Answer 1

Use the coefficient of the performance of the refrigerator formula, and substitute the efficiency of the Carnot engine in it by rearranging it. The obtained value of the coefficient is made substituted in the other formula given below to obtain the value of the heat absorbed.

Useful formula:
(1) The efficiency of the Carnot engine s given by
$\eta = 1 - \dfrac{{{Q_2}}}{{{Q_1}}}$
Where $\eta$ is the efficiency of the Carnot engine, ${Q_1}$ is the heat energy absorbed by the heat engine and ${Q_2}$ is the heat energy released from the Carnot engine.
(2) The coefficient of the performance of the refrigerator,
$\alpha = \dfrac{{{Q_2}}}{{{Q_1} - {Q_2}}}$
Where $\alpha$ is the coefficient of the refrigerator.
(3) The other relation for the coefficient of the performance of the refrigerator is given as
$\alpha = \dfrac{{{Q_2}}}{w}$
Where $w$ is the work done by the Carnot engine.

Complete step by step solution:
It is given that the efficiency of the Carnot engine, $\eta = \dfrac{1}{{10}}$
The work done by the engine, $w = 10\,J$
The efficiency of the Carnot engine is,
$\eta = 1 - \dfrac{{{Q_2}}}{{{Q_1}}}$
Rearranging the above formula.
$\dfrac{{{Q_2}}}{{{Q_1}}} = 1 - \eta$ ------------------(1)
Using the formula (2)
$\alpha = \dfrac{{{Q_2}}}{{{Q_1} - {Q_2}}}$
Rearranging the above formula into the substitutable equation,
$\alpha = \dfrac{{\dfrac{{{Q_2}}}{{{Q_1}}}}}{{1 - \dfrac{{{Q_2}}}{{{Q_1}}}}}$
Substituting the equation (1) in the above equation,
$\alpha = \dfrac{{1 - \eta }}{{1 - \left( {1 - \eta } \right)}}$
$\alpha = \dfrac{{1 - \eta }}{\eta }$
Using the value of the Carnot efficiency as $\dfrac{1}{{10}} = 0.1$
$\alpha = \dfrac{{1 - 0.1}}{{0.1}}$
By performing the basic arithmetic operation,
$\alpha = 9$
Using the value of the coefficient of the refrigeration in the formula (3)
$\alpha = \dfrac{{{Q_2}}}{w}$
$9 = \dfrac{{{Q_2}}}{{10}}$
${Q_2} = 90\,J$
Hence the heat absorbed by the engine is $90\,J$.

Thus the option (C) is correct.

Note: Remember the formula of the coefficient of the refrigeration used in the above solution. The Carnot engine works opposite to the heat engine by the process of the Carnot cycle. It transfers the heat from cooler to the warmer system, hence acting as a refrigerator.