Question

A Carnot's engine works as a refrigerator between $250K$ and $300K$. It receives $500cal$ heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is:
(A) $420J$
(B) $2100J$
(C) $772J$
(D) $2520J$

Answer 1

A Carnot engine takes an amount of heat ${Q_1}$ from the high temperature body, now $W$ amount of work is done and ${Q_2}$ amount of heat is given to the low temperature body. This whole process is carried by a Carnot engine. If the whole process is reversed, then we get a Carnot engine that works as a refrigerator. Apply the same method used in the Carnot engine and derive an expression for the Carnot refrigerator.

Complete step by step answer:
When a Carnot engine works as a refrigerator, a total of ${Q_1}$ amount of heat is taken from the cold body, $W$ amount of work is done on the system and heat transferred to the hot body ${Q_2}$.
So, we have${Q_2} = {Q_1} + W$.
As the heat ${Q_1}$ is supplied to the system as isothermal process at temperature ${T_1}$, the entropy of the system increases and the change in entropy is given by${S_2} - {S_1} = \dfrac{{{Q_1}}}{{{T_1}}}$. Note that this whole process on which a Carnot engine works is a cyclic process. Now, when the heat is given to the hot body, the entropy of the system will decrease and will be exactly equal to the negative of the change in the first process as it is a cyclic process. Hence, we have change in entropy in the isothermal process at temperature ${T_2}$ as ${S_1} - {S_2} = \dfrac{{ - {Q_2}}}{{{T_2}}} \to {S_2} - {S_1} = \dfrac{{{Q_2}}}{{{T_2}}}$.
Now, we can obtain a relation between heats and temperatures,
$\dfrac{{{Q_1}}}{{{T_1}}} = \dfrac{{{Q_2}}}{{{T_2}}} \\\ \implies \dfrac{{{Q_1}}}{{{Q_2}}} = \dfrac{{{T_1}}}{{{T_2}}} \\\$
As we have derived that${Q_2} = {Q_1} + W$, substituting this in the above expression, we get,
$\dfrac{{{Q_1}}}{{{Q_1} + W}} = \dfrac{{{T_1}}}{{{T_2}}} \\\ \implies W = {Q_1}\left( {\dfrac{{{T_2}}}{{{T_1}}} - 1} \right) \\\$
Now, from the given data in the question, heat received by the system from the cold body is ${Q_1} = 500cal$, the temperatures are ${T_2} = 300K$ (temperature of hot body) and ${T_1} = 250K$ (temperature of cold body). Therefore, substituting all the values in the work expression, we get,
$W = \left( {500} \right)\left( {\dfrac{{300}}{{250}} - 1} \right) \\\ \implies W = 100cal \\\$
Now, $1cal = 4.2J$, therefore $W = 100 \times 4.2 = 420J$.
Hence, the amount of work done in each cycle to operate the refrigerator is $420J$.

So, the correct answer is “Option A”.

Additional Information:
The actual process is as follows: The system is at $\left( {{S_1},{T_1}} \right)$, now an isothermal process is carried out which takes the system to $\left( {{S_2},{T_1}} \right)$. After this, the system is taken through an adiabatic process which brings the system to $\left( {{S_2},{T_2}} \right)$. The system is now taken to $\left( {{S_1},{T_2}} \right)$ through an isothermal process and finally after this, again an adiabatic process is carried out which brings the system back to $\left( {{S_1},{T_1}} \right)$. From here you can understand the whole process.

Note:
Remember that the whole Carnot engine works on a cyclic process as discussed above, whether normal or as a refrigerator. Also remember that how is the heat taken, heat given and work is done. Keep in mind that the temperature is always taken in kelvins and also that $1cal = 4.2joules$. Also keep in mind how we derived the expression for the relation between the heats, work done on/by the system and the temperatures of the hot and cold bodies.