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# A carnot engine, having an efficiency of $\dfrac{1}{10}$ as heat engine, is used as a refrigerator. If the work done on the system is $10J$, the amount of energy absorbed from the reservoir at low temperature is $A)\text{ }99J$ $B)\text{ }90J$$C)\text{ 1}J$$D)\text{ 100}J$  Verified
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Hint: This problem can be solved by using the direct formula for the efficiency of a refrigerator in terms of the heat absorbed from the reservoir at low temperature and the work done on the refrigerator. Also, the efficiency of a refrigerator can be written in terms of its efficiency when used as a heat engine.

Formula used: ${{\eta }_{f}}=\dfrac{1-{{\eta }_{e}}}{{{\eta }_{e}}}$
${{\eta }_{f}}=\dfrac{{{Q}_{2}}}{W}$

Let us write the formula for the efficiency of a refrigerator in terms of the heat energy absorbed from the reservoir at low temperature and the work done on the system.
The efficiency ${{\eta }_{f}}$ of a refrigerator when ${{Q}_{2}}$ amount of heat energy is taken from the reservoir at low temperature and $W$ amount of work is done on the system is given as
${{\eta }_{f}}=\dfrac{{{Q}_{2}}}{W}$ --(1)
Also, the efficiency ${{\eta }_{f}}$ of a refrigerator in terms of its efficiency ${{\eta }_{e}}$ when used as a heat engine is given by
${{\eta }_{f}}=\dfrac{1-{{\eta }_{e}}}{{{\eta }_{e}}}$ --(2)
Now, let us analyze the question.
The work done on a refrigerator system is $W=10J$.
The efficiency of the refrigerator when used as a heat engine is ${{\eta }_{e}}=\dfrac{1}{10}$.
Let the efficiency of the refrigerator be ${{\eta }_{f}}$.
Let the amount of heat energy absorbed from the reservoir at low temperature be ${{Q}_{2}}$.
Therefore, using (1), we get
${{\eta }_{f}}=\dfrac{{{Q}_{2}}}{W}$ --(3)
Also, using (2) we get
${{\eta }_{f}}=\dfrac{1-{{\eta }_{e}}}{{{\eta }_{e}}}$ --(4)
Equating (3) and (4), we get
$\dfrac{{{Q}_{2}}}{W}=\dfrac{1-{{\eta }_{e}}}{{{\eta }_{e}}}$
Putting the values of the variables in the above equation, we get
$\dfrac{{{Q}_{2}}}{10}=\dfrac{1-\dfrac{1}{10}}{\dfrac{1}{10}}=\dfrac{\dfrac{10-1}{10}}{\dfrac{1}{10}}=\dfrac{\dfrac{9}{10}}{\dfrac{1}{10}}=9$
$\therefore {{Q}_{2}}=9\times 10=90J$
Therefore, we have got the heat absorbed from the reservoir at low temperature as $90J$.

So, the correct answer is “Option B”.

Note: The relation between the efficiency of a refrigerator in terms of the efficiency when used as a heat engine comes from the fact that a heat engine and a refrigerator are essentially reverse processes of each other. In a heat engine, heat is taken from a reservoir at high temperature to do some work and then the rest is dumped into a reservoir at low temperature. On the other hand in a refrigerator, the reverse happens. Heat is taken from a reservoir at low temperature, some work is done upon the system and the heat is released into a reservoir at high temperature. From this knowledge, the efficiency of a heat engine can be written in terms of the efficiency when used as a refrigerator and vice versa.