Question

A Carnot Engine, having an efficiency of \[\eta =\dfrac{1}{10}\] as 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 a lower temperature is:
(a) 90 J
(b) 1 J
(c) 100 J
(d) 99 J

Answer 1

Hint: We have been provided with a Carnot engine that works in the reversible circle. We have given an efficiency of the Carnot cycle which will be used in the refrigerator. Therefore, use the expression of the efficiency of the Carnot cycle engine. We have been provided with the value of work done and efficiency. Then put the value and find the heat added to the system. Then use the formula of work done, and again put the value of the work done and the heat added and get the value of heat or amount of the energy absorbed from the reservoir at a lower temperature.

Formulas Used:
The efficiency of the Carnot cycle is given by
\[\eta =\dfrac{W}{{{Q}_{H}}}\]
where W is the work done and \[{{Q}_{H}}\] is the amount of heat added to the system.
Work done is given by:
\[W={{Q}_{H}}-{{Q}_{C}}\]

Complete step by step answer:
We have been given a Carnot engine whose efficiency is \[\eta =\dfrac{1}{10}\] as a heat engine which is used as a refrigerator. Thus, the efficiency of the Carnot cycle is given by
\[\eta =\dfrac{W}{{{Q}_{H}}}\]
where W is the work done and \[{{Q}_{H}}\] is the amount of heat added to the system.
Then,
\[\Rightarrow {{Q}_{H}}=\dfrac{W}{\eta }\]
We have been given the work done (W) as 10 J and \[\eta =\dfrac{1}{10}.\]
\[\Rightarrow {{Q}_{H}}=\dfrac{10}{\dfrac{1}{10}}=100J\]
Let \[{{Q}_{C}}\] be the amount of heat absorbed and \[{{Q}_{C}}\] be the amount of heat added to the system, then the work done is given by
\[W={{Q}_{H}}-{{Q}_{C}}\]
\[\Rightarrow {{Q}_{C}}={{Q}_{H}}-W\]
\[\Rightarrow {{Q}_{C}}=100-10\]
\[\Rightarrow {{Q}_{C}}=90J\]
Hence, the amount of energy absorbed from the reservoir at the lower temperature is 90 J.
Therefore, option (a) is the right answer.

Additional Information:
In 1824, Carnot first introduced the theory of thermodynamics, a simple engine process which is known as the Carnot cycle. The heat engines are used to convert heat into mechanical work, Carnot also conceived a theoretical engine which is free from all the defects of the practical engine. Its efficiency is maximum and it is also an ideal heat engine consisting of source, sink, insulated stand and working substance.
Carnot’s ideal engine works in a reversible cycle, i.e. the working substance is brought back to its initial state at the end of each cycle. By using g as the working substance, Carnot’s cycle can be described.
The efficiency of the Carnot engine is given as,
\[\eta =\dfrac{\text{Amount of heat converted into work}}{\text{Amount of heat absorbed from the source}}\]

Note:
Carnot’s engine is a perfectly reversible engine. If this engine is operated in the reverse direction, then in this process, the amount of \[{{Q}_{2}}\] is absorbed (whereas in Carnot \[{{Q}_{1}}\] is absorbed and \[{{Q}_{2}}\] is rejected.) from the sink at the lower temperature \[{{T}_{2}}\] and an amount of heat rejected is \[{{Q}_{1}}\] to the source at a higher temperature \[{{T}_{1}}\] then the heat flows from a body at a lower temperature to a body at higher temperature by doing the external work on the working substance and it works as the refrigerator. The coefficient of performance of the refrigerator is defined as the reciprocal of the efficiency which is given as,
\[\beta =\dfrac{1}{\eta }=\dfrac{{{Q}_{2}}}{W}=\dfrac{{{Q}_{2}}}{{{Q}_{1}}+{{Q}_{2}}}=\dfrac{{{T}_{2}}}{{{T}_{1}}-{{T}_{2}}}\]