
The efficiency of Carnot engine when source temperature is ${T_1}$ and sink temperature is ${T_2}$ will be?
A. $\dfrac{{{T_1} - {T_2}}}{{{T_1}}}$
B. $\dfrac{{{T_2} - {T_1}}}{{{T_2}}}$
C. $\dfrac{{{T_1} - {T_2}}}{{{T_2}}}$
D. $\dfrac{{{T_1}}}{{{T_2}}}$
Answer
132k+ views
Hint:This problem is based on Carnot Engine in thermodynamics, we know that all the parameters such as temperature, heat exchange, work done, etc., vary with the given conditions of the system and surroundings hence, analyze every option given and check which option seems to be more appropriate for the given problem.
Formula Used:
The efficiency of Carnot’s Heat Engine is given as: -
${\eta _{carnot}} = 1 - \dfrac{{{T_L}}}{{{T_H}}}$
where ${T_L} = $Lower Absolute Temperature = Temperature of the Sink
and, ${T_H} = $Higher Absolute Temperature = Temperature of the source
Complete answer:
We know that, the efficiency of Carnot Heat Engine is given as: -
${\eta _{carnot}} = 1 - \dfrac{{{T_L}}}{{{T_H}}}$ … (1)
where, ${T_L} = $Lower Absolute Temperature = Temperature of Cold Reservoir
and, ${T_H} = $Higher Absolute Temperature = Temperature of Hot Reservoir
Now, by the definition of source and sink, we know that source is a kind of reservoir that supplies an infinite amount of heat and sink is a kind of reservoir that absorbs infinite amount of heat.
Therefore, we can say that Source behaves like a hot reservoir and Sink behaves like a cold reservoir.
As the temperature of the Source is ${T_1}$ and temperature of the Sink is ${T_2}$(given). Then, from eq. (1), we get
$ \Rightarrow {\eta _{carnot}} = 1 - \dfrac{{{T_2}}}{{{T_1}}}$
$ \Rightarrow {\eta _{carnot}} = \dfrac{{{T_1} - {T_2}}}{{{T_1}}}$
Thus, the efficiency of Carnot engine when source temperature is ${T_1}$ and sink temperature is ${T_2}$ will be $\dfrac{{{T_1} - {T_2}}}{{{T_1}}}$ .
Hence, the correct option is (A) $\dfrac{{{T_1} - {T_2}}}{{{T_1}}}$.
Thus, the correct option is A.
Note:Since this is a multiple-choice question (derivation-based) hence, it is essential that given conditions are analyzed very carefully to give an accurate solution. While writing an answer to this kind of numerical problem, always keep in mind to use the mathematical proven relations to find the solution.
Formula Used:
The efficiency of Carnot’s Heat Engine is given as: -
${\eta _{carnot}} = 1 - \dfrac{{{T_L}}}{{{T_H}}}$
where ${T_L} = $Lower Absolute Temperature = Temperature of the Sink
and, ${T_H} = $Higher Absolute Temperature = Temperature of the source
Complete answer:
We know that, the efficiency of Carnot Heat Engine is given as: -
${\eta _{carnot}} = 1 - \dfrac{{{T_L}}}{{{T_H}}}$ … (1)
where, ${T_L} = $Lower Absolute Temperature = Temperature of Cold Reservoir
and, ${T_H} = $Higher Absolute Temperature = Temperature of Hot Reservoir
Now, by the definition of source and sink, we know that source is a kind of reservoir that supplies an infinite amount of heat and sink is a kind of reservoir that absorbs infinite amount of heat.
Therefore, we can say that Source behaves like a hot reservoir and Sink behaves like a cold reservoir.
As the temperature of the Source is ${T_1}$ and temperature of the Sink is ${T_2}$(given). Then, from eq. (1), we get
$ \Rightarrow {\eta _{carnot}} = 1 - \dfrac{{{T_2}}}{{{T_1}}}$
$ \Rightarrow {\eta _{carnot}} = \dfrac{{{T_1} - {T_2}}}{{{T_1}}}$
Thus, the efficiency of Carnot engine when source temperature is ${T_1}$ and sink temperature is ${T_2}$ will be $\dfrac{{{T_1} - {T_2}}}{{{T_1}}}$ .
Hence, the correct option is (A) $\dfrac{{{T_1} - {T_2}}}{{{T_1}}}$.
Thus, the correct option is A.
Note:Since this is a multiple-choice question (derivation-based) hence, it is essential that given conditions are analyzed very carefully to give an accurate solution. While writing an answer to this kind of numerical problem, always keep in mind to use the mathematical proven relations to find the solution.
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