
An ideal heat engine working between temperatures ${T_1}$ and ${T_2}$ has an efficiency $\eta $ . The new efficiency if the temperatures of both the source and sink are doubled, will be
A. $\dfrac{\eta }{2}$
B. $\eta $
C. $2\eta $
D. $3\eta $
Answer
163.2k+ 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
We know that the Source behaves like a hot reservoir and the Sink behaves like a cold reservoir.
An ideal heat Engine works between the temperatures ${T_1}$ and ${T_2}$ has an efficiency $\eta $ (given). Then, from eq. (1), we get
$ \Rightarrow {\eta _{carnot}} = 1 - \dfrac{{{T_2}}}{{{T_1}}} = \eta $
$ \Rightarrow \eta = \dfrac{{{T_1} - {T_2}}}{{{T_1}}}$
Now, if both the temperatures get doubled, then new efficiency (say $\eta '$ ) will be: -
$\eta ' = \dfrac{{2{T_1} - 2{T_2}}}{{2{T_1}}}$
$ \Rightarrow \eta ' = \dfrac{{2\left( {{T_1} - {T_2}} \right)}}{{2{T_1}}} = \dfrac{{{T_1} - {T_2}}}{{{T_1}}} = \eta $
Thus, the new efficiency of an ideal heat engine when both the temperatures get doubled will remain the same i.e., $\eta $ .
Hence, the correct option is (B) $\eta $ .
Thus, the correct option is B.
Note:Since this is a multiple-choice question (conceptual-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 conceptual 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
We know that the Source behaves like a hot reservoir and the Sink behaves like a cold reservoir.
An ideal heat Engine works between the temperatures ${T_1}$ and ${T_2}$ has an efficiency $\eta $ (given). Then, from eq. (1), we get
$ \Rightarrow {\eta _{carnot}} = 1 - \dfrac{{{T_2}}}{{{T_1}}} = \eta $
$ \Rightarrow \eta = \dfrac{{{T_1} - {T_2}}}{{{T_1}}}$
Now, if both the temperatures get doubled, then new efficiency (say $\eta '$ ) will be: -
$\eta ' = \dfrac{{2{T_1} - 2{T_2}}}{{2{T_1}}}$
$ \Rightarrow \eta ' = \dfrac{{2\left( {{T_1} - {T_2}} \right)}}{{2{T_1}}} = \dfrac{{{T_1} - {T_2}}}{{{T_1}}} = \eta $
Thus, the new efficiency of an ideal heat engine when both the temperatures get doubled will remain the same i.e., $\eta $ .
Hence, the correct option is (B) $\eta $ .
Thus, the correct option is B.
Note:Since this is a multiple-choice question (conceptual-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 conceptual problem, always keep in mind to use the mathematical proven relations to find the solution.
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