
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
232.5k+ 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.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

