Question

In a Carnot engine, when ${T_2} = {0^ \circ }C$ and ${T_1} = {200^ \circ }C$ , its efficiency is ${\eta _1}$​ and when ${T_1} = {0^ \circ }C$ and ${T_2} = - {200^ \circ }C$ , Its efficiency is ${\eta _2}$ ​, then what is $\dfrac{{{\eta _1}}}{{{\eta _2}}}$ ​.
A. $0.577$
B. $0.733$
C. $0.638$
D. Cannot be Calculated

Answer 1

Hint:This problem is based on Carnot Engine in a thermodynamic system, 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, use the scientific formula of calculating efficiency ${\eta _{carnot}} = 1 - \dfrac{{{T_L}}}{{{T_H}}}$ to give the required ratio in 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’s Engine is given as: -
${\eta _{carnot}} = 1 - \dfrac{{{T_L}}}{{{T_H}}}$
where ${T_L} = {T_2} = $ Lower Absolute Temperature = Temperature of Cold Reservoir
and, ${T_H} = {T_1} = $Higher Absolute Temperature = Temperature of Hot Reservoir
Case 1. ${T_1} = {200^ \circ }C = 473K$ and ${T_2} = {0^ \circ }C = 273K$ (given) $\left( {^ \circ C + 273 = K} \right)$
The efficiency of Carnot Engine in case 1 will be: -
${\eta _1} = 1 - \dfrac{{{T_2}}}{{{T_1}}} = 1 - \dfrac{{273}}{{473}}$
$ \Rightarrow {\eta _1} = \dfrac{{200}}{{473}}$ … (1)
Case 2. ${T_1} = {0^ \circ }C = 273K$ and ${T_L} = - {200^ \circ }C = 73K$ (given)
The efficiency of Carnot Engine in case 2 will be: -
${\eta _2} = 1 - \dfrac{{{T_2}}}{{{T_1}}} = 1 - \dfrac{{73}}{{273}}$
$ \Rightarrow {\eta _2} = \dfrac{{200}}{{273}}$ … (2)
The ratio of efficiencies in two cases is: -
Divide eq. (1) by (2), we get
$\dfrac{{{\eta _1}}}{{{\eta _2}}} = \dfrac{{\dfrac{{200}}{{473}}}}{{\dfrac{{200}}{{273}}}} = \dfrac{{273}}{{473}}$
$ \Rightarrow \dfrac{{{\eta _1}}}{{{\eta _2}}} = \dfrac{{273}}{{473}} = 0.577$
Thus, For the same Carnot engine in two different conditions, $\dfrac{{{\eta _1}}}{{{\eta _2}}}$is $0.577$
Hence, the correct option is (A) $0.577$.



Thus, the correct option is A.



Note:Since this is a multiple-choice question (numerical-based), it is essential that given conditions are analyzed 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.