Question

A Carnot's engine is made to work between ${200^ \circ }C$ and ${0^ \circ }C$ first and then between ${0^ \circ }C$ and $ - {200^ \circ }C$ . The ratio of efficiencies of the engine in the two cases is:
A. $1.73:1$
B. $1:1.73$
C. $1:1$
D. $1:2$

Answer 1

Hint:In the case, when the problem is based on Carnot Engine in a thermodynamic system, we know that all the parameters such as pressure, 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 solution 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 the Carnot Engine is given as: -
${\eta _{carnot}} = 1 - \dfrac{{{T_L}}}{{{T_H}}}$
where ${T_L} = $Lower Absolute Temperature = Temperature of Cold Reservoir
and, ${T_H} = $Higher Absolute Temperature = Temperature of Hot Reservoir
Case 1. ${T_H} = {200^ \circ }C = 473K$ and ${T_L} = {0^ \circ }C = 273K$ (given) $\left( {^ \circ C + 273 = K} \right)$
The efficiency of Carnot Engine in case 1 will be: -
${\eta _1} = 1 - \dfrac{{{T_L}}}{{{T_H}}} = 1 - \dfrac{{273}}{{473}}$
$ \Rightarrow {\eta _1} = \dfrac{{200}}{{473}}$ … (1)
Case 2. ${T_H} = {0^ \circ }C = 273K$ and ${T_L} = - {200^ \circ }C = 73K$ (given)
The efficiency of Carnot Engine in case 2 will be: -
${\eta _2} = 1 - \dfrac{{{T_L}}}{{{T_H}}} = 1 - \dfrac{{73}}{{273}}$
$ \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}} = \dfrac{1}{{1.73}}$
Thus, the ratio of efficiencies of the engine in the two cases is ${\eta _1}:{\eta _2} = 1:1.73$
Hence, the correct option is (B) $1:1.73$.


Thus, the correct option is B.



Note:Since this is a multiple-choice question (numerical-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.