A Carnot engine operates between ${227^ \circ }C$ and ${27^ \circ }C$ . Efficiency of the engine will be
A. $\dfrac{1}{3}$
B. $\dfrac{2}{5}$
C. $\dfrac{3}{4}$
D. $\dfrac{3}{5}$

Hint:This problem is based on Carnot’s Engine Efficiency 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 ${\eta _{carnot}} = 1 - \dfrac{{{T_L}}}{{{T_H}}}$ to find the solution 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 Heat 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} = $Higher Absolute Temperature = Temperature of Hot Reservoir
Carnot Engine is operating between temperatures${T_H} = {227^ \circ }C = 500K$ and ${T_L} = {27^ \circ }C = 300K$ (given) $\left( {^ \circ C + 273 = K} \right)$
The efficiency of the engine can be calculated as: -
$\eta = 1 - \dfrac{{{T_L}}}{{{T_H}}} = 1 - \dfrac{{300}}{{500}}$
$ \Rightarrow \eta = 1 - \dfrac{3}{5} = \dfrac{2}{5}$
Thus, the efficiency of the given Carnot’s Engine is $\dfrac{2}{5}$.
Hence, the correct option is (B) $\dfrac{2}{5}$.



Thus, the correct option is B.



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.