Question

The temperature of sink of Carnot engine is ${27^ \circ }C$ . Efficiency of engine is $25\% $ . Then temperature of source is
A. ${227^ \circ }C$
B. ${327^ \circ }C$
C. ${127^ \circ }C$
D. ${27^ \circ }C$

Answer 1

Hint:In the case, when a 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 of efficiency ${\eta _{carnot}} = 1 - \dfrac{{{T_L}}}{{{T_H}}}$ to calculate the temperature of the Source 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:
The temperature of the Cold Reservoir (Sink)${T_L} = {27^ \circ }C = 300K$ (given) $\left( {^ \circ C + 273 = K} \right)$
And, the efficiency of the Carnot heat engine ${\eta _{carnot}} = 25\% = 0.25$ (given)
The Source temperature can be evaluated as: -
$ \Rightarrow 0.25 = 1 - \dfrac{{300}}{{{T_H}}}$
$ \Rightarrow \Rightarrow \dfrac{{300}}{{{T_H}}} = 1 - 0.25 = 0.75$
On further calculating, we get
$ \Rightarrow {T_H} = \dfrac{{300}}{{0.75}} = 400K = {127^ \circ }C$
Thus, the temperature of the source of Carnot's Engine is ${127^ \circ }C$.
Hence, the correct option is (C) ${127^ \circ }C$.


Thus, the correct option is C.



Note:
Value should be there in SI systems. Since this is a problem of 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.