Question

The temperature of reservoir of Carnot's engine operating with an efficiency of $70\% $ is $1000K$ . The temperature of its sink is
A. $300K$
B. $400K$
C. $500K$
D. $700K$

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 Sink 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 the Sink
and, ${T_H} = $Higher Absolute Temperature = Temperature of the source
The temperature of the Reservoir (Source)${T_H} = 1000K$ (given)
And, the efficiency of the Carnot heat engine ${\eta _{carnot}} = 70\% = 0.7$ (given)
The temperature of the Sink can be calculated as: -
$ \Rightarrow 0.7 = 1 - \dfrac{{{T_L}}}{{1000}}$
$ \Rightarrow \dfrac{{{T_L}}}{{1000}} = 1 - 0.7 = 0.3$
On further calculating, we get
$ \Rightarrow {T_L} = 0.3 \times 1000 = 300K$
Thus, the temperature of the sink of the given Carnot’s Engine is $300K$.
Hence, the correct option is (A) $300K$.



Thus, the correct option is SA.



Note: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.