Question

Carnot’s engine takes in a thousand kilocalories of heat from a reservoir at $ 627^\circ C $ and exhausts it to a sink at $ 27^\circ C $ . What is its efficiency?

Answer 1

Hint The efficiency of a Carnot’s engine is given by the ratio of the work done to the heat supplied. It can also be written in the terms of the temperature of the source and the sink. Using that formula and substituting the temperatures in the Kelvin scale, we get the efficiency.

Formula Used: In this solution we will be using the following formula,
 $\Rightarrow \eta = \dfrac{{{T_1} - {T_2}}}{{{T_1}}} $
Where $ \eta $ is the efficiency of the engine, $ {T_1} $ is the temperature of the source and $ {T_2} $ is the temperature of the sink.

Complete step by step answer
The efficiency of a Carnot’s engine is theoretically the maximum efficiency that one can get when the heat engine is operating between 2 given temperatures.
If we consider the temperature of the source as $ {T_1} $ and the temperature of the sink as $ {T_2} $ , where both the temperatures are in Kelvin scale, then the efficiency of the Carnot’s engine is given by the formula,
 $\Rightarrow \eta = \dfrac{{{T_1} - {T_2}}}{{{T_1}}} $
In the question we are given that the temperature of the source is $ 627^\circ C $ . So we need to convert it to Kelvin. Hence, $ {T_1} = \left( {627 + 273} \right)K $ , where on adding the values we get
 $\Rightarrow {T_1} = 900K $
And we are given that the temperature of the sink is $ 27^\circ C $ . So we need to convert it to Kelvin again. Hence, $ {T_2} = \left( {27 + 273} \right)K $ , where on adding the values we get
 $\Rightarrow {T_2} = 300K $
So substituting these values in the equation for the efficiency we get,
 $\Rightarrow \eta = \dfrac{{900 - 300}}{{900}} $
Therefore we get on the numerator,
 $\Rightarrow \eta = \dfrac{{600}}{{900}} $
On doing the division we have,
 $\Rightarrow \eta = 0.667 $
To find the percentage efficiency, we multiply it by $ 100\% $
So we get,
 $\Rightarrow \eta = 0.667 \times 100\% = 66.7\% $
Therefore, we get the efficiency of the Carnot’s engine as, $ \eta = 66.7\% $.

Note
The Carnot’s engine is a theoretical engine that operates on the Carnot’s cycle. A heat engine goes through a process of the thermodynamic cycle, where it may perform work on its surroundings. A thermodynamic cycle occurs when a system is taken through a series of different states to finally return to its initial state.