Question

A Carnot engine efficiency 40% and temperature of sink 300 K to increase up to 60 %. Calculate the change in temperature of the source.

Answer 1

In this question, firstly we will find the initial temperature of the source and then we will find the new temperature of the source by applying the condition of increasing the efficiency to 60%.

Complete step by step answer:
Given:
The efficiency of Carnot cycle ${\eta _1} = 40\%$.
The temperature of the sink ${T_2} = 300\;{\rm{K}}$.
The efficiency is increased to ${\eta _2} = 60\%$.
Let the initial source temperature be ${T_1}$.
We will now apply the formula for efficiency:
${\eta _1} = 1 - \dfrac{{{T_2}}}{{{T_1}}}$
We will now substitute the given values.
$\Rightarrow 40\% = 1 - \dfrac{{300}}{{{T_1}}}$
$\begin{array}{l} \Rightarrow \dfrac{{40}}{{100}} = 1 - \dfrac{{300}}{{{T_1}}}\\\ \Rightarrow 6{T_1} = 3000\\\ \Rightarrow {T_1} = 500K \end{array}$
Now the efficiency has increased to 60%.
${\eta _2} = 1 - \dfrac{{{T_2}}}{{{T_3}}}$
Here, $T_3$ is the new source temperature.
$\dfrac{{60}}{{100}} = 1 - \dfrac{{300}}{{{T_3}}}$
$\begin{array}{l} \Rightarrow 6{T_3} = 10{T_3} - 3000\\\ \Rightarrow {T_3} = 750K \end{array}$
Therefore, the change in temperature of the source is:
$\begin{array}{l} \Delta T = {T_1} - {T_3}\\\ = 750K - 500K\\\ = 250K \end{array}$

Therefore, the change in temperature of the source is 250 K.

Note: The possible mistakes that one can make in this kind of problem is the confusion between the temperatures ${T_1}$ and ${T_2}$. We need to take care of the definitions of temperatures ${T_1}$ and ${T_2}$. Note that ${T_2}$ refers to the temperature of the sink ${T_1}$ refers to the temperature of the source.