Question

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

Answer 1

As we all know that the efficiency of Carnot engine is $n = \dfrac{{{T_1} - {T_2}}}{{{T_1}}}$

Complete step by step solution:
Given: Carnot engine efficiency, ${n_1} = 40\%$
Temperature of sink, ${T_2} = 300K$
Efficiency increased to ${n_2} = 60\%$

To find:
Change in temperature of the source, now ${T_1} = ?$
We know efficiency of Carnot engine is
$n = \dfrac{{{T_1} - {T_2}}}{{{T_1}}}$
$\Rightarrow \,\,\,\dfrac{{40}}{{100}} = \dfrac{{{T_1} - 300}}{{{T_1}}}$
$= 4{T_1} = 10T - 3000$
$\Rightarrow \,\,\,6{T_1} = 3000$
$\Rightarrow \,\,\,{T_1} = 500K$ is the temperature of the source initially.
Now to increase efficiency to 60%
${n_1} = \dfrac{{{T_1} - {T_2}}}{{{T_1}}}$
$\dfrac{{60}}{{100}} = \dfrac{{{T_1} = 300}}{{{T_1}}}$
$\Rightarrow \,\,\,6{T_1} = 10{T_1} - 3000$
$\Rightarrow \,\,\,{T_1} = 750K$ is the new source temperature.

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

Note: If we want to find the change in temperature of the source we have to use the same formula and method.