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

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Last updated date: 20th Jun 2024
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Answer
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Hint: 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.