Question

The efficiency of a Carnot heat engine working between two temperatures is \[60\% \]. If the temperature of the source alone is decreased by \[100{\text{K}}\], the efficiency becomes \[40\% \]. Find the temperature of the source & sink.

Answer 1

Hint:We are asked to find out the temperature of source and the sink. First, recall the formula for finding the efficiency of a Carnot engine operating between a source and sink. Use this formula to form equations for both given efficiencies. The temperature of the sink remains the same in both cases, use this fact to find the required answer.

Complete step by step answer:
Given, efficiency of a Carnot heat engine is \[{\eta _1} = 60\% \].
When the temperature of the source is decreased by \[100{\text{K}}\], its efficiency is \[{\eta _2} = 40\% \].

We have the formula for efficiency of a Carnot engine operating between a source and a sink as,
\[\eta = 1 - \dfrac{{{T_{{\text{sink}}}}}}{{{T_{{\text{source}}}}}}\] (i)
where \[{T_{{\text{sink}}}}\] is the temperature of the sink and \[{T_{{\text{source}}}}\] is the temperature of the source.
Here, we have two cases and we will use this formula for both the cases to form equations so that we can find out the value of temperature of source and sink.

For case (1) when the efficiency is \[{\eta _1} = 60\% \].Let the temperature of the source be \[{T_1}\].Let the temperature of the sink be \[{T_2}\].
For case (2) when the efficiency is \[{\eta _2} = 40\% \].The temperature of the source decreased by \[100{\text{K}}\], that is we will have source temperature as, \[{T_1}^\prime = {T_1} - 100\]. The temperature of the sink remains the same, so it will be \[{T_2}\].

Case (1):
Here, \[{T_{{\text{sink}}}} = {T_2}\], \[{T_{{\text{source}}}} = {T_1}\] and \[{\eta _1} = 60\% \]
Putting these values in equation (i), we get
\[60\% = 1 - \dfrac{{{T_2}}}{{{T_1}}}\]
\[ \Rightarrow \dfrac{{60}}{{100}} = 1 - \dfrac{{{T_2}}}{{{T_1}}}\]
\[ \Rightarrow \dfrac{{{T_2}}}{{{T_1}}} = 1 - \dfrac{{60}}{{100}}\]
\[ \Rightarrow {T_2} = {T_1}\dfrac{{40}}{{100}}\] (ii)

Case (2):
Here, \[{T_{{\text{sink}}}} = {T_2}\], \[{T_{{\text{source}}}} = {T_1} - 100\] and \[{\eta _2} = 40\% \]
Putting these values in equation (i), we get
\[40\% = 1 - \dfrac{{{T_2}}}{{{T_1} - 100}}\]
\[ \Rightarrow \dfrac{{40}}{{100}} = 1 - \dfrac{{{T_2}}}{{{T_1} - 100}}\]
\[ \Rightarrow \dfrac{{{T_2}}}{{{T_1} - 100}} = 1 - \dfrac{{40}}{{100}}\]
\[ \Rightarrow \dfrac{{{T_2}}}{{{T_1} - 100}} = \dfrac{{60}}{{100}}\]
\[ \Rightarrow {T_2} = \left( {{T_1} - 100} \right)\dfrac{{60}}{{100}}\] (iii)

Equating equations (ii) and (iii), we get
\[{T_1}\dfrac{{40}}{{100}} = \left( {{T_1} - 100} \right)\dfrac{{60}}{{100}}\]
\[ \Rightarrow {T_1} \times 4 = \left( {{T_1} - 100} \right) \times 6\]
\[ \Rightarrow 4{T_1} = 6{T_1} - 600\]
\[ \Rightarrow 2{T_1} = 600\]
\[ \Rightarrow {T_1} = 300{\text{K}}\]
The temperature of the source obtained is \[{T_1} = 300{\text{K}}\].
Putting this value in equation (ii), we get
\[{T_2} = 300 \times \dfrac{{40}}{{100}}\]
\[ \therefore {T_2} = 120{\text{K}}\]
The temperature of the sink obtained is \[{T_2} = 120{\text{K}}\].

Therefore, the temperature of the source and sink are \[300{\text{K}}\] and \[120{\text{K}}\] respectively.

Note:Carnot engine is a theoretical thermodynamic cycle, it is the maximum efficiency obtained when the heat engine is operated between two temperatures. The reservoir whose temperature is high is known as source and the reservoir whose temperature is low is known as sink.