Question

Efficiency of engine is ${\eta _1}$ at ${T_1} = 200^\circ {\text{C}}$ and ${T_2} = 0^\circ {\text{C}}$ and ${\eta _2}$ at ${T_2} = - 200^\circ {\text{C}}$ . Find the ratio of $\dfrac{{{\eta _1}}}{{{\eta _2}}}$ .
(A) $1.00$
(B) $0.721$
(C) $0.577$
(D) $0.34$

Answer 1

First of all, we will convert the given temperatures into Kelvin scale. After that we will find the efficiencies for the two different cases by directly substituting the required values. Lastly, we will find the ratio of the found efficiencies.

Complete step by step solution:
In the given question, we are supplied with the following data:
In the first case,
The initial temperature is ${T_1} = 200^\circ {\text{C}}$ while the final temperature is ${T_2} = 0^\circ {\text{C}}$ .
The efficiency of the engine is ${\eta _1}$ .
In the second case,
The initial temperature is ${T_2} = 0^\circ {\text{C}}$ while the final temperature is ${T_2} = - 200^\circ {\text{C}}$ .
The efficiency of the engine is ${\eta _2}$ .
We are asked to find the ratio of $\dfrac{{{\eta _1}}}{{{\eta _2}}}$ .
To begin with we will find the temperatures in Kelvin scale. We know that the engine operates in a range in temperatures i.e. there will be one hot temperature and cold temperature for each case. After that we will substitute the required values to find efficiency. Now, let us begin.
For the first case,
We will find the cold temperature:
${T_{{\text{cold}}}} = 0^\circ {\text{C}} \\\ \Rightarrow {T_{{\text{cold}}}} = \left( {0 + 273} \right)\,{\text{K}} \\\ \Rightarrow {T_{{\text{cold}}}} = 273\,{\text{K}}$
Again, we will find the hot temperature:
${T_{{\text{hot}}}} = 200^\circ {\text{C}} \\\ \Rightarrow {T_{{\text{hot}}}} = \left( {200 + 273} \right)\,{\text{K}} \\\ \Rightarrow {T_{{\text{hot}}}} = 473\,{\text{K}}$
So, now, we will find the efficiency:
${\eta _1} = 1 - \dfrac{{{T_{{\text{cold}}}}}}{{{T_{{\text{hot}}}}}}$ …… (1)
Where,
${\eta _1}$ indicates the efficiency of the engine for the first case.
${T_{{\text{cold}}}}$ indicates the low temperature.
${T_{{\text{hot}}}}$ indicates the high temperature.
Now, we substitute the required values in the equation (1) and we get:
${\eta _1} = 1 - \dfrac{{273}}{{473}} \\\ \Rightarrow {\eta _1} = 1 - 0.577 \\\ \Rightarrow {\eta _1} = 0.423$
Therefore, the efficiency for the first case is found to be $0.423$ .
For the second case,
We will find the cold temperature:
${T_{{\text{hot}}}} = 0^\circ {\text{C}} \\\ \Rightarrow {T_{{\text{hot}}}} = \left( {0 + 273} \right)\,{\text{K}} \\\ \Rightarrow {T_{{\text{hot}}}} = 273\,{\text{K}}$
Again, we will find the hot temperature:
${T_{{\text{cold}}}} = - 200^\circ {\text{C}} \\\ \Rightarrow {T_{{\text{cold}}}} = \left( { - 200 + 273} \right)\,{\text{K}} \\\ \Rightarrow {T_{{\text{cold}}}} = 73\,{\text{K}}$
So, now, we will find the efficiency:
${\eta _2} = 1 - \dfrac{{{T_{{\text{cold}}}}}}{{{T_{{\text{hot}}}}}}$ …… (1)
Where,
${\eta _2}$ indicates the efficiency of the engine for the first case.
${T_{{\text{cold}}}}$ indicates the low temperature.
${T_{{\text{hot}}}}$ indicates the high temperature.
Now, we substitute the required values in the equation (1) and we get:
${\eta _2} = 1 - \dfrac{{73}}{{273}} \\\ \Rightarrow {\eta _2} = 1 - 0.267 \\\ \Rightarrow {\eta _2} = 0.733$
Therefore, the efficiency for the first case is found to be $0.733$ .
Now, the ratio of $\dfrac{{{\eta _1}}}{{{\eta _2}}}$ is:
$\dfrac{{{\eta _1}}}{{{\eta _2}}} = \dfrac{{0.423}}{{0.733}} \\\ \therefore \dfrac{{{\eta _1}}}{{{\eta _2}}} = 0.577$
Hence, the ratio is found to be $0.577$ .

The correct option is (C).

Note: While solving this problem, remember that while calculating the efficiency, we should be careful about the hot and cold temperature. Most of the students seem to have confusion regarding this.Efficiency is the ratio of difference in temperature to the
temperature at higher level. It is nearly impossible to have an engine with $100\%$ efficiency.