Question

An heat engine absorbs heat at \[{327^o}C\] and exhausts heat at \[{127^o}C\]. The maximum amount of work performed by the engine in joule per kilo calorie is \[X\]. Find \[\dfrac{X}{5}\].

Answer 1

Hint: For solving this question we have to consider the concepts of heat engine, we have to use temperature in the standard unit of kelvin here. With the help of efficiency formula and work done by heat engine we will determine the value of X. Mainly, one must calculate efficiency and work done using efficiency.

Formula used:
1. \[\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}\]
Where, \[\eta \] is the efficiency of the heat engine, \[{T_1}\] is the temperature at which it absorbs and \[{T_2}\] is the temperature at which it exhausts.
2. \[W = Q\eta \]
Where, \[W\]is work done by the heat engine and \[Q\]is the source heat in kilo calorie.

Complete answer:
Let us begin with the conversion of the temperature into standard units such as
\[{T_1} = {327^o}C = (327 + 273)K = 600K\]
\[{T_2} = {127^o}C = (127 + 273)K = 400K\]

So, let’s begin with calculating the efficiency of the heat engine, we have
\[\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}\]

Let us substitute all the given values in the formula above, we get
\[ \Rightarrow \eta = 1 - \dfrac{{400K}}{{600K}}\]
\[ \Rightarrow \eta = 1 - \dfrac{2}{3} = 1 - 0.66 = 0.34\]
\[ \Rightarrow \eta = 0.34\]

Now, we have to find work for per kilo calorie, for that we have a formula for work done as below:
\[W = Q\eta \]
But, the source heat is \[Q = H = 1kcal\]
Also, remember that
\[1kcal = 4.2 \times {10^3}joule\]

Let us put all these values in the formula for work done for calculating total work.
\[ \Rightarrow W = 4.2 \times {10^3}joule \times 0.34\]
\[ \Rightarrow W = {\rm{1}}{\rm{.428}} \times {10^3}joule\]

But here work done is given in the form of \[X\]. Therefore,
\[X = W = {\rm{1}}{\rm{.428}} \times {10^3}joule\]
\[X = 1428J\]

Therefore, \[\dfrac{X}{5}\] is given by:
\[\dfrac{X}{5} = \dfrac{{1428J}}{5} = 285.6J\]

So, the answer is \[285.6J\].

Note: Here, the question is designed in such a way that there is just one concept that has been used to work efficiently. We have to recall all the important points from the heat engine and apply it over here.