One reversible heat engine operates between 1000 K and T K and another reversible heat engine operates between T and 400 K. If both heat engine have same heat input and output, the value of T is

Concept:

Efficiency of a heat engine is given by

\({\rm{\eta }} = 1 - \frac{{{{\rm{Q}}_{{\rm{rejected}}}}}}{{{{\rm{Q}}_{{\rm{supplied}}}}}}\)

If two heat engines have the same heat input (heat supplied) and heat output (heat rejected) then efficiency is also the same.

For a reversible heat engine:

\(\frac{{{{\rm{Q}}_2}}}{{{\rm{}}{{\rm{Q}}_{1{\rm{}}}}}} = \frac{{{T_2}}}{{{T_1}}}\)

So the efficiency is

\({\rm{\eta }} = 1 - \frac{{{{\rm{T}}_{{\rm{2}}}}}}{{{{\rm{T}}_{{\rm{1}}}}}}\)

Let two engines operating in series in which the first heat engine operating between temperature T_H1 and T_H2  and the second heat engine operating between temperature T_H2 and T_H3.

The efficiency of engine 1 is given by:

\({\rm{\eta }} = 1 - \frac{{{{\rm{T}}_{{\rm{H2}}}}}}{{{{\rm{T}}_{{\rm{H1}}}}}}\)

The efficiency of engine 2 is given by:

\({\rm{\eta }} = 1 - \frac{{{{\rm{T}}_{{\rm{H3}}}}}}{{{{\rm{T}}_{{\rm{H2}}}}}}\)

Since both have the same efficiencies.

\(1 - \frac{{{T_{H2}}}}{{{T_{H1}}}} = 1 - \frac{{{T_{H3}}}}{{{T_{H2}}}}\)

\( \frac{{{T_{H2}}}}{{{T_{H1}}}} = \frac{{{T_{H3}}}}{{{T_{H2}}}}\)

\(T_{H2}=\sqrt{T_{H1}T_{H3}}\)

Calculation:

Given:

TH1 = 1000 K, TH2 = T  and TH3 = 400 K     

\(T_{H2}=\sqrt{T_{H1}T_{H3}}\)  

\(T_{H2}=\sqrt{1000\times400}\)

TH2 = 632.455  K.