Correct Answer - Option 2 : 632.5 K
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 TH1 and TH2 and the second heat engine operating between temperature TH2 and TH3.
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.