A Carnot engine operates with\({T_L} = {\bf{20^\circ C}}\)and has an efficiency of 25%. By how many kelvins should the high operating temperature\({T_H}\)be increased to achieve an efficiency of 35%?

The efficiency of the Carnot engine is inversely proportional to the high-temperature reservoir. In this problem, the difference in the high temperatures of the two different Carnot engines having different efficiencies and same lower temperatures will result in the increment of the high-temperature.

This difference in temperatures is dependent on the efficiencies of two Carnot engines and common low-temperature values.

The given data can be listed as

• The efficiency of the first Carnot engine is\({e_1} = 25\% \left( {\frac{1}{{100}}} \right) = 0.25\).
• The efficiency of the second Carnot engine is\({e_2} = 35\% \left( {\frac{1}{{100}}} \right) = 0.35\).
• The cold reservoir temperature is \({T_{\rm{L}}} = \left( {20^\circ {\rm{C}} + {\rm{273}}} \right){\rm{ K}} = 293{\rm{ K}}\).

The efficiency of the first Carnot engine can be expressed as

\(\begin{array}{c}{e_1} = \left( {1 - \frac{{{T_{\rm{L}}}}}{{{T_{{{\rm{H}}_1}}}}}} \right)\\\frac{{{T_{\rm{L}}}}}{{{T_{{{\rm{H}}_1}}}}} = 1 - {e_1}\end{array}\)

\({T_{{{\rm{H}}_1}}} = \frac{{{T_{\rm{L}}}}}{{\left( {1 - {e_1}} \right)}}\). … (i)

Here,\({T_{{{\rm{H}}_1}}}\)is the first intake temperature.

The efficiency of the second Carnot engine can be expressed as

\(\begin{array}{c}{e_2} = \left( {1 - \frac{{{T_{\rm{L}}}}}{{{T_{{{\rm{H}}_2}}}}}} \right)\\\frac{{{T_{\rm{L}}}}}{{{T_{{{\rm{H}}_2}}}}} = 1 - {e_2}\end{array}\)

\({T_{{{\rm{H}}_2}}} = \frac{{{T_{\rm{L}}}}}{{\left( {1 - {e_2}} \right)}}\). … (ii)

Here,\({T_{{{\rm{H}}_2}}}\)is the second intake temperature, and\({T_{\rm{L}}}\)is the same for both Carnot engines.

Subtracting equation (i) from equation (ii) to obtain the increment in the high temperature,

\(\begin{array}{c}{T_{{{\rm{H}}_2}}} - {T_{{{\rm{H}}_1}}} = \frac{{{T_{\rm{L}}}}}{{\left( {1 - {e_2}} \right)}} - \frac{{{T_{\rm{L}}}}}{{\left( {1 - {e_1}} \right)}}\\ = {T_{\rm{L}}}\left( {\frac{1}{{\left( {1 - {e_2}} \right)}} - \frac{1}{{\left( {1 - {e_1}} \right)}}} \right)\end{array}\).

Substituting the values in the above equation,

\(\begin{array}{c}{T_{{{\rm{H}}_2}}} - {T_{{{\rm{H}}_1}}} = 293{\rm{ K}}\left( {\frac{1}{{\left( {1 - 0.35} \right)}} - \frac{1}{{\left( {1 - 0.25} \right)}}} \right)\\ = 293{\rm{ K}}\left( {1.538 - 1.333} \right)\\ = 293{\rm{ K}} \times 0.205\\ = 60.1{\rm{ K}}\end{array}\).

Thus, the high temperature should be increased by \(60.1{\rm{ K}}\) to achieve an efficiency of 35%.