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Question:(II) A heat engine exhausts its heat at 340°C and has a Carnot efficiency of 36%. What exhaust temperature would enable it to achieve a Carnot efficiency of 42%?

Short Answer

Expert verified

The exhaust temperature is \(555.5\;{\rm{K}}\).

Step by step solution

01

Given Data

The temperature is \({T_0} = 340^\circ {\rm{C}}\).

The Carnot efficiency is \({\eta _{\rm{C}}} = 36\% \).

The new Carnot efficiency is \({\eta '_{\rm{C}}} = 42\% \).

02

Understanding adiabatic and isothermal process

In this problem, firstly find the intake temperature by using the initial Carnot efficiency. Now, evaluate the exhaust temperature for new Carnot efficiency.

03

Calculation of intake temperature from Carnot efficiency

The relation of efficiency is given by,

\({\eta _{\rm{C}}} = 1 - \frac{{{T_0}}}{{{T_{\rm{i}}}}}\)

Here, \({T_{\rm{i}}}\) is the intake temperature.

On plugging the values in the above relation.

\(\begin{aligned}{c}0.36 &= 1 - \left( {\frac{{\left( {340^\circ {\rm{C}} + 273} \right)\;{\rm{K}}}}{{{T_{\rm{i}}}}}} \right)\\{T_{\rm{i}}} &= 957.8\;{\rm{K}}\end{aligned}\)

04

Calculation of exhaust temperature from new Carnot efficiency

The relation of efficiency is given by,

\({\eta '_{\rm{C}}} = 1 - \frac{{{T_0}^\prime }}{{{T_{\rm{i}}}}}\)

Here, \({T'_0}\) is the exhaust temperature.

On plugging the values in the above relation.

\(\begin{aligned}{c}0.42 &= 1 - \left( {\frac{{{T_{\rm{0}}}^\prime }}{{957.8\;{\rm{K}}}}} \right)\\{T_{\rm{0}}}^\prime &= 555.5\;{\rm{K}}\end{aligned}\)

Thus, \({T_{\rm{0}}}^\prime = 555.5\;{\rm{K}}\) is the required temperature.

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An ideal monatomic gas expands slowly to twice its volume (1) isothermally; (2) adiabatically; (3) isobarically. Plot each on a PV diagram. In which process \(\Delta U\) is the greatest, and in which is \(\Delta U\) the least? In which is W the greatest and the least? In which is Q the greatest and the least?

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