91Ó°ÊÓ

The temperature of engine is \({T_1} = 580^\circ {\rm{C}}\).

The Carnot efficiency is \(\eta = 22\% \).

The ideal efficiency is \({\eta _{\rm{i}}} = 42\% \).

In this problem, use the Carnot efficiency to find the temperature at exhaust. Again, recalculate the temperature of the heat source with the help of new ideal efficiency.

The relation of Carnot efficiency is given by,

\(\eta = \left( {1 - \frac{{{T_2}}}{{{T_1}}}} \right)\)

Here, \({T_2}\) is the temperature at exhaust.

On plugging the values in the above relation.

\(\begin{aligned}{c}0.22 &= \left( {1 - \frac{{{T_2}}}{{\left( {580^\circ {\rm{C}} + 273} \right)\;{\rm{K}}}}} \right)\\{T_2} &= 665.3\;{\rm{K}}\end{aligned}\)

The relation to find temperature is given by,

\({\eta _{\rm{i}}} = \left( {1 - \frac{{{T_2}}}{{{{T'}_1}}}} \right)\)

Here, \({T'_1}\) is the temperature of heat source.

On plugging the values in the above relation.

\(\begin{aligned}{c}0.42 &= \left( {1 - \frac{{665.3\;{\rm{K}}}}{{{T_1}^\prime }}} \right)\\{T_1}^\prime &= 1147.06\;{\rm{K}}\end{aligned}\)

Thus, \({T_1}^\prime = 1147.06\;{\rm{K}}\) is the required rate of heat output.