Question

Short answer

The temperature of heat source is \(1147.06\;{\rm{K}}\).

Step-by-step solution

Given Data

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

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

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

Understanding Carnot efficiency

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.

Calculation of exhaust temperature of engine

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.

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

Calculation of temperature of the heat source

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.