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Expert-verified(a) What is the hot reservoir temperature of a Carnot engine that has an efficiency of \({\bf{42}}{\bf{.0\% }}\)and a cold reservoir temperature of \({\bf{27}}{\bf{.0}}\;{\bf{^\circ C}}\)? (b) What must the hot reservoir temperature be for a real heat engine that achieves \({\bf{0}}{\bf{.700}}\) of the maximum efficiency, but still has an efficiency of \({\bf{42}}{\bf{.0\% }}\) (and a cold reservoir at \({\bf{27}}{\bf{.0}}\;{\bf{^\circ C}}\))? (c) Does your answer imply practical limits to the efficiency of car gasoline engines?
The temperature of the hot reservoir is \(244.24\;^\circ {\rm{C}}\) for Carnot engine. The hot reservoir temperature for real heat engine with \(0.70\) of maximum (Carnot) efficiency is \(477\;^\circ {\rm{C}}\).Yes, our result imply practical limits as temperature of hot reservoir is very high for practical applications.
We calculate temperature of hot reservoir by using the formula for efficiency of Carnot engine when efficiency and temperature of cold reservoir is given. We further calculate temperature of hot reservoir of real engine. We find that this temperature is very high for practical purposes.
Efficiency of Carnot engine\( = 42\% = 0.42\)
Temperature of cold reservoir \({T_c} = 27\;^\circ {\rm{C}} = 300\;{\rm{K}}\)
Efficiency of Carnot engine \(h = 1 - \frac{{{T_c}}}{{{T_h}}}\)
Here,
\(\eta \) - efficiency of engine.
\({T_c}\) - temperature of cold reservoir.
\({T_h}\)- temperature of hot reservoir.
Temperature of hot reservoir is calculated as
\(\begin{aligned}{}h &= 1 - \frac{{{T_c}}}{{{T_h}}}\\{\rm{0}}{\rm{.42}} &= {\rm{1}} - \frac{{{\rm{300}}\;{\rm{K}}}}{{{T_h}}}\\\frac{{{\rm{300}}\;{\rm{K}}}}{{{T_h}}} &= {\rm{1}} - {\rm{0}}{\rm{.42}}\end{aligned}\)
\(\begin{aligned}{}{T_h} &= \frac{{{\rm{300}}\;{\rm{K}}}}{{{\rm{0}}{\rm{.58}}}}\\ &= {\rm{517}}{\rm{.24 K}}\\ &= {\rm{244}}{\rm{.24}}{{\rm{ }}^{\rm{o}}}{\rm{C}}\end{aligned}\)
Efficiency of real engine \( = 0.7\) times efficiency of Carnot engine
Efficiency of real engine \( = 42\% = 0.42\)
Efficiency of Carnot engine
\(\begin{aligned}{}h &= {\rm{0}}{\rm{.42}}/{\rm{0}}{\rm{.7}}\\ &= {\rm{0}}{\rm{.6}}\\ &= {\rm{60\% }}\end{aligned}\)
Temperature of hot reservoir is calculated as
\(\begin{aligned}{}h &= 1 - \frac{{{T_c}}}{{{T_h}}}\\0.6 &= 1 - \frac{{300\;{\rm{K}}}}{{{T_h}}}\\\frac{{300\;{\rm{K}}}}{{{T_h}}} &= 1 - 0.6\end{aligned}\)
\(\begin{aligned}{}{T_h} &= \frac{{300\;{\rm{K}}}}{{0.4}}\\ &= 750 {\rm{K}}\\ &= 477{\;^{\rm{o}}}{\rm{C}}\end{aligned}\)
Therefore, the temperature of hot reservoir is \({\rm{244}}{\rm{.24}}{{\rm{ }}^{\rm{o}}}{\rm{C}}\)for Carnot engine. The hot reservoir temperature for real heat engine with 0.7 of maximum (Carnot) efficiency is\(477{\;^{\rm{o}}}{\rm{C}}\).Yes, our result imply practical limits as temperature of hot reservoir is very high for practical applications as this temperature cannot be sustained in engines.
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