Question: A car engine whose output power is 135 hp operates at about 15% efficiency. Assume the engine’s water temperature of 85°C is its cold-temperature (exhaust) reservoir and 495°C is its thermal “intake” temperature (the temperature of the exploding gas–air mixture). (a) What is the ratio of its efficiency relative to its maximum possible (Carnot) efficiency? (b) Estimate how much power (in watts) goes into moving the car, and how much heat, in joules and in kcal, is exhausted to the air in 1.0 h.

Efficiency is a dimensionless number that estimates the performance of a device using thermal energy.

The output power is \(P = 135\;{\rm{hp}}\).

The actual efficiency is \({\eta _{{\rm{actual}}}} = 15\% = 0.15\).

The lower temperature is \({T_{\rm{L}}} = 85^\circ {\rm{C}}\).

The higher temperature is \({T_{\rm{H}}} = 495^\circ {\rm{C}}\).

The time is \(t = 1.0\;{\rm{h}}\).

(a)

The efficiency of the ideal cycle or Carnot cycle is calculated below:

\(\begin{aligned}{c}{\eta _{{\rm{ideal}}}} &= 1 - \frac{{{T_{\rm{L}}}}}{{{T_{\rm{H}}}}}\\{\eta _{{\rm{ideal}}}} &= 1 - \frac{{\left( {\left( {85^\circ {\rm{C + 273}}} \right)\;{\rm{K}}} \right)}}{{\left( {\left( {495^\circ {\rm{C + 273}}} \right)\;{\rm{K}}} \right)}}\\{\eta _{{\rm{ideal}}}} &= 0.534\end{aligned}\)

The ratio of efficiency of the actual cycle and the ideal cycle is calculated below:

\(\begin{aligned}{l}R &= \frac{{{\eta _{{\rm{actual}}}}}}{{{\eta _{{\rm{ideal}}}}}}\\R &= \frac{{0.15}}{{0.534}}\\R &= 0.281\end{aligned}\)

Thus, the ratio of efficiency of the actual cycle and the ideal cycle is \(0.281\).

(b)

The work done by the engine is calculated below:

\(\begin{aligned}{l}W &= Pt\\W &= \left( {135\;{\rm{hp}}} \right)\left( {\frac{{746\;{\rm{W}}}}{{1\;{\rm{hp}}}}} \right)\left( {1\;{\rm{h}}} \right)\left( {\frac{{3600\;{\rm{s}}}}{{1\;{\rm{h}}}}} \right)\\W &= 3.63 \times {10^8}\;{\rm{J}}\end{aligned}\)

The expression for the heat given to the system is

\({Q_{\rm{H}}} = W + {Q_{\rm{L}}}\).

Here, \({Q_{\rm{L}}}\) is the heat rejected by the system.

The heat rejected by the engine can be calculated using the following expression.

\(\begin{aligned}{c}{\eta _{{\rm{actual}}}} &= \frac{W}{{{Q_{\rm{H}}}}}\\{\eta _{{\rm{actual}}}} &= \frac{W}{{W + {Q_{\rm{L}}}}}\\W + {Q_{\rm{L}}} &= \frac{W}{{{\eta _{{\rm{actual}}}}}}\\{Q_{\rm{L}}} &= \frac{W}{{{\eta _{{\rm{actual}}}}}} - W\\{Q_{\rm{L}}} &= W\left( {\frac{1}{{{\eta _{{\rm{actual}}}}}} - 1} \right)\end{aligned}\)

Substitute the value in the above equation.

\(\begin{aligned}{l}{Q_{\rm{L}}} &= \left( {3.63 \times {{10}^8}\;{\rm{J}}} \right)\left( {\frac{1}{{0.15}} - 1} \right)\\{Q_{\rm{L}}} &= \left( {2.1 \times {{10}^9}\;{\rm{J}}} \right)\left( {\frac{{2.38 \times {{10}^{ - 4}}\;{\rm{kcal}}}}{{1\;{\rm{J}}}}} \right)\\{Q_{\rm{L}}} &= 4.99 \times {10^5}\;{\rm{kcal}} \approx {\rm{5}}{\rm{.0}}\;{\rm{kcal}}\end{aligned}\)

Thus, the heat rejected by the engine is \(2.1 \times {10^9}\;{\rm{J}}\) or \({\rm{5}}{\rm{.0}}\;{\rm{kcal}}\).