Question:(I) A heat engine’s high temperature\({T_{\bf{H}}}\)could be ambient temperature, because liquid nitrogen at 77 K could be\({T_{\bf{L}}}\)and is cheap. What would be the efficiency of a Carnot engine that made use of heat transferred from air at room temperature (293 K) to the liquid nitrogen “fuel”(Fig.15–25)?

The given data can be listed below as:

• The high temperature of the heat engine is\({T_{\rm{H}}} = 293{\rm{ K}}\).
• The cold temperature of the liquid nitrogen is \({T_{\rm{L}}} = 77{\rm{ K}}\).

The ambient temperature is considered the high temperature of the heat engine.

The liquid nitrogen temperature is considered the cold temperature reservoir.

The heat is being transferred from the ambient air to the liquid nitrogen. The Carnot engine utilizes the high temperature of the heat engine to transfer heat to the fuel.

The efficiency of the Carnot engine can be expressed as:

\(e = \left( {1 - \frac{{{T_{\rm{L}}}}}{{{T_{\rm{H}}}}}} \right)\)

Substitute the values in the above equation.

\(\begin{aligned}{c}e &= \left( {1 - \frac{{{\rm{77 K}}}}{{{\rm{293 K}}}}} \right)\\ &= 0.7372\\ &= 0.7372 \times 100\% \\ &= 73.72\% \end{aligned}\)

Thus, the efficiency of the Carnot engine is \(73.72\% \).