Question

A particular car does work at the rate of about\({\bf{7}}{\bf{.0}}\;{\bf{kJ/s}}\)when traveling at a steady\({\bf{21}}{\bf{.8}}\;{\bf{m/s}}\)along a level road. This is the work done against friction. The car can travel 17 km on 1.0 L of gasoline at this speed (about 40 mi/gal). What is the minimum value for\({{\bf{T}}_{\bf{H}}}\)if\({{\bf{T}}_{\bf{L}}}\)is 25°C? The energy available from 1.0 L of gas is\({\bf{3}}{\bf{.2 \times 1}}{{\bf{0}}{\bf{7}}}\;{\bf{J}}\).

Short answer

The minimum value of temperature \({T_{\rm{H}}}\) is \(86\circ {\rm{C}}\).

Step-by-step solution

Understanding the efficiency of the engine

The ratio of the work done by the engine to the heat input at high temperature is termed the efficiency of the engine.

The expression for the efficiency is given as:

\(e = \frac{W}{{{Q_{\rm{H}}}}} = 1 - \frac{{{T_{\rm{L}}}}}{{{T_{\rm{H}}}}}\) … (i)

Here, W is the work done and \({Q_{\rm{H}}}\) is the heat input at high temperature.

Given data

The rate of doing work is \(\frac{W}{t} = 7\;{\rm{kJ/s}}\).

The speed of the car is \(v = 21.8\;{\rm{m/s}}\).

The car can travel a distance of about \(d = 17\,{\rm{km}}\).

The energy available is \(E = 3.2 \times {107}\,{\rm{J/L}}\).

The lower temperature is \({T_{\rm{L}}} = 25\circ {\rm{C}}\).

Evaluation of heat input from gasoline in the engine

The relation for the rate of heat input is given by:

\(\frac{Q}{t} = E \times v\)

Substitute the values in the above expression.

\(\begin{array}{l}\frac{Q}{t} = \left( {3.2 \times {{10}7}\,{\rm{J/L}} \times \frac{{1\;{\rm{L}}}}{{17000\;{\rm{m}}}}} \right) \times \left( {21.8\;{\rm{m/s}}} \right)\\\frac{Q}{t} = 41035.2\;{\rm{W}}\end{array}\)

Evaluation of the temperature of the engine

The relation for the efficiency is given by:

\(\begin{array}{c}\eta = \frac{{\left( {\frac{W}{t}} \right)}}{{\left( {\frac{Q}{t}} \right)}}\\1 - \frac{{{T_{\rm{L}}}}}{{{T_{\rm{H}}}}} = \frac{{\left( {\frac{W}{t}} \right)}}{{\left( {\frac{Q}{t}} \right)}}\end{array}\)

Substitute the values in the above expression.

\(\begin{array}{c}\left( {1 - \frac{{\left( {25\circ {\rm{C}} + 273} \right)\;{\rm{K}}}}{{{T_{\rm{H}}}}}} \right) = \frac{{\left( {7 \times {{10}3}\;{\rm{J/s}}} \right)}}{{\left( {41035.2\;{\rm{W}}} \right)}}\\{T_{\rm{H}}} = 359\;{\rm{K}}\\{T_{\rm{H}}} = \left( {359 - 273} \right)\circ C\\{T_{\rm{H}}} = 86\circ {\rm{C}}\end{array}\)

Thus, the minimum value of temperature\({T_{\rm{H}}}\)is\(86\circ {\rm{C}}\).