The efficiency is \(\eta = 0.22\).

The energy delivered by engine is \(W = 180\;{\rm{J}}\).

The revolution per minute for engine is \(n = 1500\;{\rm{rpm}}\).

The energy content of gasoline is \({Q_{\rm{g}}} = 130\;{\rm{MJ/gal}}\).

The power can be determined by using the standard relation of power and work done. The power is equivalent to the work done by the engine in a unit time.

The relation of work done per second is given by,

\(\begin{aligned}{l}\frac{W}{t} &= \left( {180\;\frac{{\rm{J}}}{{{\rm{cycle}} \cdot {\rm{cylinder}}}}} \right)\left( {4\;{\rm{cylinders}}} \right)\left( {25\;\frac{{{\rm{cycles}}}}{{\rm{s}}}} \right)\\\frac{W}{t} &= 18000\;{\rm{J/s}}\end{aligned}\)

Thus,\(\frac{W}{t} = 18000\;{\rm{J/s}}\) is the required work per second.

The relation of efficiency is given by,

\(\begin{aligned}{l}\eta &= \frac{W}{Q}\\Q &= \frac{W}{\eta }\\\frac{Q}{t} &= \frac{W}{t}\left( {\frac{1}{\eta }} \right)\end{aligned}\)

Here, \(\frac{Q}{t}\) is the rate of heat output and W is the work done.

On plugging the values in the above relation.

\(\begin{aligned}{l}\frac{Q}{t} &= \left( {18000\;{\rm{J/s}}} \right)\left( {\frac{1}{{0.22}}} \right)\\\frac{Q}{t} &= 8.1 \times {10^4}\;{\rm{J/s}}\end{aligned}\)

Thus, \(\frac{Q}{t} = 8.1 \times {10^4}\;{\rm{J/s}}\) is the required rate of heat output.

The relation to find the time is given by,

\(t = \frac{{{Q_{\rm{g}}}}}{{\left( {\frac{Q}{t}} \right)}}\)

On plugging the values in the above relation.

\(\begin{aligned}{l}t &= \left( {\frac{{130\;{\rm{MJ/gal}} \times \frac{{{{10}^6}\;{\rm{J/gal}}}}{{1\;{\rm{MJ/gal}}}}}}{{8.1 \times {{10}^4}\;{\rm{J/s}}}}} \right)\\t &= 1604.9\;{\rm{s}}\end{aligned}\)

Thus, \(t = 1604.9\;{\rm{s}}\) is the required time.