Answer
20 percent.
Work Step by Step
Calculate the input power from the gasoline, and the useful power.
$$Q_H/t=\frac{3.0\times10^4 kcal}{gal}\frac{1 gal}{41km}\frac{110km}{3600s}\frac{4186J}{kcal}=9.359\times10^4 J/s$$
$$W/t=(25hp)(746W/hp)=1.865\times10^4 J/s$$
The efficiency is given by equation 15–4a.
$$e=\frac{W}{Q_H}=\frac{W/t}{Q_H/t}=\frac{1.865\times10^4 J/s}{9.359\times10^4 J/s}=0.199\approx 20\%$$
