The burning of gasoline in a car releases about 3.0 x 10^4 kcal/gal. If a car averages 41 km/gal when driving 90 km/h, which requires 25 hp, what is the efficiency of the engine under those conditions? I think I have an answer, but I'm not sure if it's correct. 1 horsepower = 746 watts 25 hp = 18650 watts 1 kcal = 4186 joules (watts / sec) e = W / Qh = 1 - (Ql / Qh) 1 gal / 41 km at 91 km / h means that that 2.26 gallons will be used in one hour. Since no temperatues are given I don't think that the 1 - Ql / Qh is needed... Converting 30000 kcal / gal to watts I get 34883 joules/sec. Since 2.26 gal are used, I multiplied the above by 2.26 to get 78836. Dividing 18650 watts (from hp) by 78836 I got an efficiency of 23.46%.