The burning of gas in a car produces about 12,600 kJ/gallon of energy. If a car averages 35 km/gal when driving 90 km/hr, which requires 18.5 kW, what is the efficiency of the engine under those conditions?

Question

...please help?

Answer 1

Assume you traveled for one hour (3600sec).

You used 90/35 gallons. This was in energy 90/35 * 12600kJ worth of chem energy.

However, you used 18.5KW*3600sec, or 18.5*3600 kjoules.

efficiency = (90*12600)/(35*18.5*3600)

I get just under 50 percent efficiency. You may want to convert the decimal efficiency I cited to percent.