Asked by Lindsay
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?
...please help?
...please help?
Answers
Answered by
bobpursley
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.
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.
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