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?(adsbygoogle = window.adsbygoogle || []).push({});

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%.

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# Thermodynamics - efficiency question

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