The burning of gasoline in a car releases about $3.0 \times 10^4$ kcal/gal. If a car averages 41 km/gal when driving 110 km/h, which requires 25 hp, what is the efficiency of the engine under those conditions?
Added by Richard H.
Step 1
Given: - Heat released by burning gasoline = $3.0 \times 10^4$ kcal/gal - Car averages 41 km/gal - Car is driving at 110 km/h - Power required = 25 hp First, convert the heat released by burning gasoline to joules: $3.0 \times 10^4$ kcal/gal = $3.0 \times 10^4 Show more…
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The burning of gasoline in a car releases about $3.0 \times 10^{4} \mathrm{kcal} / \mathrm{gal}$ . If a car averages 41 $\mathrm{km} / \mathrm{gal}$ when driving 90 $\mathrm{km} / \mathrm{h}$ , which requires 25 $\mathrm{hp}$ , what is the efficiency of the engine under those conditions?
(II) The burning of gasoline in a car releases about $3.0 \times 10^{4}$ kcal $/$ gal. If a car averages 38 $\mathrm{km} / \mathrm{gal}$ when driving $95 \mathrm{km} / \mathrm{h},$ which requires $25 \mathrm{hp},$ what is the efficiency of the engine under those conditions?
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