a) A typical small passenger electric vehicle has a range of 200 km on an electric charge of 33 kWh. Convert this number into J/km (Joules per kilometer traveled). b) A comparably sized gasoline powered vehicle to (a) is rated as 25 miles per gallon. Convert this number into J/km (Joules per kilometer traveled). c) If gasoline costs $10.00 per gallon, and electricity $0.30 per kWh what is the difference in the cost of fuel per kilometer travelled between the electric vehicle in (a) and the gasoline powered vehicle in (b)?
A.
To convert the energy consumption from kilowatt-hours per kilometer to joules per kilometer, we need to multiply the kilowatt-hours by a conversion factor of 3,600,000, which is the number of joules in one kilowatt-hour.
First, let's calculate the energy consumption per kilometer:
Energy consumption per km = 33 kWh / 200 km = 0.165 kWh/km
Next, we can convert this to joules per kilometer:
Energy consumption per km = 0.165 kWh/km * 3,600,000 J/kWh.
Energy consumption per km = 594,000 J/km
Therefore, the small passenger electric vehicle has an energy consumption of 594,000 J/km.
B.)
b) To convert the fuel consumption of the gasoline-powered vehicle from miles per gallon to joules per kilometer, we can use the following conversion factors:
1 mile = 1.609344 kilometers
1 gallon = 3.78541178 liters
1 liter of gasoline contains about 34.2 megajoules (MJ) of energy
Therefore, the fuel consumption of the gasoline-powered vehicle in J/km is:
(1 gallon / 25 miles) * (3.78541178 liters / 1 gallon) * (34.2 MJ / 1 liter) / (1.609344 km / 1 mile) = 0.0827 MJ/km = 82,700 J/km
Step by step
Solved in 2 steps
