In metric terms, a typical automobile averages 5.48 km/L of gasoline burned

Question

In metric terms, a typical automobile averages 5.48 km/L of gasoline burned. Gasoline has a heat of combustion of 48 kJ/g and a density of 0.68 g/mL. How much energy is consumed in driving an automobile 3.7 km?

Answer 1

3.7 km x 1 L / 5.48 km = 0.6752 L of gas required

Since heat of combustion is given in kJ/g, we need to convert the 0.6752 liters to grams:

0.6752 L x 1000 ml /L x 0.68 g / ml = 459.1 g of gas required

Use the heat of combustion to find the amount of energy consumed:

459.1 g x 48 kJ / g = 22,039 kJ = 22,000 kJ of energy needed (2 sig. figs.)

Answer 2

Volume of gasoline needed: 3.7km/5.48km/L

Mass of gasoline needed: 0.68/mL/[(3.7km/5.48km/L)x1000]

Energy needed: 48KJ/g/{0.68/mL/[(3.7km/5.48km/L)x1000]}