Question

In walking a kilometer, you use about 100 kJ of energy. This energy comes from the oxidation of foods, which is about 30% efficient. How much energy do you save by walking 5.21 kilometers instead of driving a car that gets 9.59 km L^-1 of gasoline (22.6 miles per gallon)? Take the density of gasoline to be 0.680 g mL^-1 and its enthalpy of combustion to be -48.0 kJ g^-1.

_____ kJ

Answer 1

E = 100 kJ energy / km

30% efficient...

enregy for 5.21 km

Total E required = 100*5.21 = 521 kJ in total required

so for a 30% efficeincy:

521/0.3 = 1736.6666 kJ required to burn

Energy required for walking ( eating) = 1736.6666 kJ

Energy required for driving:

5 km --> 5 / 9.59 km/L = 0.5213 liters of gas required

so

mass = D*V = 512.3 mL * 0.680 g /mL = 348.364 g of gasoline

E = 348.364 g * 48 kJ/g =16721.472 kJ

Energy saved = Edriving gas- Ewaliking = 16721.472-1736.6666 = 14984.8054 kJ saved