I don't know what you want.
If the woman uses 96 kJ and the efficiency is 35%, she actually used 96/0.35 or about 274 kJ for the 1 km. (you can do that more accurately)
If the car gets 8.7 km/L, then to go 1 km you will use 1/8.7 = ? L gasoline.
That x 1000 = mL gasoline
Convert to g by mass = volume gasoline x density gasoline = ?
You have 49 kJ released for every g so g x 49 kJ gives energy used to go 1 km.
Compare those two values however you're supposed to do so. The comparison part is the part I don't understand.
A woman expends 96 kJ of energy in walking a kilometer. The energy is supplied by the metabolic breakdown of food intake and has a 35 % efficiency.
If the woman drives a car over the same distance, how much energy is used if the car gets 8.7 km per liter of gasoline? The density of gasoline is 0.71 g/mL, and its enthalpy of combustion is 49 kJ/g.
Compare the efficiencies of the two processes. kJ (drivings)/kJ (walking)
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