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)
To calculate the energy used while driving the car, we need to consider the energy content of the gasoline, the efficiency of the car's engine, and the distance traveled.
1. First, let's calculate the energy content of the gasoline used for driving:
- Given that the density of gasoline is 0.71 g/mL, we can calculate the mass of gasoline needed to cover 1 kilometer by dividing the distance traveled (1 km) by the fuel efficiency (8.7 km/L):
Mass of gasoline = 1 km / 8.7 km/L = 0.1149 L * 0.71 g/mL = 0.0816 g
- Now, we can calculate the energy content of this amount of gasoline by multiplying the mass of gasoline by the enthalpy of combustion:
Energy content = 0.0816 g * 49 kJ/g = 3.9984 kJ
2. Next, let's calculate the energy used while walking:
- The woman expends 96 kJ of energy in walking 1 kilometer, but her metabolic breakdown of food intake has an efficiency of 35%.
- To find the total energy intake, we divide the energy expended by the efficiency:
Energy intake = 96 kJ / 0.35 = 274.2857 kJ
Now, to compare the efficiencies of the two processes:
- Efficiency of driving = Energy used while driving / Energy intake while walking
= 3.9984 kJ / 274.2857 kJ
= 0.0146
- Efficiency of walking = Energy intake while walking / Energy intake while walking
= 274.2857 kJ / 274.2857 kJ
= 1.0000
Therefore, the efficiency of driving (0.0146) is much lower than the efficiency of walking (1.0000).
To calculate the energy used when the woman drives a car over the same distance, we need to consider the energy content of the gasoline and the efficiency of the car.
First, let's calculate the energy content of gasoline per liter:
Energy content of gasoline = Density of gasoline * Enthalpy of combustion
Energy content of gasoline = 0.71 g/mL * 49 kJ/g
Next, let's calculate the energy used when the car drives a kilometer:
Energy used by the car = Energy content of gasoline / Fuel efficiency of the car
Energy used by the car = (0.71 g/mL * 49 kJ/g) / 8.7 km per liter
Now, let's compare the efficiencies of the two processes:
Efficiency of walking = Energy expended in walking / Energy supplied by food intake
Efficiency of driving = Energy used by the car / Energy content of gasoline
Efficiency ratio = Efficiency of driving / Efficiency of walking
Efficiency ratio = (Energy used by the car / Energy content of gasoline) / (Energy expended in walking / Energy supplied by food intake)
I will now calculate the efficiency ratio.
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.