Question

In walking 1 km, 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 1 km instead of driving a car that gets $\mathbf{8}\mathbf{.}\mathbf{0}{\mathbf{kmL}}^{\mathbf{-}\mathbf{1}}$ gasoline (19 miles/gal)? The density of gasoline is$\mathbf{0}\mathbf{.}\mathbf{68}{\mathbf{gcm}}^{\mathbf{-}\mathbf{3}}$ and its enthalpy of combustion is $\mathbf{-}\mathbf{48}{\mathbf{kJg}}^{\mathbf{-}\mathbf{1}}$.

Short answer

The amount of energy saved by walking is 3800 kJ.

Step-by-step solution

Given Information

The amount of energy used in walking 1 km is 100 kJ.

The amount of gasoline is $8.0{\mathrm{kmL}}^{-1}$.

The density of gasoline is $0.68{\mathrm{gcm}}^{-3}$.

The enthalpy of combustion is $-48{\mathrm{kJg}}^{-1}$.

Energy

The capability to do work is termed energy. Numerous energy forms are there on the Earth. Some kinds of energy include light energy and wind energy.

The energy saved by walking 1 km

The energy that is released during the combustion of gasoline and fueling the car;

$\begin{array}{c}{\mathrm{E}}_1=1\mathrm{km}\times \frac{1\mathrm{Lgasoline}}{8\mathrm{km}}\times \frac{0.68\mathrm{kg}}{1\mathrm{L}}\times \frac{48\times {10}^3}{1\mathrm{kg}}\\ =4080\mathrm{kJ}\end{array}$

The energy released in the oxidation of food is about,

$\begin{array}{c}{\mathrm{E}}_2=30\%\mathrm{efficient}\\ =\left(\frac{100}{0.3}\right)\mathrm{kJ}\\ =330\mathrm{kJ}\end{array}$

The amount of energy saved can be found as:

$\begin{array}{c}\mathrm{\Delta E}={\mathrm{E}}_1-{\mathrm{E}}_2\\ =4080\mathrm{kJ}-330\mathrm{kJ}\\ =3800\mathrm{kJ}\end{array}$

Hence, the amount of energy saved by walking is 3800 kJ.