Question

What mass of water fell on the town in Problem 7? Water has a density of${}^{\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{\times }\mathbf{}\mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}}$.

Short answer

The mass of water that fell on the town is ${}^{\mathbf{1}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathbf{}\mathbf{kg}}$.

Step-by-step solution

Given data

Area of the town is ${}^{\mathbf{26}\mathbf{}{\mathbf{km}}^{\mathbf{2}}}$

Fall of rain is ${}^{\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{inch}}$

The density of water is ${}^{\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}}$.

Understanding the unit conversion

Unit conversion is the conversion between different units of measure for the same quantity, usually through multiplier conversion factors.

The expression for density is given as:

$\mathbf{density}\mathbf{=}\frac{\mathbf{mass}}{\mathbf{volume}}$ … (i)

Determination of the volume of the water

By converting volume into m3 and using it in relation to mass, density, and volume, you can find the mass of water that fell on the town.

From problem 7, the volume of water that fell on the town is,

$$\begin{gathered} \mathbf{V}\mathbf{}\mathbf{=}\mathbf{(}\mathbf{26}\mathbf{}{\mathbf{km}}^{\mathbf{2}}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{}\mathbf{inch}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{(}\mathbf{26}\mathbf{}{\mathbf{km}}^{\mathbf{2}}\mathbf{)}\mathbf{.}{\mathbf{\left(}\frac{\mathbf{1000}\mathbf{}\mathbf{m}}{\mathbf{1}\mathbf{}\mathbf{km}}\mathbf{\right)}}^{\mathbf{2}}\mathbf{.}\mathbf{}\mathbf{(}\mathbf{2}\mathbf{}\mathbf{inch}\mathbf{)}\mathbf{.}\mathbf{}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{0254}}{\mathbf{1}\mathbf{}\mathbf{inch}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{(}\mathbf{26}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{}{\mathbf{m}}^{\mathbf{2}}\mathbf{)}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{0508}\mathbf{}\mathbf{m}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{}{\mathbf{m}}^{\mathbf{3}} \end{gathered}$$

Thus, the volume of the water is ${}^{\mathbf{1}\mathbf{.}\mathbf{3}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{}{\mathbf{m}}^{\mathbf{3}}}$.

Determination of the mass of the water

Rearrange the equation (i) for mass.

$\mathbf{Mass}\mathbf{=}\mathbf{Density}\mathbf{\times }\mathbf{Voulme}$

Substitute the given values to calculate the mass.

$$\begin{gathered} \mathbf{Mass}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{}\frac{\mathbf{kg}}{{\mathbf{m}}^{\mathbf{3}}}\mathbf{\times }\mathbf{1}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{}{\mathbf{m}}^{\mathbf{3}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathbf{}\mathbf{kg} \end{gathered}$$

Thus, the mass of the water that fell on the town is${}^{\mathbf{1}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathbf{}\mathbf{kg}}$.