Chapter 1, Problem 59E

Interpretation Introduction

Interpretation:

The total mass of mercury in the lake in kilogram unit has to be calculated.

To calculate: The total mass of mercury in the lake in kilograms unit

Answer to Problem 59E

The total mass of mercury in the lake in kilogram unit is $\text{62}\text{.95}\text{×}{\text{10}}^{\text{4}}\text{kg}\text{.}$

Explanation of Solution

Given data:

$\begin{array}{l}\text{The amount mercury in a polluted lake is 0}\text{.4}\text{μg}\text{Hg/mL}\\ \text{0}\text{.4}\text{μg/mL}\text{=}\text{4}\text{×}{\text{10}}^{\text{-7}}\text{kg/L}\\ {\text{The surface area and average depth of the lake is 100mi}}^{\text{2}}\text{ and 20 ft respectively}\text{. }\end{array}$

General unit conversions

$\begin{array}{l}\text{1}\text{mile = 1}\text{.61 km}\\ \text{1 kilometer = 1000 meters}\\ \text{1 feet = 0}\text{.3048 meters}\\ \text{1}{\text{m}}^{\text{3}}\text{=}\text{1000}\text{L}\end{array}$

Calculate the surface area of the lake in square meter unit

$\begin{array}{l}\text{100}{\text{mi}}^{\text{2}}\text{×}\frac{{\left(\text{1}\text{.61}\text{km}\right)}^{\text{2}}}{\text{1}{\text{mi}}^{\text{2}}}\text{×}\frac{{\left(\text{1000}\text{m}\right)}^{\text{2}}}{\text{1}{\text{km}}^{\text{2}}}\text{=}\left(\text{100}\text{×}\text{2}\text{.58}\text{×}\text{1000000}\right){\text{m}}^{\text{2}}\\ \text{=}\text{258}\text{×}{\text{10}}^{\text{6}}{\text{m}}^{\text{2}}\end{array}$

Calculate the average depth of the lake in meter unit

$\text{20}\text{ft}\text{×}\frac{\text{0}\text{.3048}\text{m}}{\text{1}\text{ft}}\text{=}\text{6}\text{.1}\text{m}$

Calculate the volume of the lake

$\begin{array}{l}\text{Volume}\text{=}\text{surface}\text{area}\text{×}\text{depth}\\ \text{=}\text{258}\text{×}{\text{10}}^{\text{6}}{\text{m}}^{\text{2}}\text{×}\text{6}\text{.1}\text{m}\\ \text{=}\text{15,738}\text{×}{\text{10}}^{\text{5}}{\text{m}}^{\text{3}}\\ \text{The}\text{total}\text{volume}\text{=}\text{15,738}\text{×}{\text{10}}^{\text{5}}{\text{m}}^{\text{3}}\text{×}\frac{\text{1000}\text{L}}{\text{1}{\text{m}}^{\text{3}}}\\ \text{=}\text{15,738}\text{×}{\text{10}}^{\text{8}}\text{L}\end{array}$

Calculate the total amount of mercury

$\begin{array}{l}\text{mass}\text{=}\text{concentration}\text{×}\text{volume}\\ \text{=}\frac{\text{4}\text{×}{\text{10}}^{\text{-7}}\text{kg}}{\text{1}\text{L}}\text{×}\text{15,738}\text{×}{\text{10}}^{\text{8}}\text{L}\\ \text{=}\text{62}\text{.95}\text{×}{\text{10}}^{\text{4}}\text{kg}\end{array}$

Conclusion

The total mass of mercury in the lake in kilograms unit was calculated.