Chapter 1, Problem 116QRT

(a)

Interpretation Introduction

Interpretation:

From the given information, the volume of seawater that would have to be processed to obtain the required mass of gold has to be calculated.

Explanation of Solution

Given,

Debt is $\text{\$28}\text{.8}\text{×}{\text{10}}^{\text{6}}$.

The value of gold is $\text{\$21}\text{.25/troy}\text{oz}\text{(1}\text{troy}\text{oz}\text{=}\text{31}\text{.103}\text{g)}$.

The gold concentration in sea water is $\text{0}\text{.15}\text{mg}\text{gold/ton}\text{sea}\text{water}\text{(1}\text{ton}\text{=}\text{2000}\text{lb)}$.

The density of sea water is $\text{1}\text{.03}{\text{g/cm}}^{\text{3}}$.

The volume of sea water to be processed is

  $\begin{array}{l}\text{\$28}\text{.8}\text{×}{\text{10}}^{\text{6}}\text{×}\left(\frac{\text{1}\text{troy}\text{oz}}{\text{\$21}\text{.25}}\right)\text{×}\left(\frac{\text{31}\text{.103}\text{g}\text{gold}}{\text{1}\text{troy}\text{oz}}\right)\text{×}\left(\frac{\text{1}\text{mg}}{{\text{10}}^{\text{-3}}\text{g}}\right)\text{×}\left(\frac{\text{1}\text{ton}\text{sea}\text{water}}{\text{0}\text{.15}\text{mg}\text{gold}}\right)\text{×}\left(\frac{\text{2000}\text{lb}}{\text{1}\text{ton}}\right)\\ \text{×}\left(\frac{\text{453}\text{.59}\text{g}}{\text{1}\text{lb}}\right)\text{×}\left(\frac{\text{1}{\text{cm}}^{\text{3}}}{\text{1}\text{.03}\text{g}}\right)\text{×}{\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)}^{\text{3}}\text{×}{\left(\frac{\text{1}\text{km}}{\text{1000}\text{m}}\right)}^{\text{3}}\text{=}\text{250}{\text{km}}^{\text{3}}\text{sea}\text{water}\end{array}$

The volume of sea water to be processed to obtain the required mass of gold is $\text{250}{\text{km}}^{\text{3}}$.

(b)

Interpretation Introduction

Interpretation:

An Olympic-sized swimming pool is $\text{50}\text{m}\text{×}\text{25}\text{m}\text{×}\text{2}\text{.0}\text{m}$.  The number of Olympic-sized swimming pools required to hold the volume of sea water that is calculated in part (a) has to be calculated.

Explanation of Solution

The pool volume that is given in meters is converted into ${\text{km}}^{\text{3}}$.

  $\begin{array}{l}\text{=}\text{(50}\text{m)}\text{×}\left(\text{25}\text{m}\right)\text{×}\left(\text{2}\text{.0}\text{m}\right)\text{×}{\left(\frac{\text{1}\text{km}}{\text{1000}\text{m}}\right)}^{\text{3}}\\ \\ \text{=}\text{2}\text{.5}\text{×}{\text{10}}^{\text{-6}}{\text{km}}^{\text{3}}\end{array}$

The number of pools required is

  $\text{250}{\text{km}}^{\text{3}}\text{×}\left(\frac{\text{1}\text{pool}}{\text{2}\text{.5}\text{×}{\text{10}}^{\text{-6}}{\text{km}}^{\text{3}}}\right)\text{=}\text{1}\text{×}{\text{10}}^{\text{8}}\text{pools}$

The number of Olympic-sized pools required is $\text{1}\text{×}{\text{10}}^{\text{8}}\text{pools}$.