Chapter 1, Problem 110QRT

Interpretation Introduction

Interpretation:

From the given information in the question statement, the approximate age of oceans has to be calculated and the reasonableness of the answer that the age of Earth is $\text{4}\text{.5}\text{×}{\text{10}}^{\text{9}}$ years has to be given.

Explanation of Solution

Given,

The diameter of Earth as $\text{8,000}\text{mi}$.

The surface of the earth is calculated by ${\text{4πr}}^{\text{2}}$.

  $\begin{array}{l}\text{=}\text{4}\text{×}\text{3}\text{.1415926}\text{×}{\left(\frac{\text{8000}\text{mi}}{\text{2}}\right)}^{\text{2}}{\text{(8000}\text{mi/2)}}^{\text{2}}\\ \\ \text{=}\text{2}\text{.0}\text{×}{\text{10}}^{\text{8}}{\text{mi}}^{\text{2}}\end{array}$

The Earth is covered by oceans up to $\text{67\%}$.  The surface area of earth covered by oceans is

  $\begin{array}{l}\text{=}\text{0}\text{.67}\text{×}\text{2}\text{.0}\text{×}{\text{10}}^{\text{8}}{\text{mi}}^{\text{2}}\\ \\ \text{=}\text{1}\text{.3}\text{×}{\text{10}}^{\text{8}}{\text{mi}}^{\text{2}}\end{array}$

The depth of the oceans is given as $\text{1}\text{mi}$.  The volume of the oceans is calculated as

  $\begin{array}{l}\text{Volume}\text{=}\text{surface}\text{area}\text{×}\text{depth}\\ \\ \text{=}\text{1}\text{.3}\text{×}{\text{10}}^{\text{8}}{\text{mi}}^{\text{2}}\text{×}\text{1}\text{mi}\\ \\ \text{=}\text{1}\text{.3}\text{×}{\text{10}}^{\text{8}}{\text{mi}}^{\text{3}}\end{array}$

Conversion of ${\text{mi}}^{\text{3}}$ into ${\text{cm}}^{\text{3}}$.

  $\begin{array}{l}\text{Volume}\text{=}\text{1}\text{.3}\text{×}{\text{10}}^{\text{8}}{\text{mi}}^{\text{3}}\text{×}{\left(\frac{\text{5280}\text{ft}}{\text{1}\text{mi}}\right)}^{\text{3}}\text{×}{\left(\frac{\text{12}\text{in}}{\text{1}\text{ft}}\right)}^{\text{3}}\text{×}{\left(\frac{\text{2}\text{.54}\text{cm}}{\text{1}\text{in}}\right)}^{\text{3}}\\ \\ \text{=}\text{5}\text{.6}\text{×}{\text{10}}^{\text{23}}{\text{cm}}^{\text{3}}\end{array}$

The mass of sea water and $\text{NaCl}$ is calculated as

  $\begin{array}{l}\text{=}\text{5}\text{.6}\text{×}{\text{10}}^{\text{23}}{\text{cm}}^{\text{3}}\text{×}\frac{\text{1}\text{.03}\text{g}\text{sea}\text{water}}{\text{1}{\text{cm}}^{\text{3}}}\text{×}\frac{\text{3}\text{g}\text{NaCl}}{\text{100}\text{g}\text{sea}\text{water}}\\ \\ \text{=}\text{1}\text{.7}\text{×}{\text{10}}^{\text{22}}\text{g}\text{NaCl}\end{array}$

The rate of time elapsed is

  $\begin{array}{l}\text{=}\text{1}\text{.7}\text{×}{\text{10}}^{\text{22}}\text{g}\text{NaCl}\text{×}\frac{\text{1}\text{year}}{\text{2}\text{×}{\text{10}}^{\text{16}}\text{g}\text{NaCl}}\\ \\ \text{=}\text{8}\text{.7}\text{×}{\text{10}}^{\text{5}}\text{year}\cong \text{9}\text{×}{\text{10}}^{\text{5}}\text{year}\end{array}$

The time elapsed is less than the given age of oceans.  The addition of $\text{NaCl}$ to oceans is constant throughout these years.