Chapter 1, Problem 1.94QE

Interpretation Introduction

Interpretation:

A package of aluminum foil with an area of $\text{75}{\text{ft}}^{\text{2}}$ weighs $\text{12}\text{ounces}\text{avdp}$.  The density of aluminum is $\text{2}\text{.70}{\text{/cm}}^{\text{3}}$.  The average thickness of the foil has to be calculated in nanometers.

Explanation of Solution

Given, area of aluminum foil is $\text{75}{\text{ft}}^{\text{2}}$, mass is $\text{12}\text{ounces}\text{avdp}$ and the density of aluminum is $\text{2}\text{.70}{\text{/cm}}^{\text{3}}$.

The mass of the aluminum foil is converted to grams from ounces.

  $\begin{array}{l}\text{1}\text{oz}\text{avdp}\text{=}\text{28}\text{.35}\text{g}\\ \\ \text{mass}\text{=}\text{12}\text{oz}\text{avdp}\text{×}\left(\frac{\text{28}\text{.35}\text{g}}{\text{1}\text{oz}\text{avdp}}\right)\\ \text{=}\text{340}\text{.2}\text{g}\end{array}$

From density and mass, the volume of the aluminum foil is calculated.

  $\begin{array}{l}\text{Volume}\text{=}\frac{\text{mass}}{\text{Density}}\\ \text{=}\frac{\text{340}\text{.2}\text{g}}{\text{2}\text{.70}{\text{g/cm}}^{\text{3}}}\\ \text{=}\text{126}{\text{cm}}^{\text{3}}\end{array}$

The given area of foil is converted from square feet to square centimeter.

  $\begin{array}{l}\text{1}\text{ft}\text{=}\text{12}\text{in}\\ \text{1}\text{in}\text{=}\text{2}\text{.54}\text{cm}\\ \\ \text{Area}\text{=}\text{75}{\text{ft}}^{\text{2}}\text{×}{\left(\frac{\text{12}\text{in}}{\text{1}\text{ft}}\right)}^{\text{2}}\text{×}{\left(\frac{\text{2}\text{.54}\text{cm}}{\text{1}\text{in}}\right)}^{\text{2}}\\ \text{=}\text{75}{\text{ft}}^{\text{2}}\text{×}\left(\frac{\text{144}{\text{in}}^{\text{2}}}{\text{1}{\text{ft}}^{\text{2}}}\right)\text{×}\left(\frac{\text{6}\text{.45}{\text{cm}}^{\text{2}}}{\text{1}{\text{in}}^{\text{2}}}\right)\\ \text{=}\text{69660}{\text{cm}}^{\text{2}}\end{array}$

The thickness can be calculated as

  $\begin{array}{l}\text{volume}\text{=}\text{thichness}\text{×}\text{area}\\ \text{thickness}\text{=}\frac{\text{volume}}{\text{Area}}\\ \\ \text{thickness}\text{=}\frac{\text{126}{\text{cm}}^{\text{3}}}{\text{69660}{\text{cm}}^{\text{2}}}\\ \text{=}\text{0}\text{.0181}\text{cm}\\ \\ \text{1}\text{cm}\text{=}{\text{10}}^{\text{-2}}\text{m}\\ \text{1}\text{nm}\text{=}{\text{10}}^{\text{-9}}\text{m}\\ \\ \text{thickness}\text{=}\text{0}\text{.0181}\text{cm}\text{×}\left(\frac{{\text{10}}^{\text{-2}}\text{m}}{\text{1}\text{cm}}\right)\text{×}\left(\frac{\text{1}\text{nm}}{{\text{10}}^{\text{-9}}\text{m}}\right)\\ \text{=}\text{1}\text{.81}\text{×}{\text{10}}^{\text{4}}\text{nm}\end{array}$