Chapter 1, Problem 1.65P

a)

Interpretation Introduction

Interpretation:Value of gold in double eagle before and after 1933 price change is to be determined.

Concept introduction:Conversion factor is ratio that is widely used to describe relation between two different units. In order to convert one unit into another, original unit is multiplied with proper conversion factor to yield new unit. This is done as mentioned below.

  $\text{Final unit}=\left(\cancel{\text{Original unit}}\right)\left(\frac{\text{Final unit}}{\cancel{\text{Original unit}}}\right)$

Answer to Problem 1.65P

Values of gold in double eagle before and after 1933 price change are $\$19.3518$ and $\$33.8657$ respectively.

Explanation of Solution

Since double eagle contains $90.0\text{ \%}$ gold by mass, amount of gold in double eagle can be calculated as follows:

  $\begin{array}{c}\text{Amount of gold}=\left(\frac{90\%}{100\text{ \%}}\right)\left(33.436\text{ g}\right)\\ =30.092\text{ g}\end{array}$

This amount is to be converted into $\text{troy oz}$ . Conversion factor for this is as follows:

  $\begin{array}{c}\text{Amount of gold}=\left(30.092\text{ g}\right)\left(\frac{1\text{ troy oz}}{31.1\text{ g}}\right)\\ =0.96759\text{ troy oz}\end{array}$

The value of gold before 1933 can be calculated as follows:

  $\begin{array}{c}\text{Value of gold before 1933}=\left(0.96759\text{ troy oz}\right)\left(\frac{\$20.00}{1\text{ troy oz}}\right)\\ =\$19.3518\end{array}$

The value of gold after 1933 can be calculated as follows:

  $\begin{array}{c}\text{Value of gold before 1933}=\left(0.96759\text{ troy oz}\right)\left(\frac{\$35.00}{1\text{ troy oz}}\right)\\ =\$33.8657\end{array}$

b)

Interpretation Introduction

Interpretation:Number of coins that can be made from $50\text{ troy oz}$ of gold is to be determined.

Concept introduction:Conversion factor is ratio that is widely used to describe relation between two different units. In order to convert one unit into another, original unit is multiplied with proper conversion factor to yield new unit. This is done as mentioned below.

  $\text{Final unit}=\left(\cancel{\text{Original unit}}\right)\left(\frac{\text{Final unit}}{\cancel{\text{Original unit}}}\right)$

Answer to Problem 1.65P

52 coins can be made from $50\text{ troy oz}$ of gold.

Explanation of Solution

Since double eagle contains $90.0\text{ \%}$ gold by mass, amount of gold in double eagle can be calculated as follows:

  $\begin{array}{c}\text{Amount of gold}=\left(\frac{90\%}{100\text{ \%}}\right)\left(33.436\text{ g}\right)\\ =30.092\text{ g}\end{array}$

This amount is to be converted into $\text{troy oz}$ . Conversion factor for this is as follows:

  $\begin{array}{c}\text{Amount of gold}=\left(30.092\text{ g}\right)\left(\frac{1\text{ troy oz}}{31.1\text{ g}}\right)\\ =0.96759\text{ troy oz}\end{array}$

Since $0.96759\text{ troy oz}$ of gold make one coin, number of coins formed by $50\text{ troy oz}$ of gold can be calculated as follows:

  $\begin{array}{c}\text{Number of coins}=\left(\text{50 troy oz}\right)\left(\frac{1\text{ coin}}{0.96759\text{ troy oz}}\right)\\ =51.7\text{ coins}\\ \approx \text{52 coins}\end{array}$

c)

Interpretation Introduction

Interpretation:Number of coins that can be made from $\text{2}{\text{.00 in}}^3$ of gold is to be determined.

Concept introduction:Density is defined as mass per unit volume. It is denoted by $d$ . The expression to calculate density is as follows:

  $d=\frac{M}{V}$

Where,

• $d$is the density.
• $M$is mass.
• $V$is volume.

Answer to Problem 1.65P

21 coins can be made from $\text{2}{\text{.00 in}}^3$ of gold.

Explanation of Solution

The expression to calculate density of gold is as follows:

  $d=\frac{M}{V}$

Where,

• $d$is the density of gold.
• $M$is mass of gold.
• $V$is volume of gold.

Rearrange above equation for $M$ .

  $M=dV$

The value of $d$ is $\text{19}{\text{.3 g/cm}}^{\text{3}}$ .

The value of $V$ is $\text{2}{\text{.00 in}}^3$ .

Substitute the values in above equation.

  $\begin{array}{c}M=dV\\ =\left(19.3{\text{ g/cm}}^3\right)\left(2.00{\text{ in}}^3\right)\left(\frac{16.3871{\text{ cm}}^3}{1{\text{ in}}^3}\right)\\ =632.54206\text{ g}\end{array}$

This amount is to be converted into $\text{troy oz}$ . Conversion factor for this is as follows:

  $\begin{array}{c}\text{Amount of gold}=\left(632.54206\text{ g}\right)\left(\frac{1\text{ troy oz}}{31.1\text{ g}}\right)\\ =20.339\text{ troy oz}\end{array}$

Since $0.96759\text{ troy oz}$ of gold make one coin, number of coins formed by $20.339\text{ troy oz}$ of gold can be calculated as follows:

  $\begin{array}{c}\text{Number of coins}=\left(20.339\text{ troy oz}\right)\left(\frac{1\text{ coin}}{0.96759\text{ troy oz}}\right)\\ =21.02\text{ coins}\\ \approx 21\text{ coins}\end{array}$