Problem 5 - College Physics

Question

How big is a million dollars? At the time this problem was written, the price of gold was about \(\$ 1239\) per ounce, while that of platinum was about \(\$ 1508\) an ounce. The "ounce" in this case is the troy ounce, which is equal to 31.1035 g. The more familiar avoirdupois ounce is equal to 28.35 g.) The density of gold is 19.3 \(\mathrm{g} / \mathrm{cm}^{3}\) and that of platinum is 21.4 \(\mathrm{g} / \mathrm{cm}^{3} .\) (a) If you find a spherical gold nugget worth 1.00 million dollars, what would be its diameter? (b) How much would a platinum nugget of this size be worth?

Step-by-Step Solution

Answer

The gold nugget's diameter is 13.52 cm. The platinum nugget would be worth $1,350,563.

Step 1: Determine the Mass of Gold

First, we calculate the mass of gold that is worth 1 million dollars. Given that the price of gold is $1239 per ounce, we can set up the equation: \( \\frac{1,000,000}{1239} = x \, \text{ounces} \\). This results in \( x = 807.10 \, \text{ounces} \\). Given that 1 troy ounce equals 31.1035 grams, convert the mass of gold from ounces to grams: \( \807.10 \, \text{ounces} \times 31.1035 \, \text{g/ounce} = 25100.37 \, \text{g}\).

Step 2: Calculate the Volume of Gold

Use the density of gold to find its volume. The formula for density is \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \). Rearrange to find Volume: \( \text{Volume} = \frac{\text{Mass}}{\text{Density}} \).Substitute the known values: \( \text{Volume} = \frac{25100.37}{19.3} \approx 1300.02 \, \text{cm}^{3}\).

Step 3: Find the Diameter of the Gold Nugget

Since the nugget is a sphere, use the formula for the volume of a sphere: \( V = \frac{4}{3} \pi r^3 \). Solve for radius \( r \) using:\( r = \left(\frac{3V}{4\pi}\right)^{\frac{1}{3}} \).Substitute \( V = 1300.02 \, \text{cm}^3 \):\( r = \left(\frac{3 \times 1300.02}{4\pi}\right)^{\frac{1}{3}} \approx 6.76 \, \text{cm} \).The diameter is \( 2 \times 6.76 = 13.52 \, \text{cm} \).

Step 4: Determine the Value of a Platinum Nugget with the Same Size

First, calculate the volume of a platinum nugget using the previously calculated diameter. Since the volumes are equal:\( \text{Volume}_{\text{platinum}} = \text{Volume}_{\text{gold}} = 1300.02 \, \text{cm}^3 \).Using the density of platinum, calculate the mass:\( \text{Mass}_{\text{platinum}} = \text{Density}_{\text{platinum}} \times \text{Volume}_{\text{platinum}} = 21.4 \times 1300.02 = 27842.43 \, \text{g} \).Convert the mass to ounces:\( \frac{27842.43}{31.1035} = 895.39 \, \text{ounces} \).Now, calculate the value:\( \text{Value of platinum nugget} = 895.39 \times 1508 = 1,350,563 \, \text{dollars} \).

Key Concepts

Density Calculations

Density tells us how much mass is in a certain volume of a substance. It's like figuring out how crowded a space is with matter. The formula used is simple: density = mass/volume.
In our exercise, we know the density of gold is given as 19.3 g/cm³. By using the density formula, we find how much space the gold takes up based on its mass. Once we know the mass, we can calculate the volume by rearranging the formula to: volume = mass/density.
This step is crucial because it sets the stage for figuring out the size of our gold nugget. Understanding density helps in determining how compact a material is without directly seeing its size.

Volume of a Sphere

The volume of a sphere is a bit like measuring the space inside a ball. It's calculated using the formula: \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the radius of the sphere. Knowing this formula is important when we need to find the total space a spherical object occupies.
In the step-by-step solution, calculating the volume of the gold nugget starts by using the mass and density of gold. We use the formula to find its volume and then solve for the radius.

• A simple rearrangement, \( r = \left(\frac{3V}{4\pi}\right)^{\frac{1}{3}} \), is used to find the radius from the known volume.
• Finally, the diameter is twice the radius, which gives the complete dimension of the nugget.

Understanding the concept of volume is essential for determining the practical size of real-world objects.

Troy Ounce Conversion

Troy ounces are a special unit used for precious metals like gold and platinum. They are different from the more commonly known avoirdupois ounces. In metric terms, one troy ounce equals 31.1035 grams.
For our calculations, converting ounces into grams ensures that all measurements are consistent, allowing us to use metric-based formulas effectively.
In this exercise, the conversion allows us to translate the value of gold into a measurable mass, which we can then use to find volume and, subsequently, size. Understanding the conversion between troy and metric measurements is essential for dealing with precious metals.

Gold and Platinum Valuation

Valuation of precious metals involves determining how much they are worth in terms of mass and current market price. In the exercise, we explored the value of gold and platinum by converting their weight into monetary value based on the price per ounce. Here are the key steps:

• First, calculate the total mass of the metal using either density or conversion from monetary value.
• Convert this mass into troy ounces to align with market valuation metrics.
• Finally, multiply the troy ounces by the current price per ounce to find the total value.

This method allows us to understand exactly how much a precious metal nugget is worth based on its physical characteristics and market conditions. This understanding helps in making informed assessments and investments in precious metals.