Problem 103 - Chemistry The Central Science In Si Units

Question

A U.S. 1-cent coin (a penny) has a diameter of \(19 \mathrm{~mm}\) and a thickness of \(1.5 \mathrm{~mm}\). Assume the coin is made of pure copper, whose density and approximate market price are \(8.9 \mathrm{~g} / \mathrm{cm}^{3}\) and \(\$ 2.40\) per pound, respectively. Calculate the value of the copper in the coin, assuming its thickness is uniform.

Step-by-Step Solution

Answer

The value of the copper in a penny is approximately \( \$0.020 \).

Step 1: Convert Measurements to Centimeters

First, convert the diameter and thickness of the penny from millimeters to centimeters. Divide the dimensions in millimeters by 10: \(19 \text{ mm} = 1.9 \text{ cm}\) and \(1.5 \text{ mm} = 0.15 \text{ cm}\).

Step 2: Calculate Volume of the Penny

The penny is a cylinder, so use the formula for the volume of a cylinder: \[ V = \pi r^2 h \]where \(r\) is the radius and \(h\) is the height. Here, the radius \(r = 0.95 \text{ cm}\) (since the diameter is \(1.9 \text{ cm}\)), and the height \(h = 0.15 \text{ cm}\). Calculate:\[ V = \pi (0.95)^2 (0.15) \approx 0.425 \text{ cm}^3\].

Step 3: Calculate Mass of the Copper

Using the volume from Step 2 and the density of copper (\(8.9 \text{ g/cm}^3\)), find the mass of the copper in the penny: \[ \text{mass} = \text{density} \times \text{volume} = 8.9 \times 0.425 \approx 3.7825 \text{ g}\].

Step 4: Convert Mass to Pounds

Convert the mass from grams to pounds, noting that \(1 \text{ pound} = 453.592 \text{ g}\): \[ \text{mass in pounds} = \frac{3.7825}{453.592} \approx 0.00834 \text{ pounds}\].

Step 5: Calculate the Value of the Copper

Multiply the weight of the copper in pounds by the market price per pound: \[ \text{value} = 0.00834 \times 2.40 \approx 0.020 \text{ USD}\].

Key Concepts

Cylinder Volume Calculation

To calculate the volume of a cylinder, you need the formula: \[ V = \pi r^2 h \]. Here, \( r \) is the radius of the cylinder's base, and \( h \) is the height (or thickness) of the cylinder. In the case of a penny, the diameter is given, so you first find the radius by dividing the diameter by 2. Once you have the radius, plug it into the formula along with the height to compute the volume. This calculation is necessary as it helps us determine how much space the penny occupies, a crucial first step in finding the amount of copper it contains.

Density and Mass Conversion

Density helps us relate the volume of a substance to its mass. The formula used is \( \text{mass} = \text{density} \times \text{volume} \). Essentially, density tells us how much mass is contained within a certain volume. For the penny, density allows us to calculate its mass based on its volume. Knowing the mass is critical because it links us directly to the raw material's value — copper, in this case. Thus, starting with the calculated volume of the penny, we multiply by the density of copper to get the mass in grams.

Currency Value Assessment

Next, we want to understand the actual monetary worth of the copper in the penny. We begin by converting the mass from grams to a more convenient unit for market pricing, which is pounds. The conversion factor between grams and pounds is that 1 pound equals approximately 453.592 grams. Once the mass in pounds is known, the value of copper can be calculated by multiplying it by the market price per pound. This conversion provides a practical way to assess the material's worth, bridging the gap between raw material and financial valuation.

Dimensional Unit Conversion

Converting measurements between units is a fundamental skill. For example, converting millimeters to centimeters involves dividing by 10, since there are 10 millimeters in a centimeter. This conversion is crucial because it ensures consistency in units when applying mathematical formulas, such as those for volume calculations. Without converting to a common unit system, calculations could become prone to errors, as the units would mismatched. Consistent units allow us to accurately compute the volume, mass, and ultimately the value of the materials involved.