Problem 104 - Chemistry and Chemical Reactivity

Question

A spherical steel ball has a mass of \(3.475 \mathrm{g}\) and a diameter of \(9.40 \mathrm{mm} .\) What is the density of the steel? [The volume of a sphere \(\left.=(4 / 3) \pi r^{3} \text { where } r=\text { radius. }\right]\)

Step-by-Step Solution

Answer

The density of the steel is approximately 7.99 g/cm³.

Step 1: Calculate the Radius

First, find the radius of the spherical steel ball. Given the diameter is \(9.40 \text{ mm}\), the radius \(r\) is half of that. \( r = \frac{9.40}{2} = 4.70 \text{ mm} \).

Step 2: Convert Units for Radius

Convert the radius from millimeters to centimeters as density is typically expressed in g/cm³. \( r = 4.70 \text{ mm} = 0.470 \text{ cm} \).

Step 3: Calculate the Volume of the Sphere

Utilize the sphere volume formula \( V = \frac{4}{3} \pi r^3 \) to find the volume. Calculate \( V = \frac{4}{3} \pi (0.470)^3 \approx 0.435 cm^3 \).

Step 4: Calculate Density

Density \( \rho \) is given by the formula \( \rho = \frac{m}{V} \), where \(m\) is the mass and \(V\) is the volume. Substitute \( m = 3.475 \text{ g} \) and \( V = 0.435 \text{ cm}^3 \). \( \rho = \frac{3.475}{0.435} \approx 7.99 \text{ g/cm}^3 \).

Key Concepts

Spherical Volume Calculation

When dealing with objects that are spherical in shape, calculating their volume involves a specific formula. The volume of a sphere is determined by the equation: \[ V = \frac{4}{3} \pi r^3 \]where \(V\) stands for volume and \(r\) for the radius of the sphere. To use this formula correctly, it is important to first determine the radius of the sphere. The radius is half the diameter, so if a sphere has a given diameter, you can find the radius by dividing this value by 2. After obtaining the radius, substituting it back into the formula allows you to calculate the volume. In our example, with a diameter of \(9.40\, \text{mm}\), the radius \(r\) is calculated as \(\frac{9.40}{2} = 4.70\, \text{mm}\). Converting this radius to centimeters makes the volume calculation in typical measurement units possible, resulting in a volume of approximately \(0.435\, \text{cm}^3\). This approach ensures that we have accurately calculated the space the sphere occupies, which is essential in various practical applications, such as determining material quantities or designing spherical objects.

Unit Conversion

Unit conversion is a crucial step when dealing with problems in physics and engineering as it ensures all measurements align with standard units. In density calculations especially, you typically use the unit grams per cubic centimeter (g/cm³). For the spherical steel ball, the given diameter was in millimeters. However, converting it into centimeters is necessary for consistency with the density formula units. To convert millimeters to centimeters, simply divide the value by 10 because one centimeter is equal to 10 millimeters. This approach changed the radius from \(4.70\, \text{mm}\) to \(0.470\, \text{cm}\).By standardizing units, comparing different quantities becomes manageable, and calculations like density are more straightforward. Always verifying that all metric units align in any computation helps avoid unnecessary mistakes and ensures the reliability of results.

Mass and Volume Relationship

The relationship between mass, volume, and density is fundamental in understanding material properties. This relationship is usually expressed using the formula:\[ \rho = \frac{m}{V} \]where \(\rho\) is the density, \(m\) is the mass of the object, and \(V\) is its volume.In our example, the steel ball's mass is \(3.475\, \text{g}\), and the previously calculated volume is \(0.435\, \text{cm}^3\). Plugging these values into the density formula gives us:\[ \rho = \frac{3.475}{0.435} \approx 7.99\, \text{g/cm}^3 \]This calculation reveals the density of the material making up the ball.Understanding this relationship is vital for practical applications such as material selection in manufacturing or predicting how substances will interact, based on their densities. By conceptualizing how mass and volume correlate to produce density, you improve your ability to analyze material characteristics efficiently.