Question

A sheet of aluminum (Al) foil has a total area of \(1.000 \mathrm{ft}^{2}\) and a mass of \(3.636 \mathrm{~g}\). What is the thickness of the foil in millimeters? (Density of \(\left.\mathrm{Al}=2.699 \mathrm{~g} / \mathrm{cm}^{3} .\right)\)

Short answer

The thickness of the aluminum foil is \(0.014 \, mm\).

Step-by-step solution

Convert the Units of Area from Square Feet to Square Centimeters

Given that \(1 \, \text{foot} = 30.48 \, \text{cm}\), the conversion can be calculated by \(1.000 \, \text{ft}^{2} = 1.000 \times (30.48)^2 \, \text{cm}^{2} = 929.0304 \, \text{cm}^{2}\)

Substitute Values into Density Formula

Density (\(ρ\)) is defined as mass (\(m\)) divided by volume (\(V\)). We therefore can formulate \(ρ = m / V \). As we are looking for the thickness (which is a part of the volume), we understand the volume of a sheet as the product of area (\(A\)) and thickness (\(t\)). Therefore the equation can then be rearranged to \(t = m/(ρ \cdot A)\).

Calculation of Thickness

Inserting the values to our formula we get: \(t = 3.636\, g / (2.699 \, g/cm^{3} \cdot 929.0304 \,cm^{2}) = 0.0014 \,cm\)

Convert the Result to Millimeters

Now that we have our solution in centimeters, we convert it to millimeters. Knowing that \(1 \, cm = 10 \, mm\), we get the final result of \(0.014 \, mm\).

Key concepts

Unit Conversion

Understanding unit conversion is fundamental when working with measurements in different systems. In this exercise, the area of the aluminum foil is given in square feet, which needs conversion to square centimeters to match the density's units.

• The conversion factor for lengths is: \(1 \, \text{foot} = 30.48 \, \text{cm}\).
• Therefore, the area in square feet can be converted to square centimeters by squaring the length conversion factor: \(1.000 \, \text{ft}^{2} = (30.48)^2 \, \text{cm}^{2} = 929.03 \, \text{cm}^{2}\).

Conversions like these are common in science to ensure consistent measurements and calculations. Always remember to square the conversion factor when dealing with area conversions.

Aluminum Foil

Aluminum foil is a thin sheet of metal commonly used in the kitchen. However, in scientific exercises like this one, we explore its properties in more detail.

• Aluminum has a given density of \(2.699 \, \text{g/cm}^3\), crucial for calculating the volume and subsequently the thickness of the foil.
• Understanding its density helps relate mass and volume, allowing us to determine how thick a sheet is based on its mass and area.

The practical application of knowing how to manipulate these properties extends to fields like engineering and materials science.

Thickness Calculation

Calculating thickness involves understanding how mass, volume, and area relate to each other. In this problem, we use density to find the thickness of the aluminum foil.
First, recall that density \( \rho \) is mass divided by volume:\[\rho = \frac{m}{V}\]Since volume for a flat sheet can be expressed as the area \( A \) times thickness \( t \), the formula becomes:\[t = \frac{m}{\rho \cdot A}\]Substitute the given values:

• Mass \( m = 3.636 \, \text{g} \)
• Density \( \rho = 2.699 \, \text{g/cm}^3 \)
• Area \( A = 929.0304 \, \text{cm}^2 \)

Calculate the thickness:\[t = \frac{3.636}{2.699 \times 929.0304} = 0.0014 \, \text{cm}\]Finally, convert it to millimeters by knowing that \(1 \, \text{cm} = 10 \, \text{mm}\), yielding a final thickness of \(0.014 \, \text{mm}\). This procedure exemplifies how fundamental concepts intertwine to solve real-world measurement challenges.