Chapter 1: Problem 88
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
Expert verified
The thickness of the aluminum foil is 0.0145 mm.
Step by step solution
01
Convert the Area from Square Feet to Square Centimeters
First, we need to convert the given area from square feet to square centimeters. We know that
1 square foot = 929.0304 square centimeters. Therefore,
1.000 square feet = 929.0304 square centimeters.
02
Use Density to Find Volume
Next, use the density formula to find the volume of the aluminum foil. The density formula is:\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}}. \]Rearranging it gives:\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}}. \]The mass is 3.636 g and the density is 2.699 g/cm³. Thus, the volume \( V \) is:\[ V = \frac{3.636 \text{ g}}{2.699 \text{ g/cm}^3} \approx 1.347 \text{ cm}^3. \]
03
Calculate the Thickness in Centimeters
The aluminum foil is a thin flat object, so its volume can also be expressed as:\[ \text{Volume} = \text{Area} \times \text{Thickness}. \]We rearrange this to find thickness:\[ \text{Thickness} = \frac{\text{Volume}}{\text{Area}}. \]Using the calculated volume and converted area, we have:\[ \text{Thickness} = \frac{1.347 \text{ cm}^3}{929.0304 \text{ cm}^2} \approx 0.00145 \text{ cm}. \]
04
Convert Thickness to Millimeters
Finally, convert the thickness from centimeters to millimeters, knowing that 1 cm = 10 mm:\[ \text{Thickness} = 0.00145 \text{ cm} \times 10 \frac{\text{mm}}{\text{cm}} = 0.0145 \text{ mm}. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Unit Conversion
To solve problems involving different measurement units, mastering unit conversion is essential.
Understanding that different unit systems are used in various contexts helps in making accurate calculations. For example, square feet and square centimeters are units for measuring area.
Key Conversion Facts
- 1 square foot = 929.0304 square centimeters
- 1 centimeter = 10 millimeters
Volume Calculation
Volume calculation is a fundamental step when you have a given mass and need to find dimensions such as thickness. This is especially true when working with objects like sheets or foils.The Density Formula and Rearrangement
The formula for density connects mass and volume and is written as:\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}}. \]To solve for volume, rearrange the formula as:\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}}. \]This simple rearrangement allows you to find the volume from known quantities of mass and density. For example, with the aluminum foil, given a mass of 3.636 g and density of 2.699 g/cm³, we calculated the volume as approximately 1.347 cm³.This step is crucial for determining dimensions like thickness from given material properties.
The formula for density connects mass and volume and is written as:\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}}. \]To solve for volume, rearrange the formula as:\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}}. \]This simple rearrangement allows you to find the volume from known quantities of mass and density. For example, with the aluminum foil, given a mass of 3.636 g and density of 2.699 g/cm³, we calculated the volume as approximately 1.347 cm³.This step is crucial for determining dimensions like thickness from given material properties.
Thickness Measurement
Thickness measurement comes in handy when dealing with thin materials like aluminum foil.To find thickness using volume, you can use the formula:Volume Relationship
For thin objects, you can express volume as:\[ \text{Volume} = \text{Area} \times \text{Thickness}. \]To solve for thickness, rearrange as:\[ \text{Thickness} = \frac{\text{Volume}}{\text{Area}}. \]This expression helps you relate the three dimensions to find thickness if area and volume are known. In the exercise, with a volume of approximately 1.347 cm³ and an area of 929.0304 cm², the calculated thickness was about 0.00145 cm. This relationship aids in measuring tiny dimensions of materials effectively.
For thin objects, you can express volume as:\[ \text{Volume} = \text{Area} \times \text{Thickness}. \]To solve for thickness, rearrange as:\[ \text{Thickness} = \frac{\text{Volume}}{\text{Area}}. \]This expression helps you relate the three dimensions to find thickness if area and volume are known. In the exercise, with a volume of approximately 1.347 cm³ and an area of 929.0304 cm², the calculated thickness was about 0.00145 cm. This relationship aids in measuring tiny dimensions of materials effectively.
Aluminum Properties
Aluminum is a popular metal known for its useful properties and applications.
Understanding its properties is essential for applications in industries like packaging, automotive, and aerospace.
General Properties
- Lightweight yet strong
- Density: 2.699 g/cm³
- Corrosion-resistant