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Chapter 1: Problem 75

A package of aluminum foil contains \(50 \mathrm{ft}^{2}\) of foil, which weighs approximately \(8.0\) oz. Aluminum has a density of \(2.70 \mathrm{~g} / \mathrm{cm}^{3}\). What is the approximate thickness of the foil in millimeters?

Short Answer

Expert verified
The approximate thickness of the aluminum foil is 0.181 millimeters.

Step by step solution

01

Convert weight from ounces to grams

We are given the weight of the aluminum foil in ounces (8.0 oz). To convert this weight to grams, we use the conversion factor 1 oz = 28.35 g. Weight in grams = (8.0 oz) 脳 (28.35 g/1 oz) = 226.8 g
02

Calculate the volume of the foil using the density formula

We are given the density of aluminum as 2.70 g/cm鲁. We can use the formula Density = Mass/Volume to find the volume of the foil. Rearranging the formula to solve for the volume, Volume = Mass/Density Volume = 226.8 g / 2.70 g/cm鲁 鈮� 84 cm鲁
03

Calculate the thickness of the foil using the area and volume

We are given that the package contains 50 ft虏 of foil. We need to convert this area to cm虏 to match the units we've been using: Area = 50 ft虏 脳 (30.48 cm/ft) 脳 (30.48 cm/ft) 鈮� 4645.15 cm虏 Now, we can calculate the thickness of the foil using the formula: Thickness = Volume/Area Thickness 鈮� 84 cm鲁 / 4645.15 cm虏 鈮� 0.0181 cm Finally, we need to convert the thickness from centimeters to millimeters: Thickness = 0.0181 cm 脳 (10 mm/1 cm) 鈮� 0.181 mm The approximate thickness of the aluminum foil is 0.181 millimeters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Unit Conversion
Understanding unit conversion is essential when working with measurements in different systems. In this exercise, we start with the package weight in ounces and then transform this into grams. This is important as the density given for aluminum is in grams per cubic centimeter. Knowing that 1 ounce equals 28.35 grams allows us to perform this conversion easily.
  • Original weight: 8.0 ounces
  • Converted to grams: \(8.0 \times 28.35 = 226.8\) grams
Using consistent units simplifies the calculations as we progress through solving the problem.
Density Formula
The density formula is fundamental to finding relationships between mass, volume, and density. It is defined as \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \). In this context, we aim to find the volume of the aluminum foil. Given the mass (226.8 grams) and the density (2.70 grams per cubic centimeter), we rearrange the formula to solve for volume:
  • \( \text{Volume} = \frac{\text{Mass}}{\text{Density}} \)
  • Substituting known values: \( \frac{226.8}{2.70} \approx 84\) cubic centimeters
This volume will help us further in determining the thickness of the aluminum foil.
Volume Calculation
Once the volume is found, the next step is calculating how the volume spreads across the surface area of the foil. In our problem, we have a large area of foil, noted as 50 square feet. To use this in further equations, we must convert it to square centimeters:
  • 1 foot = 30.48 centimeters
  • So, \(50 \text{ ft}^2 = 50 \times 30.48 \times 30.48 \approx 4645.15\) square centimeters
Calculating with consistent units aids in precise mathematical assessments, making the problem easier to follow.
Area Conversion
Finally, understanding the conversion of area and the importance of converting volume to thickness is crucial. With both the total area in the correct units and the volume determined, we can find the thickness:
  • \( \text{Thickness} = \frac{\text{Volume}}{\text{Area}} \)
  • \( \text{Thickness} = \frac{84}{4645.15} \approx 0.0181\) centimeters
Converting centimeters to millimeters (since 1 cm = 10 mm) gives:
  • \(0.0181 \times 10 \approx 0.181\) millimeters
Therefore, the approximate thickness of the aluminum foil is carefully determined through these conversions and calculations.

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Most popular questions from this chapter

A cylindrical rod formed from silicon is \(16.8 \mathrm{~cm}\) long and has a mass of \(2.17 \mathrm{~kg}\). The density of silicon is \(2.33 \mathrm{~g} / \mathrm{cm}^{3}\). What is the diameter of the cylinder? (The volume of a cylinder is given by \(\pi r^{2} h\), where \(r\) is the radius, and \(h\) is its length.)

Suppose you decide to define your own temperature scale using the freezing point \(\left(-11.5^{\circ} \mathrm{C}\right)\) and boiling point \(\left(197.6^{\circ} \mathrm{C}\right)\) of ethylene glycol. If you set the freezing point as \(0^{\circ} \mathrm{G}\) and the boiling point as \(100^{\circ} \mathrm{G}\), what is the freezing point of water on this new scale?

The density of air at ordinary atmospheric pressure and \(25^{\circ} \mathrm{C}\) is \(1.19 \mathrm{~g} / \mathrm{L}\). What is the mass, in kilograms, of the air in a room that measures \(12.5 \times 15.5 \times 8.0 \mathrm{ft}\) ?

Using your knowledge of metric units, English units, and the information on the back inside cover, write down the conversion factors needed to convert (a) \(\mu \mathrm{m}\) to \(\mathrm{mm},(\mathrm{b}) \mathrm{ms}\) tons, \((\mathrm{c}) \mathrm{mi}\) to \(\mathrm{km},(\mathrm{d}) \mathrm{ft}^{3}\) to \(\mathrm{L}\).

Three spheres of equal size are composed of aluminum (density \(=2.70 \mathrm{~g} / \mathrm{cm}^{3}\) ), silver (density \(\left.=10.49 \mathrm{~g} / \mathrm{cm}^{3}\right)\) and nickel (density \(\left.=8.90 \mathrm{~g} / \mathrm{cm}^{3}\right)\). List the spheres from lightest to heaviest.
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