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Chapter 1: Problem 86

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 0.0181 millimeters.

Step by step solution

01

Convert weight from ounces to grams

To apply the density formula, we need to convert the weight of the aluminum foil from ounces to grams. Recall the conversion factor: 1 oz ≈ 28.35 g Given that the package weighs 8.0 oz, let's convert it to grams: Weight in grams = 8.0 oz * 28.35 g/oz ≈ 226.8 g Now the weight of aluminum foil is approximately 226.8 g.
02

Calculate the volume of the aluminum foil

With the density (2.70 g/cm³) and weight (226.8 g) of the aluminum foil, we can now calculate its volume using the density formula: Density = mass/volume; therefore, Volume = mass/density. Volume = 226.8 g / 2.70 g/cm³ ≈ 84.0 cm³ Now we have the volume of the aluminum foil, approximately 84.0 cm³.
03

Calculate the thickness of the foil in centimeters

We are given the area of the package as 50 ft². To calculate the thickness of the foil in centimeters, let's first convert the area to square centimeters: 1 ft² ≈ 929.0 cm² Area in cm² = 50 ft² * 929.0 cm²/ft² ≈ 46450 cm² Now, since the volume = area x thickness, we can find the thickness: Thickness = Volume / Area Thickness = 84.0 cm³ / 46450 cm² ≈ 0.00181 cm The thickness of the aluminum foil is approximately 0.00181 cm.
04

Convert the thickness to millimeters

Now, let's convert the thickness from centimeters to millimeters: 1 cm = 10 mm Thickness in mm = 0.00181 cm * 10 mm/cm ≈ 0.0181 mm The approximate thickness of the aluminum foil is 0.0181 millimeters.

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