Final answer:
Calculating the thickness of the aluminum foil involves converting the weight from ounces to grams, the area from square feet to square centimeters, and then using the density of aluminum to find the volume. The thickness is then obtained by dividing the volume by the area, which results in a thickness of approximately 0.017 mm.
Step-by-step explanation:
To calculate the approximate thickness of the aluminum foil in millimeters, we'll need to use the given density of aluminum, 2.70 g/cm3, and the weight and area of the foil from the package. We know the weight is 7.5 oz, which we need to convert to grams. There are approximately 28.35 grams in an ounce, so the weight in grams is 7.5 oz × 28.35 g/oz = 212.625 g.
Next, we convert the area from square feet to square centimeters. There are 929.03 cm2 in a square foot, so for 50 ft2, the area in cm2 is 50 × 929.03 cm2/ft2 = 46451.5 cm2.
Now with the mass (m) and the area (A), we can calculate the volume (V) because density (ρ) equals mass divided by volume (ρ = m / V). The volume of the foil in cubic centimeters is V = m / ρ = 212.625 g / 2.70 g/cm3 ≈ 78.75 cm3.
Finally, to find the thickness (T), we use the volume and the area (V = A × T). The thickness in centimeters is T = V / A ≈ 78.75 cm3 / 46451.5 cm2 ≈ 0.001695 cm, and converting to millimeters (1 cm = 10 mm) gives us approximately 0.01695 mm, or about 0.017 mm as the approximate thickness of the foil.