Question

The aluminum foil on a certain roll has a total area of `18.5` `m^2` and a mass of `1275` `g`. Using a density of `2.7` `g` per cubic centimeter for aluminum, what is the thickness in millimeters of the aluminum foil?

Answer 1

Since aluminum has a density of `2.7 g//(cm)^3`, `1275g` of aluminum has a volume of `472.2 cm^3`. This is calculated by the formula linking density to mass and volume, `density = mass//volume`.

Given that `volume = area xx depth`, where volume of the aluminum foil is `472.2cm^3` and the area of the foil is (`18.5xx10^3`)`cm^2`, it follows that the depth, or the thickness, of the foil should be equal to `volume//area`.

The volume and the area should all be expressed in the same units, which is why I converted the `m^2` to `cm^2`. As you require the thickness in `mm`, the volume and area should be converted to `mm^3` and `mm^2` respectively.

So you have an aluminum foil of volume (`472.2xx10^3`)`mm^3` and area (`18.5xx10^4`)`mm^2`. You need the thickness of such a foil, so you substitute the values in the formula `depth = volume//area`, which follows as `depth = (472.2xx10^3)//(18.5xx10^4)` `=> 2.55 mm` `~~ 2.6mm`