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, determine the thickness in millimeters of the aluminum foil.

Answer 1

The thickness in millimeters of the is 0.02552.

Step-by-step explanation:

Area of an aluminum foil = A =
`18.5 m^2=185,000 cm^2`

`1 m^2=10000 cm^2`

Mass of the assuming foil = 1275 g

Volume of the aluminum foil = V

Thickness of the assuming foil = h

Density of the aluminum foil =
`2.7 g/cm^3`

`Density=(Mass)/(Volume)`

`2.7 g/cm^3=(1275 g)/(V)`

`V = 472.222 cm^3`

Volume = Area × Depth

`V=A* h`

`h=(472.222 cm^3)/(185,000 cm^2)=0.002552 cm`

1 cm = 10 mm

h = 0.02552 mm

The thickness in millimeters of the is 0.02552.

Answer 2

We use formula, the volume of aluminium

`Volume = (mass)/(density) \\\\ Area * thickness = (mass)/(density) \\\\  thickness = (mass)/( Area * density)`

Given,
`m = 1275 \ g`,
`Area = 18.5 m^2 =18.5 * 10^4 mm^2` and
`density =  2.7 \ g/cm^3 = 2.7 * 10^(-3)  g/mm^3`

Substituting these values, we get

`thickness = (1275 \ g)/( 18.5 * 10^4 mm^2 * 2.7 * 10^(-3)  g/mm^3) =  2.55 \ mm`.

Thus, the thickness of the aluminium foil is 2.55 mm.