A package of aluminum foil contains 50 ft2 of foil, which weighs approximately 9.0 oz . Aluminum has a density of 2.70 g/cm3. What is the approximate thickness

Question

A package of aluminum foil contains 50 ft2 of foil, which weighs approximately 9.0 oz . Aluminum has a density of 2.70 g/cm3. What is the approximate thickness of the foil in millimeters?

Answer 1

v = lwh, h = v/lw

A package of aluminum foil contains 50 ft2 of foil, which weighs approximately 9.0 oz . Aluminum has a density of 2.70 g/cm3

area = lw = 50 ft^2

So, now we have h = v/50 ft^2

9.0 oz x 28.3495g/1 oz = 255.1455 g

 

1 cm^3/2.70 g  x  255.1455 g = 94.5 cm^3

94.5 cm^3 x (1 in/2.54cm)^3 = 5.8 in^3

5.8 in^3  x  (1ft/12 in)^3  = .003337 ft^3

h = vol/50 ft^2  = .003337ft^3/50 ft^2 =

6.67 x 10^-5 ft thick.

(6.67 x 10^-5 ft)  x (12 in/1 ft)  x (2.54cm/1in)  x (10 mm/cm) =.0203 mm

I did this the hard way.  Instead, change 50 ft^2 into cm^2 to accomplish an easier set of calculations.