Question:
Find the thickness of a piece of aluminum foil?
KyLiE<3
2010-03-27 12:21:24 UTC
Length = 30 cm
Width = 30 cm
Volume = 5 mL
Mass = 4.8 g
Density = 0.96g/cm^3


How do I find the depth or height of the piece of foil?
Five answers:
Peppers_Ghost
2010-03-27 12:30:44 UTC
just plug in all the number



area = L x W = 30 x 30 = 900 cm^2



Volume = Thickness x area



thickness = V/900 (in cm)



mass = 4.8 g



denisty = mass/volume



0.96 = 4.8 / vol



0.96/4.8 = 1/Vol = 0.2 therefore Volume = 1/0.2 = 5



thickness = 5/900 = 0.00556 cm



= 0.0556 mm

= 55.6 um
That 'Guy'
2010-03-27 12:25:27 UTC
"Assuming that you haven't an instrument sensitive enough to measure the thickness of a sample of foil directly, then you will have to measure something easier to measure and then do some maths. Try any of the following:



1. Measure the length and width of a piece of aluminium foil. Look up the density of aluminium in Kaye and Laby (or, if you can find a suitable chunk of aluminium, a pair of scales and a graduated cylinder, measure it yourself; then look it up anyway to see how accurate your measurement was). Weigh your piece of foil (feel free to fold it up or crumple it into a ball to make it fit on the scales; as long as you don't tear any off, it will still weigh the same) and calculate its volume using mass / density. Divide this by the area you measured earlier to get the thickness.



2. Place several layers of foil in intimate contact with one another and measure the thickness of the stack. Divide by the number of layers to get the thickness of each layer. Or, keep adding layers and measuring; plot a graph of thickness vs. number of layers and measure the slope.



3. Use a lens to project a magnified image of the edge of a sample of foil onto a screen. Divide the thickness of the foil *in the image* by the magnification factor (remember, light travels in straight lines; so, by the properties of similar triangles, the magnification factor will be equal to the ratio of the distances of image and object from the lens) to get the thickness of the *real* foil.



Actually, if this is a coursework project, do all three (plus any more that you can think of or that other people suggest) and compare the results of each method. Where might errors creep in? What can you do to minimise them?"
anonymous
2010-03-27 12:26:57 UTC
I do not know how to find the depth or height of a piece of aluminum foil. But I do know that it has a thickness of less than 0.2 mm / 0.008 in, although much thinner gauges down to 0.006 mm are commonly used.
Abby
2016-04-15 01:38:23 UTC
0.18 oz *(1 lb/16 oz)*(454 g/lb)*(1 cm^3 / 2.70 g)*(10 mm/1 cm)^3 / [(12 in*2.54 cm/in*10mm/cm)*(15.5 in*2.54 cm/in*10 mm/cm)] I converted the oz to grams, then grams to cubic cm, which I then divided by the product of the length and width in cm. That should end up with the thickness in mm, all in one line. Someone check my work pleez.
Ladonna
2010-03-27 12:23:35 UTC
Use a ruler.


This content was originally posted on Y! Answers, a Q&A website that shut down in 2021.
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