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Thickness of final plating on U.S. Lincoln 2010 To Date Shield Cents
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<p>[QUOTE="messydesk, post: 2850029, member: 1765"]If a current cent, weighing 2.5 grams is 2.5% copper by weight, then it contains .0625 grams of copper.</p><p><br /></p><p>The surface area if a cent (19.05mm diameter, 1.52 mm thick) is 661 mm^2.</p><p><br /></p><p>661 * density of copper (0.0027 g/mm^3) * thickness of copper (mm) should equal 0.0625 g. If you solve for the thickness, it comes out to 35 microns, assuming there is no copper in the zinc substrate.[/QUOTE]</p><p><br /></p>
[QUOTE="messydesk, post: 2850029, member: 1765"]If a current cent, weighing 2.5 grams is 2.5% copper by weight, then it contains .0625 grams of copper. The surface area if a cent (19.05mm diameter, 1.52 mm thick) is 661 mm^2. 661 * density of copper (0.0027 g/mm^3) * thickness of copper (mm) should equal 0.0625 g. If you solve for the thickness, it comes out to 35 microns, assuming there is no copper in the zinc substrate.[/QUOTE]
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Thickness of final plating on U.S. Lincoln 2010 To Date Shield Cents
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