davefoc
Philosopher
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BY the way if you want to know, making such a calculation, and taking a very rough approximation, I get that a sample 2000 years old should have 78% 14C left, a sample from from 750 years ago ( from year ~1250 ) has 91% 14C left.
So you have the following :
Time period 14C left quantity per mole
Modern carbon 100% 1 carbon 14C per 1 trillion C so 6.0 10^11 per mole
0 AD carbon 78% 0,78 carbon 14C per 1 trillion C so ~4,7 10^11 per mole
Seemingly 1250 AD 91% 0,91 carbon 14C per 1 trillion C so ~5,5 10^11 per mole
Let us define the quantity Mc , Mh, Mt as follow :
Mh=historic quantity of carbon original
Mt=total sample given
Mc=contaminating foreign modern carbon
m Carbon molar mass
It is obvious that Mt=Mh+Mc and also that the relationship ebtween the carbon quantity is fixed (number carbon 14 sample=contaminated carbon 14+historical 14C)
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I used Aepervius's data and thought about the problem in my own way. My answer was:
1.6 moles of carbon contaminant for every mole of genuine sample tested.
My logic:
Ms - moles of carbon in the unadulterated sample
Mc - moles of carbon in contaminant
4.7 x 1011 * Ms = C14 atoms in the unadulterated sample
6.0 x 1011 * Mc = C14 atoms in the contaminant
5.5 x 1011 * (Mc+Ms) = C14 atoms in adulterated sample
4.7 x 1011 * Ms + 6.0 x 1011 * Mc = total C14 atoms in adulterated sample
4.7 x 1011 * Ms + 6.0 x 1011 * Mc = 5.5 x 1011 * (Mc+Ms)
solve (I hope) for Mc in terms of Ms
Mc = 1.6 Ms
The way I interpret this result is that for every gram of a 0 AD unadulterated sample there would need to be 1.6 grams of contaminant from a modern source added to look like a 1250 AD sample. If the contaminant was not modern more contaminant would be required to get the same result.
I didn't check my work by comparing it with what the experts have calculated but I knew it was a large number and that is why I thought the contaminant theory had been rejected by the shroud authenticity advocates.
Any ridicule for an error in logic or calculation will be appreciated.
ETA:
Aepervius' weird spelling of soot (does he pronounce it so that it rhymes with out and doubt? Is this some weird British thing like gaol?) got me to thinking about how much soot would be required to contaminate the sample. It is a smaller number since a much higher proportion of soot is carbon than of a textile.
I didn't find much about the composition of fabric by elements on line but assuming that fabric is made up principally of compounds consisting of glucose a rough estimate of the percentage of fabric that is carbon is 40%. This is based on glucose formula C6H12O6 atomic weight =~ 6 * 12 + 12 * 1 + 6 * 16 =~ 180.
So if the contaminant was modern soot (or sout
.64 grams of soot (.4 * 1.6) for every gram of uncontaminated 0AD sample
And that's a much smaller number. Maybe nobody noticed that about 40% of the sample by weight was soot.
And that's a much smaller number. Maybe nobody noticed that about 40% of the sample by weight was soot.
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