That is an interesting approach Eric. The actual cell consists of all the mechanical pieces in addition to the nickel wire as well as the hydrogen gas. You should expand your calculation to include these if you get a chance. The actual active wire is only .055 grams if I recall. I am not sure of the mass of the other wire, the mica, the steel rod, glass shell or hydrogen gas. I hope there is an easier way to get the answer!
Dave -----Original Message----- From: Eric Walker <eric.wal...@gmail.com> To: vortex-l <vortex-l@eskimo.com> Sent: Tue, Dec 11, 2012 12:35 am Subject: Re: [Vo]:Energy Stored Within MFMP Celani Cell I wrote: I recall that chemical binding energy maxes out at around 4eV per atom in a material; so perhaps 10eV is a safe upper limit. Have you calculated what 28 KJ would be, divided among the approximate number of atoms contained within the mass that is possibly acting as a storage device? Here's a *very* rough calculation coming from the other direction. Suppose a high capacity battery made of pure nickel could store 10eV chemical binding energy per atom. How much nickel would you need to store 28 KJ? 28 KJ = 1.75E23 eV 1.75E23 eV / 10 eV*atom^-1 = 1.75 E22 atoms 1 mol = 6.02E23 atoms 1.75E22 atoms / 6.02E23 atoms*mole^-1 = 0.029 moles nickel 0.029 moles nickel * 58.70 g*mole^-1 = 1.70 g nickel I assume that the wire they are using is far heavier than 2 grams -- a US nickel coin, which is similar in composition to constantan, is 5g. Perhaps I have done something wrong in my calculation? Eric
