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Conservation of mass for burning log 
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#1
Mar305, 12:28 PM

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Suppose I burned a log. If I collected all the products of the burning process (the smoke particles, the ashes, etc.) would they have the same exact mass as my original log? Or would they have less mass because they are at a lower energy state then the original log (Energymass equivalence)? Thanks.



#2
Mar305, 04:51 PM

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PF Gold
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They'd have a little less mass, since some of the mass has been released as heat, light etc.



#3
Mar305, 06:20 PM

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AM 


#4
Mar305, 08:19 PM

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PF Gold
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Conservation of mass for burning log
That happens only in nuclear processes. Burning a log does not convert mass into energy. This experiment has, in fact, been done many times the total mass solid ash, water, and gases, less, as Andrew Mason reminded me, the atmospheric oxygen trapped in carbon dioxide and perhaps carbon monoxide remains the same. 


#5
Mar305, 10:04 PM

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#6
Mar305, 11:24 PM

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(Btw, first post! Hey everybody! ) 


#7
Mar305, 11:29 PM

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Have a look at this, for example: http://www.ccmr.cornell.edu/education/ask/?quid=590 "As an example, the energy released in chemical reactions when an average person shovels snow for one hour amounts to a mass loss (by E=mc2) of only 10 billionths of a gram!"AM 


#8
Mar505, 06:04 PM

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There's a derivation / disucssion of the "gravitational mass" (not my wording) of an electromagnetically bound system in http://lanl.arxiv.org/abs/grqc/9909014 by Steve Carlip which dots all the i's and crosses all the t's. (The argument is presented in detail only for the weak field case and in the case where the internal velocities of the matter are nonrelativistic, but another paper is referenced to support the argument in general).
The end result is that the total "gravitational mass" of a system is the sum of the rest masses, mc^2, plus the total energy of the system E. Use is made of the virial theorem to derive this result. E is divided into two parts, potential energy U and kinetic energy T, and the nonrelativistic virial theorem states that T = U/2. 


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