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From lyman.Stanford.EDU!rjg  Sat May  4 14:15:35 1996
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Date: Sat, 4 May 1996 14:11:27 -0700 (PDT)
From: "Richard J. Green" 
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To: Jamie McCarthy 
Subject: Re: 960502: It is amazing that the world has not yet been informed of this
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On Sat, 4 May 1996, Jamie McCarthy wrote:

> Calculations by Rich Green, presented on this newsgroup, clearly
> demonstrate that at the temperatures used by the Nazi crematory ovens,
> the heat energy released by the burning tissue was more than sufficient
> to maintain the incineration and to evaporate the water in the body.
> (This article is emailed to Rich Green, who is invited to provide the
> calculations in a followup article.  Since this article is for a lay
> audience, I won't get too technical.)

I note that I have not dealt with the cooling of the ovens themselves
through convection etc.

It has been asserted in this news group that burning a body is not
exothermic.  As evidence we are told that one cannot light a hamburger
on fire with a match.  The proponents of this view do not understand the
meaning of the word exothermicity.  Exothermicity means that the
reaction releases heat.  It is also impossible to light a large log on
fire with a match; yet that log releases heat when burned.  The
difficulty in lighting the log is a kinetic issue rather than a
thermodynamic one.  Energy is needed to overcome an activation barrier.
This barrier is why one uses tinder and kindling or else lighter fluid
to start a fire.

Let's do a few calculations to prove the point.  Any errors are mine.

1) Let us first show roughly that burning a hamburger is exothermic under a
   gross approximatio under a gross approximation.  We will do a better job
   when treating the human body.

   Suppose we have a 100 g hamburger.  Suppose that 80% is water.
   Suppose that 20% is combustible.  

   Let's treat the 20% as if it were methane.  This is not quite right, but 
   it should give a correct order of magnitude.   We will fine tune when
   we address the human body.

   Let's combust the methane:

	The enthalpy of combustion of methane is -890.4 kJ/mol.

	20 g * 1 mol/16 g = 1.25 mol methane
	1.25 mol * (-890.4) = -1113 kJ

   Let's vaporize the water:

   The heat of vaporization of water at 100 C is 40.656 kJ/mol.
   The heat capacity of water (Cp) is 4.184 J/g*K

	Let's suppose our burger starts at 25 C (since it has no
	residual body heat, clearly we will be overestimating
	the amount needed for a warmer body).

	To heat the water to 100C we will need:

	80 g * 75K * 4.2 J/gK = 25200J = 25.2 kJ

	Now to vaporize this water we will need:

	80g * 1 mol/18g * 40.7 kJ/mol =  180.6 kJ

    So to vaporize all the water takes:

	180.6 kJ + 25.2 kJ = 205.8 KJ

   The total heat given off by entirely burning the burger is:

	205.8 kJ - 1117 kJ = - 911 kJ

2)  Let's extend our rough model to a human.  We will fine tune it
    afterwards.  Suppose a human weighs 75 kg.  Our burger weighed .1 kg.
	750 * (-911) = - 683,250 kJ
   Certainly a more accurate calculation can be done.  I have from M.

	A) He suggests 50-70% water may be a better estimate.
	B) He suggests 14% and 22% fat for the avearge man and woman
        C) He suggests a minimum of 16.8% proteins.

3) Let us do an underestimate: assume the body is 80% water and 14%

	Mark Van Alstine gives us an example fat:

	The combustion equation for the fat glyceryl trimyrisate, C45H86O6, 
	for example, is:

	C45H86O6(s) + 127/2 O2(g) -> 45C02(g) + 43H20(l); H = -27820 kJ/mol

The molecular mass of this fat is:

	M = 45*12 + 86 + 6*16 = 540 + 86 + 96 = 722.

So 1 g. of this fat has a heat of combustion:

	H = -27820 kJ/mol  * 1 mol/ 722 g. = -38.5 kJ/g.

Suppose that the human body is 14% fat and 80 % water.  In 100 g of a
body we have 14 g of fat that release

	14 g * (-38.5 kJ/g) = - 539 kJ when burnt. 205.8 kJ are needed
	to evaporate 80 g water.

	So every 100 g of a body releases -333 kJ ignoring the presence
	of proteins, urea etc., which also burn exothermically.

A 75 kg person releases about (750) * (-333) = 249,750 kJ. from fat
alone taking into account the necessity of evaporating water.

> Rich Green has commented that having a mass of people in the gas chamber
> would increase, rather than decrease, the diffusion rate of the gas,
> because the people would of course be moving around and creating
> turbulence.  I submit that he knows a bit more about chemistry and gas
> diffusion than Ms. Rimland does.  I submit that he probably knows more
> than our anonymous source, who offers no qualifications whatsoever.

I don't remember stating this.  It seems to me that the most important
contribution of the people, however, is to decrease the overall volume
of air (increasing the concentration of HCN) and increasing the
temperature which will increase the rate of turbulent diffusion.

Assuming homogenous diffusion is, of course, only an approximation.
Diffusion problems become intractable very quicky, and there is no point
in going to a higher level of approximation until the deniers show that
there is a real issue.


Rich Green

Richard J. Green                                   
"If it works, take it apart and find out why."  - unknown

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