My Inner Archivist is screaming for a link so I can save this for posterity!
The interview is here:
http://www.911freefall.com/2012/12/mark-basile-and-wtc-dust.html
And he speculates about where the thermite and its residue was towards the end - can't give you a time stamp right now. I think after 45 minutes. Going to bed next, so I don't want to look it up myself yet.
Iron's enthalpy of vaporization is a little over 6 kJ/g, so their energy account is already almost depleted. (2 g thermite produces 1 g iron, enthalpy of thermite is 4 kJ/g) I haven't been able to find alumina. It's quite probable that alumina begins to dissociate back to aluminum and oxygen at that temperature, which makes things complicated!
Let me double-check for accuracy...
Iron's enthalpy of vaporization is 340 kJ/mol
Standard atomic weight is 55.845 (g/mol)
So enthalpy of vaporization is (340 kJ/mol) / (55.845 g/mol) = 6,09 kJ/g.
Cool, you're right
Heat capacity is 25.1 J/mol/K = 0,449 J/g/K
Melting point is 1538 °C
So to get Iron to its melting point from room temperature, 20 °C, you expend 25.1*(1538-20) J/mol = 38.10 kJ/mol
Add to that the Heat of Fusion = 13,81 kJ/mol = 0,247 kJ/g, and you need 51.91 kJ/mol = 0.93 kJ/g to melt iron at juuust above 1538 °C
Same with
Al2O3:
Molar mass is 101.96 g/mol
Heat capacity (
Specific Heat) = 0.88 J/g/K = 89,72 J/mol/K
Melting point = 2072 °C
Heat of fusion = 1,09 kJ/g = 111,1 kJ/mol
Boiling point = 2977 °C
Heat of vaporization = ?
So we expend (2072-20) °C * 0.88 J/g/K = 1.81 kJ/g to bring Alumina to the brink of melting, and an additional 1,09 kJ/g to melt it.
1 g of stoichiometric thermite produces ideally 0.523 g iron and 0.477 g alumina.
The mix has thus an average heat capacity of (0.523*0,45 + 0,477*0.88) J/g/K = 0.655 J/g/K.
To heat it to the melting point of iron you need 1518 K * 0,655 J/g/K = 0.994 kJ/g of thermite.
Then the iron melts and eats (0.523 * 0,247) kJ/g = 0.129 kJ/g of thermite.
Then we continue to the melting point of alumina, or an additional (2072-1538) K = 534 K. Let's assume the same 0.655 J/g/K still apply - that costs 0.35 kJ/g of thermite.
Then the alumina melts and eats 0.477 * 1,09 kJ/g = 0.52 kJ/g of thermite
Then we heat up an additional (2862-2072) K = 790 K to the boiling point of iron. Still assuming a constant heat capacity of 0.655 J/g/K, that would cost 0.52 kJ/g of thermite.
So up to this point, the thermite products would have swallowed
(0.13 + 0.35 + 0.52 + 0.52) kJ/g = 1.52 kJ/g.
To now boil off the iron, we'd need (0.523 * 6,09) kJ/g of thermite = 3,2 kJ/g.
However, the thermite reaction only releases 3.96 kJ/g, and after expending 1.52 kJ/g already, we are left with only 2.44 kJ/g - enough to vaporize at most 76% of the iron - and none of the alumina, which has a higher boiling point.
So that's indeed the max temperature the thermite reaction could reach in theory.
I have ignored the fact that heat capacity is not constant, it tends to increase with temperature. Also ignored are all the losses that you encounter in practice - that you start out with surplus alumina, that you can't avoid losses due to radiation, etc.
