How would one mathematically determine the temperature (if possible) of a detonation (non-chemical) of something on a macroscopic scale; more specifically: detonations from matter and anti-matter annihilation that far-exceed the yield of any human-made weapon? If this question is ambiguous, then i apologize i can try to specify if needed.
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2 Answers
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Firstly you have to understand that when matter and antimatter collide they form gamma-ray photons and no conceivable actual matter remains if equal amount of matter and anti-matter is present. As temperature is actually just the average kinetic energy of matter(with rest-mass) it is not possible here to find the temperature inside the explosion chamber as there is no matter just photons. Now, the energy of those photons is given by the Einstein's mass-energy equivalence. So if the mass of the matter and antimatter combined is $m$, the energy released is given by: $$E=mc^2$$ Where $c$ is the speed of light. Also, you may now ask if the energy spreads it can raise the temperature of the surroundings suppose the air but the effect will be such that around 50% of the gamma rays are absorbed in 400 meters so you can actually find the temperature of the air roughly to $E=\frac{3}{2}nRT$ where $T$ is the temperature you want to find. Notice that this explanation is for a electron and anti-electron(positron) collisions. For more mathematical examples visit this answer: https://www.quora.com/What-happens-when-10kg-of-antimatter-bomb-detonates-upon-contact-with-normal-matter-on-the-surface-of-the-Earth
answered May 26, 2018 at 9:07
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$\begingroup$ So, i want to see if i understand: would this process be similar to how the initial temperature of nuclear detonations are measured ? From what i understand: a lot of the temperature is generated from the ultra-small wave length electromagnetic radiation that is produced by the detonation, consequently heating the surrounding air into a plasma; but i also have seen that extremely high atmospheric pressure is also a factor. Would the super heating be the cause of the atmospheric pressure within or is their another force at hand. Thank you !! $\endgroup$ Commented May 26, 2018 at 17:46
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$\begingroup$ You have understood correctly and yes indeed, the heating does causes the surrounding pressure to dramatically increase. $\endgroup$ Commented May 26, 2018 at 18:41
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$\begingroup$ So, if E=3/2nRT describes the internal energy of an ideal gas system, is it possible to describe it in a function that factors how the temperature of the system changes over time in the scenario of the detonation ? $\endgroup$ Commented May 27, 2018 at 4:23
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$\begingroup$ i also came across another issue: to determine n (moles), wouldn't that imply that temperature of the system in is already known given n=PV/RT? consequently meaning that finding the pressure would be impossible as well, correct ? $\endgroup$ Commented May 27, 2018 at 5:58
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$\begingroup$ We can assume we know the number of moles as we know by Avogadro’s law that in the initial standard conditions the number of moles is proportional to volume, V=24n suppose for RTP. So we’ll have no problem for ponding the number of moles. $\endgroup$ Commented May 27, 2018 at 6:17
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hypothetically speaking: if E = 10^17 joules (roughly the energy produced in the annihilation of 1 kg of matter and anti-matter and i wanted to measure the temperature of the system when the fireball has a 1 meter radius or a 4.16 m^3 volume (generously assuming it's a uniform sphere), then would the following be an accurate premise?:
E=3/2 nRT → T=2E/3nR,n R ≠0
where:
R= Gas constant 8.31 J/mol K
n = V/0.024 (decimeter^3 to meter^3 converstion)
then,
T=(2(10^18))/(3n (8.31))
T=(2x10^18)/(3(4.19/0.024)(8.31))
T=(2x10^18)/(3(174.6(8.31))
T=(2x10^18)/4,352
T≈ 4.6x10^14 kelvin
this answer is seemingly counter intuitive, but i wanted to make sure i understood correctly. Thank's a ton guys!