I was sitting during a lecture last week in one of my classes, and we were talking about chemical warfare and the history behind it. My teacher had brought up the use of Napalm during Vietnam. After a few minutes of talking about how it was used, she mentioned it burned at around 4 million degrees Fahrenheit. This seemed to be inaccurate to me, and all I've been able to find online is that it burns between 1,200 to 5,000 degrees Fahrenheit. Which of the two is accurate? Or is neither?

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    $\begingroup$ 4 million is absolutely wrong, that is for sure... $\endgroup$ – Jon Custer May 14 at 20:04
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    $\begingroup$ @JonCuster Burning napalm, core of the sun; basically the same thing right? $\endgroup$ – JMac May 14 at 20:12
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    $\begingroup$ @JMac sure would make fusion machines easier! $\endgroup$ – Jon Custer May 14 at 20:20
  • $\begingroup$ Napalm is no doubt a hydrocarbon. Look up the flame temperature of gasoline in atmospheric air, and you will be in the right ball park. $\endgroup$ – David White May 15 at 2:16
  • $\begingroup$ The temperature of the order of millions is not even found on the surface of stars. Only the cores of stars can reach to that order of temperature through a nuclear reaction. The argument of your teacher is quite naive. $\endgroup$ – Jitendra May 15 at 2:37

Your online research is correct. Napalm relies on oxygen in the atmosphere in order to burn, so it produces just an ordinary flame.

Ordinary flames reach equilibrium at less than few thousand degrees as radiation and convection carry away the limited energy output available from oxygen travelling towards the fuel.

A traditional flame is an "oxidation" process which releases energy by replacing oxygen double bonds with lower energy bonds. When the fuel is a hydrocarbon, as it is with napalm, oxygen is available only from the atmosphere. While the fuel is concentrated in a solid/liquid state, the gaseous oxygen in the atmosphere takes it time to get to the napalm. This is a major factor limiting the temperature of flames.

  • $\begingroup$ When the fuel is a hydrocarbon, as it is with napalm, the bonding energies of the hydrogen and carbon atoms are about the same as before combustion. Huh? When a hydrocarbon burns, $C-H$ bonds are broken and $C-O$ bonds and $H_2O$ are formed. The overall process is strongly exothermic. $\endgroup$ – Gert May 14 at 20:37
  • $\begingroup$ @Gert that is true. But look up the strength of O-H, C-H, O=O bonds $\endgroup$ – Paul Young May 14 at 20:41
  • $\begingroup$ Consider $CH_4$ + 2$O_2$ yielding $CO_2$ + 2$H_2O$ $\endgroup$ – Paul Young May 14 at 20:42
  • $\begingroup$ All the energy comes from breaking the oxygen double bond. The C's and H's are just as "happy" to face each other as to face an oxygen. $\endgroup$ – Paul Young May 14 at 20:44
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    $\begingroup$ Ok. I'll give it a +1. Thanks. $\endgroup$ – Gert May 14 at 22:07

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