Timeline for What is the temperature at which Napalm burns?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| May 14, 2019 at 22:07 | comment | added | Gert | Ok. I'll give it a +1. Thanks. | |
| May 14, 2019 at 21:46 | comment | added | Paul Young | tried yet again ... | |
| May 14, 2019 at 21:45 | history | edited | Paul Young | CC BY-SA 4.0 |
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| May 14, 2019 at 21:01 | comment | added | Gert | Note that the energy comes from the breaking of the oxygen double bond. Again, very ambiguous. Breaking the $OO$ bond costs energy. Despite that, the formation of $CO_2$ and and water make the process very exothermic. You are right that access to oxygen has a strong bearing on combustion speed and thus flame temperature. | |
| May 14, 2019 at 20:51 | comment | added | Paul Young | I gave it a try, check now | |
| May 14, 2019 at 20:50 | history | edited | Paul Young | CC BY-SA 4.0 |
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| May 14, 2019 at 20:48 | comment | added | Gert | Sorry but I think that last paragraph is ambiguously worded: it reads as if the reaction isn't exothermic at all. I suggest to edit it. | |
| May 14, 2019 at 20:46 | comment | added | Paul Young | @Gert - I agree. I am only making the point that the energy available to the flame is gated by the flow of oxygen. Napalm is not a material which contains its own oxidizer. | |
| May 14, 2019 at 20:44 | comment | added | Gert | The 'strength' is irrelevant here. What matters are the bonding energies. From these the specific combustion energy of Napalm can even be estimated fairly accurately. The released energy has bearing, along with many other factors, on the temperature at which it burns. | |
| May 14, 2019 at 20:44 | comment | added | Paul Young | 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. | |
| May 14, 2019 at 20:42 | comment | added | Paul Young | Consider $CH_4$ + 2$O_2$ yielding $CO_2$ + 2$H_2O$ | |
| May 14, 2019 at 20:41 | comment | added | Paul Young | @Gert that is true. But look up the strength of O-H, C-H, O=O bonds | |
| May 14, 2019 at 20:37 | comment | added | Gert | 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. | |
| May 14, 2019 at 20:18 | history | edited | Paul Young | CC BY-SA 4.0 |
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| May 14, 2019 at 20:07 | history | answered | Paul Young | CC BY-SA 4.0 |