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Question: Can the radiation emitted by fire be approximated by black body spectrum?

It has been discussed in this community that black-body spectrum mostly serves as an approximation to actual spectra of thermal light sources:

  • firstly because perceived radiation are not in thermal equilibrium (although this depends on how one defines BBR - as an equilibrium state of radiation or as a radiation emitted by a black body)
  • secondly because the real sources of radiation are usually only approximately black bodies
  • finally, the real sources of radiation are usually in a stationary state rather than in thermal equilibrium

Still, a slab of metal taken out of an oven, or even an oven itself could be treated as a black body heated to a constant temperature (which is also a strong assumption). Can we however say the same about a highly dynamic campfire?

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    $\begingroup$ Worth looking at wikipedia: "Flame color depends on several factors, the most important typically being black-body radiation and spectral band emission, with both spectral line emission and spectral line absorption playing smaller roles. In the most common type of flame, hydrocarbon flames, the most important factor determining color is oxygen supply and the extent of fuel-oxygen pre-mixing, which determines the rate of combustion and thus the temperature and reaction paths, thereby producing different color hues." $\endgroup$
    – Andrew
    Commented Jun 28, 2021 at 12:21
  • $\begingroup$ @Andrew Thanks for the link. The spectrum given is however clearly not black-body - with several marked spectral lines. I am looking for a more developed discussion (with equations even better!), since the claim that fire is BBR is rather common (see here, e.g.) $\endgroup$
    – Roger V.
    Commented Jun 28, 2021 at 12:31
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    $\begingroup$ In flame emission spectroscopy (a type of atomic spectroscopy) the color of the flame is dependent on the atom. Emission spectra due to energy level jumps of the electron. Plasma is atoms that have become ionized. This is electronic excitation whereas heat is rotational. At 1071 degC blackbody radiation becomes visible, fire is between 200-4980degC $\endgroup$
    – ChemEng
    Commented Jun 28, 2021 at 15:14
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    $\begingroup$ @Andrew I am not forcing you to write an answer :) However, I think discussing type of flame is a part of it: can we always describe a flame by black body spectrum? Or never? If only in some cases, what does it depend on? Are there several distinct types of flame or a continuous spectrum from those that can be represented as black body to those that can? $\endgroup$
    – Roger V.
    Commented Jun 29, 2021 at 7:49
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    $\begingroup$ @RogerVadim Fair enough. I am interested, and curious to follow the answers you get. I am not an expert so I don't want to attempt a real answer. But my intuition is that it's very unlikely that a fire is purely a blackbody; I would expect a thermal component of the spectrum, and a non-thermal part which encoded the specific chemical reactions that were happening in that fire. Anyway, I don't want to contribute to the noise, just noting my intuition and looking forward to detailed answers by people who actually know :) $\endgroup$
    – Andrew
    Commented Jun 29, 2021 at 19:43

5 Answers 5

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Different bits of the fire have different characteristics, the spectrum of a flame would usually consist of discrete line radiation perhaps superposed on a weaker continuum. However, the base of the fire, especially say in the cavities between any burning material would emit radiation that more closely approximates the Planck function. In this respect, a just-lit fire would have light dominated by "flame" and the spectrum would not be Planckian, but a well-established hot fire with a lot of radiation coming from its "heart" would be more blackbody-like.

The flames that you see would usually be "optically thin" - that is they are transparent to their own and other radiation. In such circumstances, the flames emit light that usually correspond to discrete transitions at particular wavelengths - about as dissimilar to the Planck function as it could be. This is how "flame tests" for the presence of different chemical elements work. If the flame was hot enough to ionise atoms you would also get some recombination continuum radiation, but as I said, most flames are optically thin, so the radiation never gets the chance to come into equilibrium with the matter at some particular temperature and you would not get the Planck spectrum. If you can see through the flame, at any wavelength, then it isn't radiating a Planck spectrum.

On the other hand, the space between the coals of a hot fire can provide a reasonable simulacrum to the cavity radiation of an ideal blackbody. Here, we are looking into an enclosed space where the radiation field has had a chance to come into equilibrium with the surrounding material. This is still just an approximation - but a clue that you are looking at something close to the Planck spectrum comes when you start to lose the ability to discern the texture or shape of the material within the cavity. That is because everything is at a similar temperature and the radiation field is starting to approach isotropy.

The above discussion by the way is often why one will notice that it isn't the flames of a fire that give off the most heat, it is the base of the fire. That is because blackbody radiation is the most efficient way that a thermal radiator can lose heat (radiatively).

EDIT:

Here are some spectra taken of flames from a simulated wildfire, compared with blackbody spectra - not very Planckian (from Boulet et al. 2011). The main peak at 2300 cm$^{-1}$ is due to radiation from CO$_2$ and CO molecules. The conclusion in this paper is that the temperature is about 1500K and the CO$_2$/CO radiation is optically thick and thus reaches the Planck function, but that the flames are optically thin at all other wavelengths. Flame spectra from a wildfire

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    $\begingroup$ Weirdly enough, your description (describing the differences between the flames and coals) made me think about my older sister's instructions about how to roast a marshmallow (when I was 9 or 10 years old) - the flames will light the marshmallow on fire, but the intense even heat from the coals will toast it up quite nicely. Who knew that Planck was involved?! $\endgroup$
    – Flydog57
    Commented Jun 29, 2021 at 20:30
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A good example of profoundly non-blackbody campfire features: the blue flame bases.

  1. they are transparent (not really black in any sense)
  2. they are blue. Their "color temperature" (the best-fit blackbody approximation of their spectrum) will be ~10 times the highest temperature one can find in a campfire.

Another (less visible, but pretty much important) non-blackbody part is the gas directly above the flames. It has pretty much linear spectrum (like gases do in general), but its characteristic lines are deep in the infrared.

On the other hand, there are at least two separate things in a campfire that look like a textbook black bodies:

  1. The burning chars. They are black in the first place and they burn at 500-700C depending on the available oxygen.

  2. The yellow parts of the flames. Here you get 800-1500C (depending on exactly where you look) blackbody radiation. The particular blackbodies in play here are the soot particles. They are pretty much black as well.

Note that by mixing two or more blackbody spectrums you will get a non-blackbody spectrum, unless their temperatures are equal or one of them pretty much dominates the final result.

p.s. obligatory XKCD: https://xkcd.com/1308/

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    $\begingroup$ @AccidentalTaylorExpansion couple of stick figures, sitting around a carbon-burning fire.(the CO/CO2 spike is quite indicative). Behind the acorn-head stickfigure is a pyramid of red, and green lights, topped with what looks to be ?gold? metallic shine.... i.e. family sitting around the fireplace, christmas tree behind them. on top of tree is golden ornament, likely the traditional star. I think their fireplace is a gas-burning one, from that curve, not wood or coal. $\endgroup$
    – PcMan
    Commented Jun 29, 2021 at 12:56
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    $\begingroup$ The top of the tree is a typical white LED. $\endgroup$
    – fraxinus
    Commented Jun 29, 2021 at 14:42
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    $\begingroup$ @lalala hardly sodium. There is no much sodium in the wood to start with. You get exactly the same yellow flame from propane, where no traceable sodium exists. On the other hand, you can alter the color temp by altering the burning conditions to rich/lean. $\endgroup$
    – fraxinus
    Commented Jun 29, 2021 at 20:52
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    $\begingroup$ @lalala an interesting fact: the usual table salt was sold as a "burning enhancer" not so long ago. It really made better-looking flames. $\endgroup$
    – fraxinus
    Commented Jun 29, 2021 at 20:54
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    $\begingroup$ @lalala You also get the same yellow when burning (iso)propanol, it produces a lot of soot. Ethanol, on the other hand, is blue. Methanol can be invisible. $\endgroup$ Commented Jun 30, 2021 at 12:48
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The most accurate and at the same time most pedantic answer is that absolutely nothing emits exact black body radiation. It is always an approximation valid at a given temperature in a given frequency range.

In the case of such loosely defined physical systems as a 'fire' or a 'campfire' any statement about their spectrum is particularly loosely defined. In this handwaving sense a (camp)fire emits black body radiation. It does not exclusively emit black body radiation, but that was not the original question.

See here for a woodfire spectrumwoodfire

The text states that the size of the CO2 peak is exaggerated for visibility. This is close enough to BBR for me.

Note that you are looking at hot glowing coal and hot gases at the same time.

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  • $\begingroup$ +1 I expected/hoped that someone comes with actual images of the spectra. It loosely looks like Planck's curve, even though the peaks are clearly deviations from it. II think one of the following figures in the linked wiki article actually shows that this is not Planck's spectrum, if fitted. $\endgroup$
    – Roger V.
    Commented Jun 28, 2021 at 15:08
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    $\begingroup$ As a general comment, graphs without any indications of zero point or scaling aren't appropriate for judging the shape of a function. But yes, I would expect the overall shape of a well-established fire (i.e. one where the spectrum is not dominated by flame, like one that has just been lit) would look a bit like a Planck function. $\endgroup$
    – ProfRob
    Commented Jun 28, 2021 at 15:12
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    $\begingroup$ @ProfRob I couldn't find a more precise graph on the entire CERN website. :-)) $\endgroup$
    – my2cts
    Commented Jun 28, 2021 at 15:14
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    $\begingroup$ Well ironically, if there is any kind of emission line poking up above the continuum, that tells you that the rest of the spectrum isn't a blackbody. Since the maximum possible height of the emission line is the Planck function value at that wavelength. The rest of the spectrum is then a "grey body" with roughly constant emissivity, so that it's shape is roughly Planckian. $\endgroup$
    – ProfRob
    Commented Jun 29, 2021 at 18:40
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    $\begingroup$ @JEB There's always a firework designer who disagrees. Perhaps you mean a campfirework designer? $\endgroup$
    – my2cts
    Commented Jun 29, 2021 at 18:59
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I think the answer depends on what is meant by 'approximated', and on how well established the fire is. If we take the two key requirements for BBR, namely that the body should be unreflective and in thermal equilibrium, I suspect that parts of a campfire would satisfy both requirements reasonably well.

When wood burns it passes through a phase in which it is charred and black before becoming a white ash residue. During that phase it is probably a good absorber of visible radiation at least.

My guess is that the radiation from a fire can be crudely considered as having two main components. One will be the emission of radiation from the chemical reactions of combustion; the other will be radiation that is not from combustion itself but arises simply from the temperature to which the charred wood has been heated by the flames. I would expect the first component to be nothing like BBR, while the second would be like it.

I have a firebowl at home, and often study the different aspects of the appearance of the fire. When the fire is well-established, the incandescence from the hot logs at its heart seems to be quite different in nature from the light from the flames. It may be an optical illusion, but the impression is given that the glow is coming from some way inside the wood, and not just off its surface. I could readily imagine that the radiation from the gaps between the heated logs would be like that from a heated cavity of graphite.

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  • $\begingroup$ Thanks, this gave me an interesting angle to think about this problem. Approximated here mostly means be described by Planck's formula. The fire is non-reflecting, but not necessarily absorbing - it may be transparent. Also, fire is very far from thermal equilibrium - I am not even sure whether it always can be considered as a steady-state. $\endgroup$
    – Roger V.
    Commented Jun 28, 2021 at 13:57
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It depends on the thickness of the fire.

Thin bodies/objects radiate less than the BB of the same $T$ - due to being "optically transparent".

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  • $\begingroup$ I think fire couldn't be black body already because there are strong temperature gradients present. Moreover, I am not sure how radiation equilibrates - most probably what we see are traces of specific radiative transitions in a few hydrocarbon molecules. $\endgroup$
    – Roger V.
    Commented Jun 28, 2021 at 12:35
  • $\begingroup$ @RogerVadim: Again, if the flame is thick enough, there are other processes that bring the radiation closer to the BB one. $\endgroup$ Commented Jun 28, 2021 at 12:39

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