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Sound is air particles vibrating (thus hitting each other to make longitudinal waves) and heat is the vibration of air molecules. Because we can only assume that heat made from fire is a higher intensity of vibration than sound (because we don't burn ourselves when we speak) why doesn't it make an extremely loud sound?

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    $\begingroup$ If your premise were correct, a glowing-hot chunk of metal would also be making a ton of noise. Does that help point you to where your assumptions about molecular energy -to - sonic frequency conversion are going wrong? $\endgroup$ – Carl Witthoft Sep 5 '18 at 12:34
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    $\begingroup$ Remember the question is why doesn't make as much noise as I initially expected it would $\endgroup$ – yolo Sep 5 '18 at 20:09
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The combustion reactions don't inherently make any sounds. But they release plenty of energy causing the nearby molecules to acquire higher random kinetic energy, which is theoretically detectable as Brownian noise. But it's not as easy as detecting a human's speaking because the noise is indistinct and the power at the eardrum is lower: a lot of the particles' velocities aren't towards the eardrum so a limited amount of energy is transferred. Unlike longitudinal sound waves, where groups of particles oscillate periodically with large amplitudes, the movement here is rather disorganized and isn't audible for all reasonable temperatures, so you're more likely to hear the sound of wood popping due to the expansion of fibres and escape of moisture and air, or sounds of wind blowing as air expands and rises rapidly.

I decided to take a shot at finding the temperature needed for the Brownian noise to be about as loud as a conversation. For $60\ \rm dB (SPL)$, with the usual reference of $2\times 10^{-5}\ \rm Pa$ and considering the ear's highest sensitivity region, we're looking for an eventual rms pressure of $2\times 10^{-2}\ \rm Pa$. Using the equation from On Minimum Audible Sound Fields by Sivian and White$$\bar{P} =\left [ \frac{8 \pi \rho k_B T} {3c} ({f_2}^3-{f_1}^3)\right ]^{1/2}$$where $\bar{P}$ is the rms (root mean square) pressure, $\rho$ is the density of air, $k_B$ is the Boltzmann constant, $T$ is the temperature, $c$ is the speed of sound in air, and $f_1$ and $f_2$ are the frequency range. Let's consider the frequency range of $0\ \rm Hz$ to $2\times 10^4\ \rm Hz$, because according to the Brownian frequency distribution, higher frequencies are negligible. Throwing in common values for all the constants, we see that for $\bar{P}=2\times 10^{-2}\ \rm Pa$, we need an unbelievable $10^8\ \rm K$, to 1 significant figure, which is hotter than the Sun's core.

It's important to not think that this implies that we're never going to hear Brownian noise. If you take some software like Audacity (or probably any sound editing tool), you can render Brownian noise and listen to it at $60\ \rm dB$. But there we have a deliberate superimposition of several frequencies' waveforms (with amplitudes according to the distribution) being played. But we won't hear the noise caused by the random kinetic energy of air molecules at a loudness comparable to normal conversations.

"Because we can only assume that heat made from fire is a higher intensity of vibration than sound" isn't really true: because of the direction of particles' oscillations/movement, longitudinal sound waves end up transferring far more energy to the ear drum.

Eventually, we're not going to detect the noise; the flows due to convection and expansion of hot air are more likely to be audible.

Sivian, L. J., and S. D. White. “On Minimum Audible Sound Fields.” The Journal of the Acoustical Society of America, vol. 4, no. 3, 1933, pp. 176–177., doi:10.1121/1.1901988.

And some cool related stuff: Difference between sound and heat at particle level

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Sep 5 '18 at 16:22
  • $\begingroup$ What is “rms” pressure? $\endgroup$ – Chase Ryan Taylor Sep 9 '18 at 10:35
  • $\begingroup$ @ChaseRyanTaylor Root Mean Square. I've never seen any convention about capitalization... RMS is just as good as rms. $\endgroup$ – user191954 Sep 9 '18 at 10:39
  • $\begingroup$ @Chair I figured it was an acronym, I just forgot about the concept $\ddot\smile$ $\endgroup$ – Chase Ryan Taylor Sep 9 '18 at 10:41
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The term "sound" is used to describe macroscopic variations in fluid pressure on a 50-microsecond to 50-millisecond time scale. When something appears to burn almost silently, such combustion will involve many individual events that increase ambient gaseous pressure, but nearly all of the effects are either so fast that the variations in pressure they cause will be nullified by averaging over a 50μs interval, or are so slow that the change in pressure over any 50ms interval will be negligible.

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The flame in a propane torch, oxyacetylene torch, or blowtorch produces significant amounts of noise. This occurs because the fuel and air mixing process is turbulent and the combustion process occurs in a turbulently-moving mass of air. The combustion is nonuniform in time and space, which produces the random white-noise "roar" of a torch.

The combustion process in a candle flame, propane BBQ burner can, or cigarette lighter does not have combustion occurring at the same time as turbulent mixing of fuel and air, and so it produces very little noise.

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    $\begingroup$ Releasing a gas from a highly compressed state (such as a propane tank) is noisy in and of itself. $\endgroup$ – Acccumulation Sep 4 '18 at 22:57
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    $\begingroup$ yes it is, but by blowing out a propane torch while it is running you will reveal the (faint) hiss of the gas orifice inside the body of the torch, and immediately hear the sound level difference. $\endgroup$ – niels nielsen Sep 4 '18 at 23:11
  • $\begingroup$ @nielsnielsen: besides the difficulties to blow out such a torch, when you do it, you might be able to generate short louder sounds by relighting it afterwards. The same noise can however even be experienced in a "normal" fire when laminar wind flow gets turbulent in the flame a big camp fire can create some noise too. $\endgroup$ – PlasmaHH Sep 5 '18 at 9:29
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    $\begingroup$ The sound of such gas torches is not really white noise, it sounds more like pink noise or even red noise. $\endgroup$ – leftaroundabout Sep 5 '18 at 14:10
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Sound is a coherent oscillation of the molecules in the air - in particular, a more-or-less synchronized compression/rarefaction (molecules coming closer than apart) that propagates through it.

That makes it, thermodynamically, effectively more like "work". That is, it is a relatively low-entropy form of energy: very orderly, or equivalently it "looks like signal", and could be described with a relatively compressible data pattern (i.e. contains statistical regularities) in a computer with enough storage, as per the information-thermodynamic entropy law of equivalence

$$H = \frac{S}{k_B \ln(2)}$$

On the other hand, the heat of a fire is just that - heat. It is maximal entropy, and molecules moving about entirely randomly (meaning no compression possible.). It "looks like noise", meaning its dynamic description will be random data, roughly, looks like noise in the sense of "noise" as in "meaningless information". Keep in mind this is not the same thing as "noisy sound waves": those are still coherent waves, it's only the amplitudes that are random or, in effect, they are only random in one dimension thus still low-entropic, though higher than a pure tone.

Thus there will be no sound from the heat of the flame alone. In particular, one can give a reasonable thermodynamic argument you cannot hear heat directly simply by noticing that to hear it, it has to do work on your eardrum to displace it from oscillatory equilibrium, and work is precisely what you cannot do with heat in this way. Were it possible to hear heat directly, i.e. for the random collisions of molecules against your eardrum to directly create an orderly motion thereof, or for random collisions to become spontaneously orderly waves on contact with air, you could essentially use the ear as a device to extract work from it for free in violation of the second law of thermodynamics.

That said, one can object that this would only apply in a thermal equilibrium situation, and a fire is far from equilibrium - simplistically, 2000 K flame temp versus 300 K ambient temp, and thus you should be able to extract work.

And it turns out, that is exactly how that a fire is, in fact, audible. The hot gas is able to undergo expansion - a concerted outward motion - thanks to the gradient, and drive convection currents within the air, thus converting heat into work (low-entropy energy), and these currents effectively become wind, and that produces an audible rushing/blowing sound in the same way that one can hear the wind from a storm. But direct audition of the random motion of molecules is still impossible as per the argument above.

(The "crackle" and "pop" of burning solid, complex organic fuels like wood are, as mentioned elsewhere, due to sudden expansion/explosion of small pockets of vapor forming within the fuel from volatiles, including water.)

ADD: Now that I think about this some more I don't think it necessarily works. There will be a random fluctuation in the central position of the eardrum that results from averaging the motion of all its molecules. This would technically be sound. It will be very small, but not necessarily zero. (In terms of entropy analysis, the entropy may be very high, but it is not $\infty$.) That said, you still will not hear anything unless your ear is in direct contact with the source, because the distribution of particles within any given small parcel of air is more or less thermal - it's a temperature gradient and so the only noise you will hear will be that at the ambient temperature around your eardrum. With @Chair's calculations, this means it will vaporize your ear (and you) long before you get high enough to hear it.

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Sound is air particles vibrating (thus hitting each other to make longitudinal waves) and heat is the vibration of air molecules.

Neither of these statements are quite accurate. The human ear contains a bunch of hairs that each have different resonant frequencies. If the air is vibrating at the resonant frequency of one of the hairs, then that hair will send a signal to the brain. So the human ear basically acts like an analog Fourier transform.

An analogy would be a swing. If you push the swing at exactly the right point of its cycle, over and over again, the swing will swing higher. If you repeatedly push the swing, but at random times, you'll be just as likely to cancel out a previous push as to reinforce it. So adding more pushes, if they're random, doesn't do much to increase the amplitude of the swing. Similarly, if there's a sound wave with a frequency such that its peak keeps hitting the hair at the same time in its cycle, then that hair will swing back and forth further. If you just have air molecules hitting the hair at random times in its cycle, then they will cancel out, and adding more collisions or increasing the magnitude of the collisions won't change that.

Note that what we perceive as "sound" is vibration of the air as a whole, not vibrations of individual molecules. Its takes a very large number of molecules moving together to create what we perceive as "sound". Molecules moving separately don't create sound.

Also, it's quite correct to say that heat is vibrations of molecules. While some heat energy is internal vibrations, it's also kinetic energy of molecules moving around (especially in gases).

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  • $\begingroup$ You need to relate this back to the question. This doesn't answer the question, it just talks about perception of sound and heat. $\endgroup$ – Aaron Stevens Sep 4 '18 at 23:28
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    $\begingroup$ The sentence before the last one should contain "not"? $\endgroup$ – Vladimir F Sep 5 '18 at 6:07
  • $\begingroup$ you could make it more clear that none of these hairs have actual contact to the air - the eardrum translating local Schallschnelle to movement of the bones contacting the inner ear is a vital part of that equation. $\endgroup$ – bukwyrm Sep 5 '18 at 12:46
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Sound is an oscillation in pressure. You might think if pressure is what we call the net force of gas molecules colliding with a surface, and those molecules are moving about randomly, then some will collide with a lot of force and some with not much force, and thus the pressure might vary.

However consider two points:

Firstly, increasing the temperature of a fixed quantity of gas in a fixed volume increases the average kinetic energy of the molecules, and thus increases pressure. But because the pressure is higher now than it was 10 minutes ago doesn't mean it's oscillating, not at a frequency high enough to be audible anyway. To be sound the pressure must be changing, and to be audible sound it must change at a frequency that's physiologically detectable by our ears, between 20 and 20,000 Hz.

Secondly, by the law of averages, the variation in the energies of individual are molecules is insignificant because an eardrum is many orders of magnitude more massive. So it's only the average collision force of many millions of collisions that determine the macroscopic forces on an ear drum.

If you want to get technical, the law of averages means the variation in pressure is really small, but not zero. If we had a pressure measuring instrument of sufficient sensitivity, we could measure a random variation in pressure just due to chance which could be called "sound". As the temperature increased the random variance would increase as there would be a wider spread between the most energetic and least energetic molecules. We don't hear it simply because it is many orders of magnitude too small a variation to be detectable by ears.

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Oddly enough, the answer is "yes, if you do it right." plasma loudspeakers, sometimes implemented via controlled flame, are an actual thing.

edit

As Jmac noted, in flame-based versions, the flame only gets the plasma going, and the plasma uses magnetism to generate the sound; not the heat. So in a sense it's a couple extra conversion steps of the flame's energy to produce the sound waves.

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    $\begingroup$ Sounds like the flame only gets the plasma going, and the plasma uses magnetism to generate the sound; not the heat. $\endgroup$ – JMac Sep 5 '18 at 13:29
  • $\begingroup$ @JMac you are correct, and I should have included that. I'll update. $\endgroup$ – Carl Witthoft Sep 5 '18 at 15:21
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    $\begingroup$ This is more of a "fun fact" than an answer. It's at best a comment. It was interesting to read, but your first paragraph is still misleading. Don't just add "the above was wrong" notes, re-write the part that was wrong or delete the answer. $\endgroup$ – Peter Cordes Sep 6 '18 at 16:11
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Apart from Chair's assertion that the air temperature would have to be enormous before the associated Brownian motion of the air molcules could be heard, it is important to note that the heat doesn't reach your ear at all. We only hear oscillating air in the ear proper; we cannot hear oscillating air far away, directly. We can see a glowing piece of metal; we can feel the infrared; but we are not exposed to air at 1000K.

That is obviously a good thing.

Why sound — a more or less oganized oscillation of air molecules — carries over long distances while the chaotic Brownian motion carries comparatively badly is a different question (and one I cannot immediately answer). But it's true for most materials, for some of them in an astonishing fashion: You can use a blowtorch on a ceramic heat shield and touch it on the other side; but its likely that banging at it with a hammer would carry through the whole space shuttle.

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