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Both thermal energy and air are propagated through vibration of particles so why sound does not heat up the air e.g loud musical instrument does not generate much heat ?

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    $\begingroup$ Note that your title & text slightly disagree (specifically the modifier "much" is present in the text but not the title). $\endgroup$ – Kyle Kanos Sep 9 '14 at 18:39
  • $\begingroup$ Related question by OP: Difference between sound and heat at particle level $\endgroup$ – BMS Sep 9 '14 at 18:51
  • $\begingroup$ @BMS For my two cents, I think there is enough distance between the two to indeed consider them as separate (but related) queries. $\endgroup$ – Bryson S. Sep 9 '14 at 18:57
  • $\begingroup$ The OP was asked, perhaps erroneously, to make this a separate question. They are very similar, however, and they exhibit a misunderstanding that is common amongst both questions. $\endgroup$ – David Hammen Sep 9 '14 at 19:12
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Put more simply: sound waves are attenuated as they propagate through air (this is more easily measured for very short wavelengths, e.g. ultrasound). This means they lose energy - which is turned into heat of the air.

The amount of heating, however, is very very small. Let's do the math. A sound wave of 120 dB (really loud) has energy of only $1 \frac{W}{m^2}$.

The attenuation of sound in air is a function of wavelength - for example, we see from http://www.sengpielaudio.com/RelativeHumidityA.gif

enter image description here

that a 10 kHz wave in air at 50% humidity is attenuated 4 dB in 30 m, or 44 dB in 330 m (which is the distance that sound travels in 1 second).

The energy lost by a 120 dB 100 kHz sound wave (that would be quite loud and obnoxiously high pitched) in the first meter is $\frac{4}{30} = 0.13dB$ which is 3%. The heat capacity of a cubic meter of air is about 1280 J/K (from Wolfram Alpha), so the temperature rise due to 30 mW of heat is $2.3 \times 10^{-5} K/s$. That is pretty hard to measure...

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    $\begingroup$ +1 Nice example of how the ideal gas approximation can only go so far. It might be useful to note that attenuation also depends to some extent on intensity. The loudest undistorted sound at 1 atmosphere pressure is 194 dB. Above that, pressure would have to go negative to have an undistorted sound. Sounds louder than this can exist, but they will be strongly distorted and hence will attenuate rapidly. Sounds close to 194 db also undergo distortion and greater attenuation than does a not-quite so violent sound wave. $\endgroup$ – David Hammen Sep 11 '14 at 14:57
  • $\begingroup$ @DavidHammen - that's a great insight! I have never thought about "really really loud" sounds like the eruption of Krakatoa (I guess when people say they measured 180 dB they were 160 km away... so how loud it really was for the people who were killed by it is not known. There were some bad jokes on Reddit about "permanent hearing loss" in this context). Clearly things go nonlinear in that regime, and the distortion and attenuation will be very significant. $\endgroup$ – Floris Sep 11 '14 at 15:29
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Sound waves do generate changes in temperature because the propagation of sound is an approximately isentropic process. Keep in mind though that changes in static temperature can very well occur without the generation of heat. Moreover, the pressure changes associated with sound waves are of such a small magnitude that the observable temperature changes are minimal (but non-zero). In fact, when Isaac Newton first attempted to derive the speed of sound, his answer was off by almost 15%, precisely because he assumed that the propagation of sound was an isothermal ($\Delta T=0$) instead of isentropic ($\Delta S=0$) process. In short, sound waves do alter the temperature of the surrounding medium, just not by that much.

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