since sound waves are propogated by causing motion of air molecules, does having a speaker play raise the temperature of the air? if so, how loud (in decibels) would i have to play white noise in my car (89 nissan, little blue four-door number) for one hour to raise the temperature one degree centigrade? oh, and i’m not in the car, and all the doors are shut and there’s no ventilation and the car is pretty airtight.
I swear to God, either you put down that bottle of acid or I’m going to have to come over there.
JB does it again. . .
Ohm’s law states the it’s I before E and sometimes W and Y. That means the battery would heat up and not the speakers. You cannot expend more energy than you have available in your power plant. So EVERYTHING would heat up EXCEPT your car unless you have it parked in the sun with windows up. This scenario logically assumes the standard Nissan speakers.
Sound can indeed heat up an enclosed space. Sonic waves can cook food at high frequencies and volume (more energy). But at audio frequencies, it would take a while.
If I remember my physics:
0 decibels equates to 10[sup]-12[/sup] W/m[sup]2[/sup]
10 decibels equates to 10[sup]-11[/sup] W/m[sup]2[/sup]
20 decibels equates to 10[sup]-10[/sup] W/m[sup]2[/sup]
…
100 decibels equates to 10[sup]-2[/sup] W/m[sup]2[/sup]
110 decibels equates to 10[sup]-1[/sup] W/m[sup]2[/sup]
120 decibels equates to 1 W/m[sup]2[/sup]
(At 120 dB, sound is starting to get painful. )
The energy passing through m[sup]2[/sup] in 1 second would heat 1 cc of water 1[sup]o[/sup]C if the energy transfer was complete. But I believe the energy transfer is very inefficient, so any heating would be slight.
As many in traffic next to a loud car stereo can attest, the energy of the stereo is hardly contained within the car. Vibrating the panels of the car takes energy (which causes propagation of the sound outside of the car). The wires and electronics of the stereo also take energy, heating up as well. (The latter would probably heat the interior of the car more efficiently than the sound waves they make. Just feel the electromagnets of your speakers after using them at high volume for a while.)
It is very inefficient. I suggest covering your car with sound-absorbing panels for maximum heat transfer 
In fact most speakers themselves are very inefficient, when you compare the power dump in (often 100+ watts in some cars) to the audio power coming out (just a few watts, as AWB pointed out). The difference is lost in heating up the speaker itself.
Arjuna34
Well, I’ve had a tendency to screw up Physics posts recently, so let’s see if I can get one right.
AWB: “The energy passing through m² in 1 second would heat 1 cc of water 1°C if the energy transfer was complete.”
If you had 1W of Power directed at 1cc of liquid H[sub]2[/sub]O for 1s, it would result in 1J of Energy, but a Joule is not enough to raise the temperature of the water 1° - for that you want a calorie, which is about 4.186 J. Does that sound correct?
Now this part is just a WAG, so if I screw it up I won’t feel so bad, but I thought that the reason that most Energy transfers were so inefficient was that a good amount of the Energy got lost as heat. Since producing heat is what you’re trying to do anyway, wouldn’t this Energy transfer be pretty efficient as it is?
I’m also worried that if you can’t hear your speakers, it’s meaningless to put a decibel level on it, because the intensity in decibels depends on how close you are to the sound source. Or are you saying, how loud in decibels would it have to be if I were (say) 1m from the speakers?
geez, you people are nitpicky (i’d put a smiley face, but i loathe them). i could have sworn that all bases were covered with that OP.
so let’s say that my car is filled with pure nitrogen, is at a resting temperature of 20 degrees centigrade, is incredibly well soundproofed, has a loudness control which is digitally controlled, which can be set to 27 decimal places, and which has as it’s focal point the exact center of the car’s people space. the engine is not on, the cab is very well insulated from heat from the speakers etc, there are no fires anywhere inside the car, my bloodtype is o+, the pressure inside the car and outside is the same, 1 atmosphere, the car itself is stationary, and i have a giant thermometer sticking out the top, which i can see from 100 feet away in my hermetically sealed bunker, with a good pair of binoculars, at say 20x magnification, and i had eggs for breakfast.
remember this is a lark and an exercise in futility. if i wanted to heat my car, i would not use the speakers. i would just crash into a pinto.
Yes it would raise the temperature.
I will calculate what I think the answer is when I get to work later.
If you’re car is soundproof, ALL the acoustical energy is converted to heat inside the car. In fact assuming a 100W mono sound system, your car battery is probably supplying 200W of power, of which maybe 100W is dissipated as heat in the power amp, 95W in the speakers (heating the coils, etc.), and 5W as actual acoustic power, which gets absorbed by the soundproofing.
DO NOT ask what happens if your perfectly-soundproof car hits a perfectly sound-reflective immovable wall at 35MPH!!
Arjuna34
Geez, jb_farley, you’re leaving out way too much stuff! What’s the volume (or mass) of Nitrogen in your car? For that matter, list all the materials that will be heated, and the mass of each. Upholstery, plastic, metal, paint, glass, Et Cetera. Also, how big is the car’s interior, so we know how far the focal point is from the speakers? I guess I could figure this all out by going off and buying a 1989 Nissan and stripping it down, but call me lazy.
An acoustical transducer produces longitudinal pressure waves within the medium it is operating in. In the case of a stereo speaker, the piston (diaphragm) produces pressure variations in air. Sure enough, these pressure variations WILL heat up the air, but the effect would be so slight that it would be difficult to measure with any degree of accuracy.
And as someone else has already pointed out, a “100 W” stereo channel will NOT produce 100 watts of acoustical energy. The efficiency rating of your typical speaker is usually around 3% to 7% (over 90% is burned-up in the voice coil).
Why bother asking you for this stuff, I thought. I can just tell you how to do it. Here goes. As you may or may not know, in Physics, temperature and heat are not the same thing, but heat and Energy are. Temperature is related to heat, however, in the following equation:
Q = mcT
Q is the heat (or energy) you need, m is the mass of the stuff you have, and T is the temperature change. In your case, you specified T = 1C°. I’ll just make up a number for m. Say you have a box 2m×2m×2m. That’s 8m[sup]3[/sup] of air, which will have a mass of about 9.8 kg. We’ll ignore the box, and say you’re just trying to heat up the air inside. Also, c is not the speed of light - it’s something called the specific heat. For air, c = 1004 J/kgC°. Plugging in all those numbers gives you a heat of 9839J.
So, you know how much heat, or Energy, you need. But how do you tell how much Power you need? You probably know that Power is just Energy over time. Fortunately, you specified the time as 1 hour, or 3600s. Dividing, we get that you need 2.7W of Power.
Now we know the Power output of the sound, and we need to get the Intensity of the sound. Intensity can be found like this:
I = P/(4×pi×r²)
where r is the distance from the speaker. If the box is 2m to a side, let’s put you right in the middle, making it 1m from the nearest wall. Then we have an Intensity of 0.217 W/m². The decibel level B is given by:
B = 10(12 + log(I))
So B would be 113 decibels. You going to give it a shot anytime soon?
*Achernar: If you had 1W of Power directed at 1cc of liquid H[sub]2[/sub]O for 1s, it would result in 1 J of Energy, but a Joule is not enough to raise the temperature of the water 1 - for that you want a calorie, which is about 4.186 J. Does that sound correct? *
Ding ding ding! You caught me. I did indeed confuse Joules and calories. But a Joule is 4.2 calories, not the other way around like you stated.
Is it? Geez, I knew I’d get something wrong. Oh well. At least I’m getting less inaccurate.
'Tisn’t. You were right in the first place: 1 calorie = 4.1868 Joule (according to the ‘units’ program on my friendly neighborhood Solaris box).
Rick
Excuse my confusion. The same site where I got my info lists two different formulas:
Quoth Achernar:
Is that the specific heat at constant pressure, or the specific heat at constant volume?
This 2.7W of acoustical power would be maybe 100W or so of audio, or maybe 50W/channel for stereo, depending on the speakers. Depending on the efficiency of the amp, you’d be burning 200 or 250W total. Of course, all that extra power to get 2.7W of sound will raise the temperature quite a bit more
Arjuna34
Mean specific heat at constant pressure c[sub]pa[/sub] = 1005 J/kgK
Mean specific heat of air at constant volume c[sub]va[/sub] = 718 J/kgK
Well, I can’t believe I forgot all my Thermodynamics already. Thanks, Chronos, for watching my back - I didn’t check which c that site was listing. Anyway, for an isovolumetric process (which we should assume in this case), I believe the correct equation is: Q = mc[sub]v[/sub]T, so that throws all the rest of my calculations off. Here are the correct values:
Q = 7036 J
P = 1.95 W
I = 0.155 W/m²
B = 112 decibels
Oh, and by the way, I’m not disagreeing with anyone else; the speaker inefficiency would contribute much more to the heating of the air than would the sound waves themselves. However, that’s old news. The OP seemed more interested in the heating properties of the sound itself. That’s why I looked only at those. Well, jb_farley, does that answer your question or not?