This is a poster in a hallway here in the Texas A&M physics building: Sort of an odd question, but an interesting one. Sound waves carry energy, and if that energy we being absorbed by your coffee because you were yelling at it, how long would you have to do so before it was piping hot?

But let's put that question aside for a minute and talk about sound levels. Most of the time you'll hear sound intensity quoted in decibels. Decibels confuse a lot of people because it's a logarithmic scale rather than a linear one. Logarithmic scales are useful for measuring quantities that can span a very wide range of values, because logarithms turn multiplication into addition. Here's what I mean by that: each increase of 10 decibels corresponds to the level of sound increasing by 10 times. So if you're having a quiet conversation at a level of 30 dB, a 40 dB sound carries 10 times the intensity of the 30 dB sound. A 50 dB sound carries 10 times the intensity of the 40 dB sound and 100 times the intensity of the 30 dB sound. A 60 dB sound is 1000 times the intensity of a 30 dB sound, and so on.

A jet aircraft taking off might output sound of some 150 dB close to its engine - how much more intense is that sound than the quiet conversation? Well, 150 dB - 30 dB = 120 dB. And 120 decibels represents 12 increases by a factor of 10, so the jet is 10^12 = one trillion times more intense than the casual conversation.

If you're on top of things you may have noticed that the 0 dB level is pretty arbitrary in this kind of scale. 0 dB doesn't represent absolute quiet, instead it's 1/10 as intense as a 10 dB sound, or 1/100 as intense as a 20 dB sound, or 10 times more intense than a -10 dB sound, or 100 times more intense than a -20 dB sound, and so on. No sound at all would be represented by -infinity dB. A little counterintuitive, but it's the price we pay for a logarithmic scale.

The 0 dB level corresponds to an arbitrarily set reference intensity of 10^-12 watts per square meter. A person shouting might reach 100 dB, which is 10^10 times more intense than the 0 dB reference level. That means the sound intensity emitted by a shouting person is about (10^10)*(10^-12 W/m^2) = 0.01 W/m^2. Which is not very much. If we generously assume the coffee cup has a cross-sectional area of 100 cm^2 and absorbs all the sound energy incident on it, it will absorb something like 0.0001 watts.

If I've done the math right this works out to something along the lines of 12 hours per gram per degree. There are probably worse ways to heat you morning coffee, but I think it's also safe to say that there are better ways too.

Here, by the way, is the coffee-heating essay the poster is promoting. (I avoided looking it up before writing this) Their presentation and numbers are a bit different than mine, but the overall answer of "it takes a very long time" is the same.

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What I've never understood is why the bel doesn't get more respect. It's always deci-mated. Seems a bit unfair to such a useful unit-less unit.

By Dale Sheldon-Hess (not verified) on 09 Jun 2010 #permalink

So 10^-12 watts/m^2 is 10^-12 joules/sec m^2. What energy (or work) would be measured and how would one measure it? Radio Shack sells sound intensity ("decibel") meters and I assume they're calibrated to other "standard" sound intensity meters but how is the "master" meter calibrated to a source measured in watts/m^2?

Wow, a whole series of questions come to mind. I was thinking of using a known output of an amplifier (in watts) and assuming some level of efficiency in converting the electrical energy to sound via a transducer ("speaker"). But, just as a ballpark set of numbers, if I have a 200 watt output amplifier on and I assume that the transducer is 50% efficient at converting electricity to sound and have it "pretty loud" (like shouting),let's say it's radiating 100 joules/second of sound energy. If I'm 2 meters away and assuming an inverse square law, the area of the sphere around the transducer is about 50 meters^2 so the power/m^2 should be about 100/50 or 2 watts/meter^2, 200 times larger than your number. I'm sure I'm missing something but what, exactly, is being measured if we say 100 dB or 0.01 watts/m^2?

It's my understanding that the plug-to-sound efficiency of a speaker is very, very low - on the order of 1% or worse for average home stereo speakers.

Conversely, relatively efficient conversion methods like thermoacoustic devices make some mind-bogglingly loud noises with a relatively small input power.

Decibels are handy for anything where the values are relative and multiplicative. For instance - I have thought for a long time that currency and stock fluctuations would make more sense on a logarithmic graph.

Thanks Matt, I didn't know that, it would certainly make the numbers align more closely, certainly within an order of magnitude. But then the rest of the 198 watts must be going to heat the speaker coil I suppose? It doesn't seem like that much heat is present (though they are warm).

Oh well, be that as it may, any insights on the other questions (what is being measured - pressure possibly?, how is it measured)?

@No. 4,

Google finance lets you set the vertical scale on stock plots to either linear or logarithmic. Basic stock market models treat returns as lognormal, not normal (i.e. the log of the price changes are normally distributed).

Re OP, I can think of worse ways to heat your coffee (step 1: stand next to cup, step 2: light self on fire), but I'm not sure I can think of a less effective one.

By Snarkyxanf (not verified) on 09 Jun 2010 #permalink

@5: Firstly, your "200W" probably isn't really 200W if it's the advertised spec of an audio amp; there's a good chance it's only 50W RMS per channel (this figure is doubled to give "peak" power and then the two channels added together).

100dB would be a very loud yell; a better reference would be a jackhammer at 1 meter distance. Long-term hearing damage starts at around 85dB.

50W into a typical home speaker would give between about 100dB and 112dB at a point 1 meter directly in front of the speaker depending on speaker sensitivity. Add 3dB for the second channel assuming you are 1 meter from both speakers. From this it should be clear that you generally don't have your volume actually turned up to the point of delivering 50W continuous, since that would not just be "loud" but "unbearably loud".

(If your amp were actually 200W RMS per channel and you ran it at that volume, you'd be talking another 6dB, so 109dB to 121dB, which is hitting the "short term hearing damage" range, with safe exposure times measured in seconds. Pain threshold is at about 130dB.)

By Andrew G. (not verified) on 09 Jun 2010 #permalink

How much power would actually get absorbed to the coffee? I would guess that a great deal of power gets reflected.

By Enok (not verified) on 12 Jun 2010 #permalink

Much of the efficiency problem of speakers is caused by their passive crossover networks. These passive devices, capacitors, resistors, etc eat up much of the power that the amplifier delivers.

Take away the passive crossover, use active crossovers and the amp can suddenly deliver much more of its rated power. Depending on the component values in consideration (in the passive crossover network) this can result in a power loss of about 50% in some cases.

So don't just blame the speaker drivers themselves. Much of the fault lies in how the speaker is designed.

Peter