Heat/Efficency of Speakers


What % of power sent to the speakers is turned to waste heat? That's the short version of my question.

I'm looking to minimize waste heat accross my stereo as my listening room is unforgiving come summer; no cooling and a computer system which cannot be relocated. I understand amplifier efficency & the classes as well as speaker efficency measured as W/db however the interplay eludes me.

Taking two hypothetical amplifiers: a Class A amplifier outputting 10W w/ 100W from the wall & a Class D outputting 200 w/ 220W draw I understand the D will be the cooler operator however this is where the discussion tends to end, D only wasting 20W vs the A amplifier's 90W. Considering appropriate speaker matches to each amp(as well as a standard HE speaker at say 95db/w), how do I determine the wattage converted sound and the watts spent as heat?

I'm asking because I was previously running a 10W tube amplifier in this room(4xel84 tubes) with 96db speakers. This was bearable in two hour doses this last summer. My friend assures me any Class D amplifier and many AB amps would have no such heating problems and says it's class not wattage that is my issue. Before I move to a different amplifier technology(and swap speakers, these voiced for SE tube partnering) I want to understand this issue fully. I'm unconcerned with power usage and only care about the heat.
redfuneral

Showing 3 responses by brownsfan

Uh, maybe I am missing something here.  According to the law of conservation of mass and energy, any energy taken from the outlet will be conserved.  That energy must either remain as energy or be converted to mass, which is extremely unlikely.  What form does the energy take?  It could be retained as potential energy (also unlikely, unless you are charging a battery) or remain as some form of kinetic energy.  When a source of audible energy stops, the sound quickly stops.  Where does it go? It does not cease to exist, but it continues in a form of energy which is not audible, i.e., heat. The OP can simplify his problem by simply selecting active components that draw the least amount of current, since all of the current that is drawn is quickly converted to heat.   Have I got something wrong here?  Its been about 44 years since my last physics class.   
Maybe I did not make my point clear.  Everything coming out of the wall ends up as heat.  Reduce current draw and you reduce heat.  Forget the speakers,  They are irrelevant.  They do not create energy, they only take the energy that is sent to them and convert it directly to heat or indirectly to heat by first creating sound waves, which then strike surfaces in rooms and is quickly converted to heat.  Energy enters the room as light, current, and body heat.  All of the energy in the room is conserved.  It does not disappear.  Reduce current draw and light entering the room and you have done what you can do.  As has been pointed out several times in this thread, a class D amp is about as good as you can do in reducing current draw.  Al, help me out here.  Am I missing something?  Isn't this just simple first and second law stuff?
Al, thanks.  All of the caveats you listed I assumed would be understood so I did not explicitly state them.  Of course, the 2nd law demands that heat moves from an area of higher temperature to one of lower temperature.  If the listener is in a poorly insulated room, and the exterior is 40 degrees lower, and he opens the window, then the "disappearance" would be significant.   I also made an assumption that everyone participating in this discussion knows that less efficient speakers draw more energy into the room in order to afford a given dB level, and that energy ends up as heat.