A different thermal problem (part II)

>>It just changes form or dissipates (entropy).

That's the point, but ¿why is HEAT the final stage of any form of energy? If the motor I feed is that of a fan, I'll need to waste some energy to move the air which passes thru blades.

If I have to cool TWO rooms, one with a 1000 watt heater and another one with a 1000 watt motor, ¿do you think I have to consider the SAME amount of BTU's in both cases?

PS: I'M DYING OF ANGER with the problem that I CAN NOT submit a comment. Grrrrrr... ~~~~~>:(
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Yes. If the rooms are closed and insulated, both the heater and the motor will dump the same amount of energy into the room. You'll need the same amount of A/C to cool it no matter what.

Heat is *decorrelated* energy. All the atoms are moving in different directions so no "work" is getting done. This is the most probable distribution of the energy and hence the energy moves from a improbable, correlated state (nice rotational energy) into a more probable, decorrelated state (random brownian motions of the atoms in the room).

The 1000W power of the motor is dissipated into heat via several mechanisms. Some of the energy is dissipated as I^2 *  R losses in the windings. Some of the energy is lost in the motors bearings. The remaining power is transferred to the fan blades and imparted to the atoms of the gasses in the room by the action of the blades. The atoms disperse into the room bringing that kinetic energy with them. They bounce off the walls and such gently heating the entire room.

Now the heater's 1000W produces a higher localized temperature initially as all 1000W are in one spot (the heating element). But say it's run for one hour and turned off. As the heating element cools, it's energy is dissipated to the surrounding air. After a period of time the heat has been dissipated to the entire room and the temp everywhere in the room is the same, up 1000W/hrs from the starting temp.

If the electric motor is run for the same amount of time with a load on it so it draws the same 1000W once equilibrium is again obtained the motor room will have the same overall temp rise as the heater room because the same amount of energy, 1000W/Hr's was dumped into the room. The motor appears to be running "cooler" because the energy from it is not as localized as the heaters energy is but in the end, the effect is the same.

Externally, an observer with a wattmeter on the powerline and a temp probe will find that after the 1000W/hr has been dumped into both rooms, the temp will have risen by the same amount in both cases. The observer will be unable to tell which room has the heater and which has the motor. How could this be otherwise? You dump 1000W/hrs into an insulated container, what difference does the internal configuration of the container make? Energy applied was same in both cases, why should ending temp NOT be the same?

Assumptions implicit in all of this is that the rooms are thermally isolated from the rest of the universe and that they both had the same starting temperature and same mass.

Of course, this relates back to another question having to do with A/C capacity for a room with computer equipment.

My point was that total electrical power input to computer equipment is equal to total heat load output from the equipment.  As with the motor example, it cannot be any other way.  Computers (as amazing as they are) neither produce or consume energy, they merely convert electrical energy into heat and happen to do some useful things (at times) as well.

Which reminds me... I used to have an APOLLO DN660 Workstation that was about the size of a dishwasher.  Now that thing was a better space heater than a computer!!

Jhance, your original answer on the other Q was 100% correct. The total draw of the computers, plus the lighting, plus (typically) 150W per person, plus the influx of heat from the windows/ceiling, etc. all must be handled by the A/C.

Yes, the total draw to the computer is eventually vented to the ambient space as *heat*. The energy "stored" in the disk drives is dissipated as heat when the system is turned off. Every joule pulled from the powerline goes into the ambient heat of the room (with the possible exception of the miniscule amount sent back out on the LAN/modem cabling.

Vikiing, a 250W heater and 250W computer throw *EXACTLY* the same amount of heat into the room. But you have to remember that the ambient temp of the computer need not be the same as the ambient temp of the heater! The heat of the heater is concentrated into the *slight* mass of the heating element. The heat of the computer is spread thru the physical mass of the whole machine plus all the air moved by the fans. The TOTAL heat dissipated is the same even tho the *temp* of the heater may be higher.

I think you're confusing temperature with heat. Temp is the instantaneous value, heat is the product of temp X mass. This is like how VOLTAGE is to WATTAGE.

Imagine a spark thrown off from a grinding wheel. The temp of the spark is several thousand degrees (white hot, no?), but the THERMAL MASS of the spark is so low that you can place your hand into a shower of sparks and only feel a tickle and not get burned.

This is why a pot of boiling water can inflict severe damage, the water has a high thermal mass even tho the temp is only 100C. Yet the spark at 1000C doesn't hurt - not enough thermal mass to matter.

Jhance: I can do you one better! When I started out in electronics, lo those many years ago, Vacuum Tubes were still found in a *LOT* of gear. We had an old hp 425 Frequency Counter that was about 10 cu ft. of tubes. I worked in the "great white north" at the time and on cold mornings we'd turn that puppy on and have the shop toasty in no time! It had a 10" diameter fan on the back and sounded like it was taxi-ing for takeoff while it revved up! :-)

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I'm learning more than I wanted to when I asked this question. Not that I don't like to learn, but I just needed a quick formula to finish my overdue work.
BTW - How does one compute the BTU blown from a politician during a campaign? There's a space heater for you.
Ever notice they never get stuck in a blizzard like we do?
vikiingAuthor Commented:
>>I think you're confusing temperature with heat. Temp is the instantaneous
>>value, heat is the product of temp X mass.

Yes; that's my mess... :(
Don't feel bad. Happens to all of us. The mind is the first thing to go and ... and ... and I can't remember the 2nd! ;-)

You think thermodynamics is confusing? Wait until you get to quantum physics!

(DOCTOR! My brain hurts!
Oi! It'll have to come out!)

I've got a real "brain breaker" on my web page - "Thoughts on the limits of Time". Visit: www.cyberchute.com/rvbus/madmark but don't blame *me* if you bite yourself in your forehead!

(I posted an answer, I assume we're ready to close this thread?)


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vikiingAuthor Commented:
Thanks for your comments, Mark.  ;)
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