Thought experiment on entropy

February 23, 2009 8:28 PM

ThermodynamicsFilter: Assume I have a perfectly insulated room, a battery with a fixed amount of power, and an air-conditioning/heating unit.....

If I use the power in the battery to power the unit to COOL the room (and assume that all the power contained in the battery is used), then obviously the extent to which the room is cooled depends on the efficiency of the AC unit. This is because you are working against entropy, and you will always lose some energy to heat. How much depends on the efficiency of the AC unit.

Now, here's the tricky question: Suppose I use all the power in the battery to HEAT the room. In this case, will the room always be heated the exact same amount no matter the efficiency of the heating unit? After all, since inefficiency will ultimately be thrown off as heat anyway, then it should make no difference, right?

(Note that the imaginary room is "perfectly insulated," so the speed with which the heat is released should not be taken into account.)

posted by zachawry to science & nature (26 comments total)

I don't know if this falls under "perfectly insulated" in your stipulations ... but you will have differences in the room materials, air flow and whatnot, in their relative levels of heat convection and conduction. While you may have 100 percent transfer of energy to heat in your imaginary scenario, you may not "heat the room" with 100 percent efficiency every single time.
posted by Cool Papa Bell at 8:39 PM on February 23

To clarify what I think I'm answering: Discharging the battery will always heat the room the same, regardless of whether you use a computer, a heater, or an open or imperfect refrigerator. True or false?


The answer is true, except for the small amount of energy will escape as electromagnetic interference (the room isn't also a faraday cage, is it?). Even if you're trying to broadcast radio waves you'll still keep around 30% of the energy. If you're running a space heater or something you'll still lose some, but I doubt it would be enough to make a measurable difference in the final temperature.
posted by aubilenon at 8:42 PM on February 23 [2 favorites has favorites]

the extent to which the room is cooled depends on the efficiency of the AC unit.

Well, first first -- and I'm sorry if this is too basic, but your premise leaves some room for misunderstanding -- the extent to which the room is cooled depends on the AC units ability to move heat out of the room. That's what AC units do, they move heat. For example, if your window air conditioner is sitting on a table in the middle of a sealed up room, it just ends up being a very elaborate heater.
posted by Chuckles at 8:49 PM on February 23

(an imperfect fridge is one that is not insulated with the same stuff your room is)
posted by aubilenon at 8:49 PM on February 23

Refridgerators are heat pumps. They do not magically make things cool; all they do is move heat from a low temperature (inside your fridge, for example) to a higher temperature (behind your fridge). This requires work (how much depends on the temp. difference and the engine efficiency). Work is provided by the battery.

If you are in a perfectly insulated room, your fridge is not going to cool it. It must have somewhere to move the heat to.

(on preview, as others have stated above). So let's look at a different question:

Divide the room in two, with your perfectly insulating walls. Stick the fridge=heat pump in the wall. Run it; now, one side will get colder, and one side will get hotter. The final temperature difference you can create between the two will depend on the amount of work available from your battery, and the efficiency of your pump.

Of course after you drain the battery, now what was your heat pump has become instead a heat-conducting path; eventually the two halves of the room will equilibrate. They will be warmer than when you started; the extra heat here is the heat due to inefficiency in the pump.

(I'm assuming you really can build a perfectly insulating room).
posted by nat at 8:59 PM on February 23

Chuckles is right. Stated differently, you can't have a perfectly insulated room and also have a functional air conditioner. An air conditioner dumps heat out the window. Sitting in the middle of the room, the air conditioner is indeed simply an elaborate heater.

That's why I tell people not to run fans in closed, empty rooms (empty of people) and expect it to cool the room -- all they do is stir the air and add heat. Now, if the room has a person in it, then the moving air helps that person's perception of the temperature, due to sweating and evaporation effects. The fan is still adding heat, though.

Another favorirte question is "what happens when I take a flashlight intoa room with perfectly mirrored walls, ceiling and floor, and turn it on?" I love that one :)

I got a rotten grade in thermodynamics because that's when I was being a lazy bastard.
posted by intermod at 9:03 PM on February 23 [1 favorite has favorites]

I feel silly about the refrigerating unit in the middle of the room thing. Of course.
Assume, as stated above, that it's installed in a wall.
posted by zachawry at 9:11 PM on February 23

The answer is no. I think you are neglecting the half of the air-conditioning system that is outside the room (and if you don't have one then [on preview: as others said above] it's not an air-conditioning system and it can't cool the room at all). An air-conditioning system being used to provide heat (aka "reverse cycle" air conditioning or a "heat pump") is more than 100% efficient because it moves the heat in from outside rather than having to generate it. Typically 3 to 4 times more heat is moved than electricity consumed. Now consider an inefficient heat pump, in fact an 0% efficiency, "broken", heat pump, using up all the energy in the battery. That energy will be dissipated as heat by the motor(s) of the unit, most of which are probably in the external part, so not heating the room at all. Even if all the heat did end up inside that's only a third of what the properly functioning system could have provided.
posted by Canard de Vasco at 9:15 PM on February 23 [1 favorite has favorites]

I feel silly about the refrigerating unit in the middle of the room thing. Of course.
Assume, as stated above, that it's installed in a wall.

Of course that causes problems too, because of air flow. Instead, let's install an ideal Peltier junction in the wall, or something.
posted by Chuckles at 9:24 PM on February 23

Couldn't you theoretically use a battery to create a sort of 'reverse refrigerator' that would cool off the outside room while moving heat into a central core? I don't think it would work very well, and once you ran out of power the temperature levels would even back out over time.

In addition to heat, sound could be emitted from the heater and maybe travel through the walls of the room. I can't think of any other ways that energy might escape beyond that, though.
posted by delmoi at 10:21 PM on February 23

Canard de Vasco has the right idea. Assuming you have a hole in the wall, you can use your refrigerator/heat pump either to heat the room or cool the room. The amount of heat energy you move in or out of the room for heating or cooling can exceed the amount of energy in the battery by a factor of 3 or 4. Given the same temperature differences, a heat pump will be slightly more efficient when heating than when cooling because the work running the compressor contributes to the heat while it is lost when cooling. A heat pump is more efficient for heating than just using a toaster-like heater.

A heat pump is commonly used to heat houses in winter and cool them in summer.
posted by JackFlash at 11:34 PM on February 23 [1 favorite has favorites]

I'm no engineer, but I think heating elements can emit both heat and light. Assuming light doesn't contribute to entropy or whatever, a light bulb could be less efficient than a stove.
posted by pwnguin at 12:50 AM on February 24

I think there's a second part of the question, which I'd rephrase like this:

Given:
- a battery with a certain amount of juice
- a perfectly sealed room
- two mechanical devices - say, a space heater and an electric motor.

If I convert all of the chemical energy from the battery into heat, using one of these two devices, will the ultimate temperature of the room be the same?

My intuition and long-ago physics remembrances tell me yes, but I'm too tired to do the google searches at the moment. Paging all physicists...
posted by chrisamiller at 1:06 AM on February 24

pwnguin: All the light that the bulb emits will bounce around off the walls. White paint has an albedo of about 70%, so every time it bounces 30% of it is absorbed and ... turned into heat! Eventually all of it will be heat.

I totally forgot about sound vibrations in my above talk about radio emissions.

Also you could use the battery to store the energy in some other way (e.g. lifting a bucket of marbles, or turning water into oxygen & hydrogen), or to charge a different battery! In those cases too you would wind up with less total heat ... for now anyway! *melodramatic chord*

If you discharge the battery, then the total stored energy in the battery minus whatever escapes the room (be it by heat, sound, radio, or gas exchange), minus any new potential energy you might have stored somewhere, is how much more heat will be in the room.
posted by aubilenon at 1:16 AM on February 24

You're kind of breaking your own question.
1. perfectly insulated room.
2. imperfectly functioning heater.

So really you are seeking to calculate the inefficiencies of the 'heater'?
posted by From Bklyn at 2:08 AM on February 24

If you're talking about a scenario where the heater is contained entirely within the perfectly insulated room, then all electric heaters are essentially 100% efficient at converting electricity to heat. As are just about all other electrically-powered things you might use in this experiment to represent an "inefficient" heater. With any of these the room will ened up at the same temperature when all the battery's energy has been converted to heat through the.

Exceptions would include things that get some of the energy out of the room in a non-heat form (radio transmitter) or store it in another form (electric winch lifting a heavy weight up to the ceiling, electrolysis unit splitting water into hydrogen and oxygen). These would not heat the room as much (as long as you don't do anything with the stored energy in the latter cases).

Now if you're talking about a heater that has some connection to the outside world, such as an "air conditioner in reverse" (a heat pump) then these can have apparent efficiencies greater than 100%. In fact that is the point of using heat pumps to heat houses--you get more heat out than energy you put in, the balance coming from the outside air which is actually cooled in the process.

If it's warm outside, a "coefficient of perfomance" of 3 to 4 isn't unusual, meaning you get 3 to 4 times as much heat supplied by a heat pump as energy put into it, so you could make the room much warmer this way. Conversely if it's really cold outside, heat pumps can have apparent efficiencies less than 100%, and they end up heating the outside air a little bit.
posted by FishBike at 6:13 AM on February 24

Here's a recent thread on Physics Forums on the same subject that I participated in. As aubilenon and others have stated, all heaters all equally "efficient" at eventually increasing the temperature of a sealed room for a given amount of power. Yes, there is an asymmetry between this example and the AC-in-the-wall example, where different models and configurations would be expected to cool the room differently. Sound reasonable?
posted by Mapes at 6:14 AM on February 24

There also seems to be another law of physics which states no matter how many times I proofread my post beforehand, a number of typos will become apparent after posting it. Can I blame the laws of thermodynamics that require entropy to increase?
posted by FishBike at 6:16 AM on February 24

So, to clarify/expand/test:

I've got the theoretical sealed room and the battery, and we're assuming that the battery is perfectly efficient, in that it puts out an identical amount of power no matter what we hook up to it.

When I set up any conventional heater, I'll get the same amount of heat out of any of them.

What if I set up an unconventional heater? Say, a giant electric motor with a giant spinning flywheel. We'll assume that the room WILL contain any vibrations/sound from the motor. If I fire that up and let it run until the battery is exhausted, and then until the flywheel stops spinning. I will have gotten an equal amount of heat out of the motor, the bearings and the air friction as if I'd just hooked up a toaster?
posted by gjc at 6:27 AM on February 24

gjc: Yes, even the unconventional heater you describe will be 100% efficient at converting all that electricity to heat. Well, minus the tiny, tiny amount that's going to come out of it as radio waves of course.
posted by FishBike at 6:39 AM on February 24

What if I set up an unconventional heater?

Amazing how tricky these questions of heat and temperature can be. There is an unconventional heater (at least one) which will give you a final temperature greater than any conventional heater, if you set it up just right.

A dehumidifier! You set it up so the water it collects flows into a bottle which is then capped so it cannot re-evaporate. Then, when everything comes to final equilibrium, final temperature is a result of all the available energy in the battery, plus the heat given off by the water when it condensed.

If this seems like a fiddly special case, consider that you could do the same trick by having the battery power a dehumidifier-like device which liquefies one of the other gases in the room, such as oxygen, though in that case there is one last piece of the argument to be made about the final state of the oxygen in the capped bottle.
posted by jamjam at 10:20 AM on February 24

jamjam: You could also freeze little blocks of dry ice, and put them in your perfect insulation, making the room hotter (but oddly, making the average temperature including the dry ice, colder, as the temperature difference represents stored energy)
posted by aubilenon at 2:43 PM on February 24

jamjam, nice.
posted by fantabulous timewaster at 2:58 PM on February 24

I think an evaporating block of dry ice would cool the room, since the phase change will "take up" energy from the room.
posted by gjc at 4:34 PM on February 24

Here's a question/observation: I was under the impression that brakes on an automobile absorb energy of the moving car through abrasion of the brake pad. While it's obvious that this does result in heat, it always seemed to me like for the large amount of mechanical energy removed (e.g. decelerating a 4000 pound mass from 60 MPH to rest) never equaled the resulting warming of the rotors, that some of that energy was dissipated in simply breaking the bonds of the compounds that comprise the pads. I know that rotors can glow red hot after extended downhill or high speed braking, but in normal everyday driving conditions it doesn't seem like the amount of heat generated is equal to the amount of mechanical energy removed. Is there some kind of phantom "energy of breaking apart stuff" that doesn't result in heat or is it really all just heat and I'm not noticing it because the rotors are able to quickly dump most of that heat to the ambient air (it is after all like a big rotating fan.)
posted by Rhomboid at 2:34 PM on February 25

Rhomboid: while it does take some energy to wear out the pads and rotors, the amount is tiny compared to what goes into the rotors as heat. Since I'm imperial-unit-challenged I'll convert your hypothetical example to SI metric units and see what a back of the envelope calculation says.

So we have a vehicle moving at 26.83 m/s with a mass of 1809 kg. Kinetic energy is 1/2 mv^2 so I make that about 651,000 J (651 kJ) of kinetic energy to be converted into brake heat.

Let's somewhat arbitrarily decide there's 20 kg of cast iron being heated (4 rotors x 5 kg each). Specific heat capacity of cast iron is 0.46 kJ/kg K so multiply that by 20 kg and I get 9.2 kJ required to heat the brakes by 1 deg K (or deg C if you like).

651 kJ divided by 9.2 kJ/K = 70.76 K. So this hypothetical panic stop heats up the brake rotors by about 70 deg K (or C). If they were around room temperature to start with, they'd be pretty close to boiling water after one stop.

This is an over-simplification assuming none of the generated heat escapes from the brake rotors and that the entire mass heats to the same temperature, but hopefully it gives some idea of the order of magnitude. There doesn't seem to be a need to account for huge amounts of non-heat energy going missing.

And if anyone finds any mistakes in the above calculations, please be advised they are intentional and I'm just checking to see who notices.
posted by FishBike at 3:46 PM on February 25