[ars]

Something is missing here. You can not generate energy just from heat you must have both a heat source and a cold sink. It's en extremely fundamental thermodynamic law.

Where is the cold source in this experiment? (I'm not saying there isn't one, just that the article makes no mention of it which is a huge omission.)

[jnhnum1]

Energy is certainly not being generated. The energy of the light output will be equal to the heat absorbed plus the electric energy input.

If you want to think of it in terms of temperatures, I'm a little fuzzy on the details, but I think in a sense empty space has a temperature associated with it based on the radiation that is present (similarly to how black body radiation is a function of temperature).

[ars]

I guess if the LED was heated more than the room where you receive the light it would work.

So you have to put this into an oven for it to work - if you just left it in a room, and expect it to also put light into that room (i.e. the device and the room are the same temperature) it would not work.

[justncase80]

So can we solve global warming by creating a bunch of gigantic LED billboards?

[ajuc]

Yeah, if we could cool some place without heating another by more than the amount of heat we cooled this place - it would breake 2nd law of thermodynamics - we could then use difference of temperatures to extract energy and we'll be in the same place, but with more energy. We could then repeat such cycle as many times as we want == infinite free energy.

[karamazov]

The heat absorbed is released as light; conservation of energy is still observed.

[ajuc]

But there was no gradient of temperatures before, and there is now - it's possible to use this gradient to produce energy. So this energy had to come from somewhere. It didn't come from electricity, because LED already has >100% efficiency (so there's more energy in the photons, than in electricity we put into the system).

So it has to come from temperature differnce between the place we cooled, and the place we heat up by photons, I think. So this place had to be cooler than the LED before experiment, so gradient of temperatures was there from the start, and we are only decreasing it.

[ars]

Conservation of energy is not the issue. It's not enough just to have energy in = energy out. In order to turn heat into any other kind of energy you MUST have a cold sink.

If you want to have black body radiation it's not enough to just have a dark area - the recipient of the light must be colder than the source.

So this would only work if the LED was heated more than the room around it. I guess if they put it in an oven, and then viewed it from a window in the colder room.

[Tuna-Fish]

> In order to turn heat into any other kind of energy you MUST have a cold sink.

No, in small enough scales this is not necessary. This is not "heat moving from a hotter object to a colder object and doing work on the way", this is "heat being directly converted to photons". When you view a patch of hot gas as individual molecules bouncing about you can transform the heat of a molecule into momentum of your target at 100% efficiency (heat is kinetic energy!). In quantum environments, the second and third law do not work like they do in the macroscopic world. The second and third are still not broken in the large because to do this kind of trickery you need to have a lot of exact information, and that has a cost in entropy. (So you'd need a maxwell's demon to scale this up.)

[ars]

No, this is just not true.

You can only convert the kinetic heat motion of a single molecule to kinetic energy in your target if your target is colder than the source! And there's your cold sink.

The efficiency is not 100% because the target sends energy back to the source since the target is moving (from heat kinetic energy).

If the target was standing still efficiency would be 100% - but that's the same as saying the target is at absolute zero and we already know that Carnot efficiency is 100% if the sink is at absolute zero, so it makes no difference that you are dealing with a single molecule.

To your second point that this is "heat being directly converted to photons" - that's exactly the definition of blackbody radiation. But the blackbody also absorbs radiation from the environment it is in, so it's not perfectly efficient either.

[pbhjpbhj]

It's simply not a full system analysis, > unity efficiency WRT input electrical energy. There's just a corresponding draw of energy from the environment that is also producing photons.

>"You can only convert the kinetic heat motion of a single molecule to kinetic energy in your target if your target is colder than the source! And there's your cold sink." //

This is counter logical. You're saying that a small fast moving body can't impart energy to a large slow moving body. Can you explain further how this works at the single molecule level?

[DougBTX]

There needs to be a cold sink since heat is just a variation on kinetic energy (small vibrations), and momentum needs to be conserved. The released photons have momentum too, and it looks like they get that from the heat (=vibrations) in the LED.

[weavejester]

It's not about conservation of energy, it's about entropy. If you have a device that can absorb heat from the environment and convert it into light without any form of cold sink, you're reducing entropy.