LED's efficiency exceeds 100%
physorg.com
physorg.com
There might be some applications in solid-state heat pump technology as well. Actually I would think that the more interesting possibility, given that the field is still maturing.
But they are energy inefficient because the laws of thermodynamics requires it. (You can make them more efficient by not cooling as low - trading temperature for volume. In an A/C that would mean instead of a 40 degree drop, do a 20 degree drop, but with twice the air flow. The net result for the living space is the same, but the efficiency is much improved.)
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.)
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).
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.
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.
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.)
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.
>"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?
It is only possible to do that if the photons are sent somewhere that is colder than the device.
> You're saying that a small fast moving body can't impart energy to a large slow moving body.
No, I'm saying it can't impart all of its energy, and the more energy the other body has the less energy it can impart. i.e. it's exactly Carnot efficiency.
Perhaps what you don't realize is that Carnot efficiency doesn't represent lost energy. It represents heat energy than can not be converted to another form, and remains in the source.
Where are you deriving this from. What force prevents me shining a light towards a hot object.
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.
> Although scientifically intriguing, the results won’t immediately result in ultra-efficient commercial LEDs since the demonstration works only for LEDs with very low input power that produce very small amounts of light.
The scientist then spend a lot of his time trying to come up with a better force field which could remain active for more time, but he stopped when he realized it was theoretically impossible (and hence, it was impossible to land in Jupiter). Meanwhile, the "experimenter" achieved the impossible after using the already created field, by turning it on-and-of very quickly and thus, preventin it from destabilizing!.
What does it have to do with this? Well, what if we had hundreds of those very dim leds? that might provide usable light.
But I agree engineers are weird animals. Often people have no idea of what they are really doing when they say "that can't be done" to one of us.
If the claim were more spectacular, we'd call this snake oil.
> [it] instead took advantage of small amounts of excess heat to emit more power than consumed. This heat arises from vibrations in the device’s atomic lattice, which occur due to entropy.
edit: I see from mikeknoop's comment they mean inside-an-oven levels: 135'C/275'F.
Heat into electricity is easy with a thermocouple. Electricity into light is easy with an LED.
And heat into light happens automatically with black body radiation.
It's not nothing though - it's pretty neat to use black body radiation to boost existing LED light output.
So you would not be cooling the room - you would be cooling the oven you put the devices into.