I agree. Note that an air conditioner has >100% efficiency if you measure the heat output of the external exchanger divided by the electricity use. When used like this for heating, it's call "heat pump" instead of a "air conditioner". https://en.wikipedia.org/wiki/Heat_pump
Here however there is the question ... where is the temperature differential? Surely it's not the case that the light produced is colder than the semiconductor. Light rays don't interact, so they wouldn't have any meaningful temperature ... I think.
Good question! I need some time to reach the answer.
Note that a air conditioner or a heat pump creates a temperature difference using energy. Here the cold thing is the LED (or whatever is the name of this device, I'll call it LED). The hot thing is ... the light!
If you have an empty box, and the walls are made by a perfect emisor/absorber that is is like a perfect black body, then inside the box you will get light emitted by the walls. After a while, the light inside the box and the walls of the box will reach equilibrium. Then you can assign a temperature to the radiation inside the box, that is equal to the temperature of the walls. Moreover, you can assign energy and entropy to the radiation. The technical name is "photon gas" https://en.wikipedia.org/wiki/Photon_gas
So the cold part is the LED, the hot part is the light. In a case of a LED like in this experiment, the radiation is not at equilibrium, and the LED is not a black-body emisor, so the calculations are more complicated.
(I can somewhat reconstruct the calculation in the article in Wikipedia if I have enough time and a small amount of cheating with Google. But the calculations of the temperature/entropy of the light emitted by the LED in this experiment are above my level.)
This light emission that cools the semiconductor is like when sweating cools you.
The part of the sweat which evaporates is formed by molecules that have a higher average energy than corresponding to the body temperature, so the body loses energy and its temperature goes down.
The photons emitted by a LED have also a higher energy than the average energy of the photons that are emitted or absorbed as thermal radiation, corresponding to the semiconductor temperature, so the LED loses energy by emitting light, so its temperature tends to go down.
The LED also receives energy from the electrical current which passes through it and the heating due to the passing of the electrical current is normally greater than the cooling due to the emitted light.
Except that in special conditions the electrical heating can be less than the cooling due to light emission, so if the LED becomes cooler than the ambient temperature, heats begins to flow into the LED from the ambient medium and this provides additional energy, which balances the energy lost by light emission.
5 comments
[ 3.1 ms ] story [ 20.8 ms ] threadKind of impressive - they take heat from the environment.
Note that a air conditioner or a heat pump creates a temperature difference using energy. Here the cold thing is the LED (or whatever is the name of this device, I'll call it LED). The hot thing is ... the light!
If you have an empty box, and the walls are made by a perfect emisor/absorber that is is like a perfect black body, then inside the box you will get light emitted by the walls. After a while, the light inside the box and the walls of the box will reach equilibrium. Then you can assign a temperature to the radiation inside the box, that is equal to the temperature of the walls. Moreover, you can assign energy and entropy to the radiation. The technical name is "photon gas" https://en.wikipedia.org/wiki/Photon_gas
There are also more details in https://www.nature.com/articles/s41598-017-01622-6
So the cold part is the LED, the hot part is the light. In a case of a LED like in this experiment, the radiation is not at equilibrium, and the LED is not a black-body emisor, so the calculations are more complicated.
(I can somewhat reconstruct the calculation in the article in Wikipedia if I have enough time and a small amount of cheating with Google. But the calculations of the temperature/entropy of the light emitted by the LED in this experiment are above my level.)
The part of the sweat which evaporates is formed by molecules that have a higher average energy than corresponding to the body temperature, so the body loses energy and its temperature goes down.
The photons emitted by a LED have also a higher energy than the average energy of the photons that are emitted or absorbed as thermal radiation, corresponding to the semiconductor temperature, so the LED loses energy by emitting light, so its temperature tends to go down.
The LED also receives energy from the electrical current which passes through it and the heating due to the passing of the electrical current is normally greater than the cooling due to the emitted light.
Except that in special conditions the electrical heating can be less than the cooling due to light emission, so if the LED becomes cooler than the ambient temperature, heats begins to flow into the LED from the ambient medium and this provides additional energy, which balances the energy lost by light emission.