3 ms·

> But if you were to heat the PV cell to that same temperature as the surface of the sun [...] Much easier to replace the sun by some other source of EM-waves

> But if you were to heat the PV cell to that same temperature as the surface of the sun [...] Much easier to replace the sun by some other source of EM-waves of suitable wave length and put it into a freezer, below the temperature of the cell. How does the wave poking some electron to a higher state remember the temperature of its source?

jbay808 2y ago
Good question. Temperature is a funny thing. When a system is far from equilibrium, the notion of temperature becomes a little unfamiliar. The thermodynamic temperature of the radiation is connected to the entropy of its power spectrum. LEDs and especially lasers emit very low-entropy light, which can be focused and heat a surface up to very high temperatures. (Sunlight focused onto a surface cannot heat that surface above the temperature of the sun). Thermodynamically speaking, a low-entropy power source like a laser -- and whatever is driving it -- must have a very high exergy, which is equivalent to behaving like a high temperature heat source, even though it might feel cold to the touch. Some more details here: https://en.wikipedia.org/wiki/Exergy#Quality_of_energy_types https://en.wikipedia.org/wiki/Exergy#Quality_of_energy_types Stored electricity's equivalence to a high-temperature heat source is one of the things that makes it so useful. It's intrinsically connected to why it takes a lot of low-grade heat to produce a small amount of electricity in the first place, and also why electric furnaces can produce such high temperatures. So while a battery can drive an LED that shines on a PV panel that generates power with everything feeling equally warm to the touch, a temperature gradient is still necessary; it's just been moved outside the boundaries of the system, to the process that distilled the entropy out of the energy that became the battery's stored charge. We can reversibly recreate this temperature gradient by driving a Carnot engine with that battery, instead of a laser.
eigenket 2y ago
> How does the wave poking some electron to a higher state remember the temperature of its source? The black-body spectrum of the source determines the amount of waves it outputs at each frequency, and in turn is determined by the temperature.
- robocat 2y ago
  I think semiconductors (such as most solar cells) do not act as ideal black bodies. Thermal excitation (what temperature required?) can move electrons into the conduction band and when the electron relaxes it emits a photon corresponding to the band gap. I looked for a good article, but didn't find anything. There was one relevant comment in the first article I skimmed that said "semiconductors do not behave remotely close to a blackbody": sounds authoritative but the comment was on stack-overflow so hard to judge its correctness!
  - eigenket 2y ago
    Oh yeah, I was talking in the context of the source being the sun. Of course other sources vary.
    - robocat 2y ago
      Right. The context you were answering is jbay's comment that "if you were to heat the PV cell to that same temperature as the surface of the sun". Sorry.

jbay808 2y ago

Good question. Temperature is a funny thing. When a system is far from equilibrium, the notion of temperature becomes a little unfamiliar. The thermodynamic temperature of the radiation is connected to the entropy of its power spectrum. LEDs and especially lasers emit very low-entropy light, which can be focused and heat a surface up to very high temperatures. (Sunlight focused onto a surface cannot heat that surface above the temperature of the sun). Thermodynamically speaking, a low-entropy power source like a laser -- and whatever is driving it -- must have a very high exergy, which is equivalent to behaving like a high temperature heat source, even though it might feel cold to the touch. Some more details here: https://en.wikipedia.org/wiki/Exergy#Quality_of_energy_types https://en.wikipedia.org/wiki/Exergy#Quality_of_energy_types Stored electricity's equivalence to a high-temperature heat source is one of the things that makes it so useful. It's intrinsically connected to why it takes a lot of low-grade heat to produce a small amount of electricity in the first place, and also why electric furnaces can produce such high temperatures. So while a battery can drive an LED that shines on a PV panel that generates power with everything feeling equally warm to the touch, a temperature gradient is still necessary; it's just been moved outside the boundaries of the system, to the process that distilled the entropy out of the energy that became the battery's stored charge. We can reversibly recreate this temperature gradient by driving a Carnot engine with that battery, instead of a laser.

eigenket 2y ago

> How does the wave poking some electron to a higher state remember the temperature of its source? The black-body spectrum of the source determines the amount of waves it outputs at each frequency, and in turn is determined by the temperature.

robocat 2y ago
I think semiconductors (such as most solar cells) do not act as ideal black bodies. Thermal excitation (what temperature required?) can move electrons into the conduction band and when the electron relaxes it emits a photon corresponding to the band gap. I looked for a good article, but didn't find anything. There was one relevant comment in the first article I skimmed that said "semiconductors do not behave remotely close to a blackbody": sounds authoritative but the comment was on stack-overflow so hard to judge its correctness!
- eigenket 2y ago
  Oh yeah, I was talking in the context of the source being the sun. Of course other sources vary.
  - robocat 2y ago
    Right. The context you were answering is jbay's comment that "if you were to heat the PV cell to that same temperature as the surface of the sun". Sorry.