# Infrared Detectors & The 2nd Law of Thermodynamics

• B
• Devin-M
In summary, the photodetector is behaving in a way that does not agree with the 2nd Law of Thermodynamics.f
As a matter of interest, today I found myself doing the experiment of measuring IR flux from a plastic beaker of hot water whilst measuring the temperature as it gradually cooled. Without thinking properly, I thought a plot of IR flux vs temperature would extrapolate back to absolute zero, but it did not, but decayed gently towards room temperature. So zero indicated flux at 300K.

As a matter of interest, today I found myself doing the experiment of measuring IR flux from a plastic beaker of hot water whilst measuring the temperature as it gradually cooled. Without thinking properly, I thought a plot of IR flux vs temperature would extrapolate back to absolute zero, but it did not, but decayed gently towards room temperature. So zero indicated flux at 300K.

Which IR wavelengths were you measuring?

Putting aside the laser and detector and black body radiation for now, can’t I make a 320k well insulated tank of water get hotter by pushing high enough electrical current through a wire leading through the tank such that the electrical power through the section of wire in the tank exceeds the power of thermal heat flow out of the tank via the same wires (ie make sure the electrical power output into the tank exceeds the thermal flow out of the tank through the wires in order to accumulate heat in the tank)?

Is this a Rhetorical question? Yes of course you can. And so...?

russ_watters
Is this a Rhetorical question? Yes of course you can. And so...?

So I have a large current from the detector… more energy is stored in the water than escapes the water, yes?

A current at zero volts delivers no energy.

russ_watters
The detector produces the current via the photovoltaic effect. Doesn’t it mean there’s voltage otherwise there wouldn’t be current?

0v Bias voltage means there’s no voltage applied on the photodiode by an outside voltage source so the current that is produced when it’s hit with photons is by the photovoltaic effect.

So how much power is being provided? What is the circuit? The current by itself is not a measure of power as I mentioned previously. It is likely feeding a very high impedance op-amp FET .

russ_watters
So far what I envision is a laser is producing some wattage of 3.5 micrometer photons. These photons generate a current in the wires coming out of the detector via the photovoltaic effect, and the electrical power (wattage) through the wire section inside the 320k tank exceeds the thermal conduction power out of the tank through that same wire.

Shining a laser at an underwater photodiode to produce power violates no laws of thermodynamics. I could run solar panels in my swimming pool if I really wanted to. Why do you wish to do this?
This is quite different from your initial scenario.

russ_watters
Shining a laser at an underwater photodiode to produce power violates no laws of thermodynamics. I could run solar panels in my swimming pool if I really wanted to. Why do you wish to do this?
This is quite different from your initial scenario.

So now I make a really big detector and point it at a 300k lake surface, to heat the 320k water with the 3.5 micrometer photon flux off the 300k surface, with no laser.

This thread has run its course. You've been repeatedly told that what you suggest isn't possible and you're chasing your tail (and getting others to chase it) making little changes to your scenario to try to keep it alive. Thread closed.