See subject, with all conditions being equal. I hope so, or my understanding of frequency and energy is corrupt. I won’t be surprised if it is.
On a heat trip, huh?
I would think so, as it’s inverse to wavelength. However, what is important is the bandwidth of the light, or it’s spectral properties. You could potentially make a fluorescent light that looks like sunlight to everybody, but the spectrum would not be equal. Lots of sharp peaks in fluorescent. More simply for bandwidth, a 1 nm wide blue 440nm might be warmer than a 1 nm wide 650 nm red, but if the red were more than 1 nm wide, it’s more energy.
You also need to consider emission of light, which is the opposite colors of absorption. A black object (absorbs most light) would be warmer than a white one, but a colored object (even a “white” one like paper) may selectively absorb red, say.
What does “wide” mean? The measured width of the actual irradiated footprint?
E.g. the graph here has three Gaussians, and the variance of each (sigma squared on the graph) varies, with a (relatively) narrow blue curve and a very wide green one. You could make multiples that are all “blue” but the wide ones would have more energy if the height is equal.
Yes, it would be hotter under the blue light than under the red.
That is, if by “equal” you mean:
- The two light bulbs emit the same number of photons.
- The paper absorbs the same percentage of red photons as it does blue photons (e.g. if it were a grey body).
Each blue photon would dump more energy into the paper than a red one would so the paper would get hotter under the blue light.
In practice both 1) or 2) would be unknown.
They will not, if both lamps are the same wattage. i.e. a red 1W lamp will produce more photons than a blue 1W lamp.
And by will not, I mean they will be equally hot (assuming the absorption is the same)
Even that is debatable, since although we define the wattage, we have not defined the temperature of the filament, and thus the spectrum of the black body radiation, nor do we define the efficiency of the lamps. As a black body become hotter its radiation at all wavelengths is greater than when cooler. It is just that the peak shifts to shorter wavelengths and thus the overall colour shifts. But the curve always encompasses the entire curve of a lower temperature.
A single lamp run at two different surface temperatures will suck different amounts of power, so it isn’t really possible to compare like with like with the same lamp. So you need two different designed lamps, with different filament designs, one that runs hotter than the other, but both using the same power. Then you end up with a smaller bluer filament compared to the cooler redder one. However the total power radiated is proportional to the fourth power of the temperature, so the smaller filament will probably be radiating more total power than the larger but cooler one. So with both more total power radiated, and a spetcrum that will always radiate more energy at every wavelength than the redder filament, the blue filament will also be radiating more total photons no matter what the wavelength.
But the above is only black-bodies. If you have something with a line emission, or filtered light, it still isn’t going to be clear.
I thought he was using LEDs or lasers, not blackbody radiation. Also assuming 1W is output power, not input.
Yeah, it isn’t clear at all. I was just taking the Joe Average assumption that a 1W lamp would be a one watt incandescent. There are so many unstated assumptions in the OP that it is hard to nail down. That of course is half the fun. “All conditions being equal” alone leaves a lot to work out.
Hmm as long as incident wattage is the same (controlled by distance or a mask or a neutral density filter) and absorption is the same, won’t they get equally hot, regardless of the spectrum? I can’t think of a reason why not.
Of course if you start with the same wattage and use filters, then what you get is anybody’s guess.
The problem here is the “all other conditions equal” in the OP, since all other conditions can’t be equal. We can make it the same number of photons, or we can make it the same power radiated, but we can’t make it both.