Re: Want to prove E=mc²? University labs should try this!

On Wed, 20 Nov 2024 0:48:23 +0000, gharnagel wrote:

On Tue, 19 Nov 2024 20:04:06 +0000, ProkaryoticCaspaseHomolog wrote:

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On Tue, 19 Nov 2024 18:04:04 +0000, gharnagel wrote:

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The 300 degree temperature rise I calculated was under such
condition, based upon 300 cm^2 of surface area radiating with
no convection.
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The point is, though, that with 3x10^9 bounces/second, it would
take much less than a second to whittle 5 W down to nothing:
1E5 bounces: 0.905Po
1E6 bounces: 0.368Po
2E6 bounces: 0.135Po
3E6 bounces: 0.050Po (in the first msec)
4E6 bounces: 0.018Po

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5 watts is approximately the energy output of an old-style
incandescent nightlight run off a Powerstat variable transformer
set to 100 volts. Even if I plugged the nightlight into 120 volt
mains, I can still hold the nightlight in my hand without it being
too uncomfortable.

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Most of the nightlight energy is in the IR (see attachment), and
the glass envelope is transparent to most of the IR.  The warmth of
the bulb represents only a fraction of the 5W.
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How did you calculate 300 C? Are you sure you
didn't mean 300 K?

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I have a GE radiation slide rule calculator from the long ago
good ol' days.  I could have written an app to do it to eight
decimals, but this is close enough to show that the whole idea
is not worth doing that.
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Anyway, the surface area is about 300 cm^2, so .0167 W/cm^2.
Emissivity of 0.3 ... Oop!  I was reading the K scale, 320K,
about 45 C.  Ya got me!  That doesn't change the fact that
the light is quickly absorbed and converted to heat.
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Your bounce attenuation calculation is a bit off.

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Yes, to be exact I should calculate a reduced power level after each
bounce.  Being a back-of-the-envelope calculation I just took it
as .999999^n.  Good enough for government work.  Not important
enough to find or write an app since there's not enough left after
1 msec to make a microbe sneeze.
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Possibly, covering the outside of the ball with a low-emissivity
coating might give a faint possibility of weighing the heat
energy in the ball.
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https://en.wikipedia.org/wiki/Low_emissivity
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Actually, they have a value of 0.03 for Al foil itself.  But
Al would melt before it was hot enough.  In fact, EVERYTHING
would melt:
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Wall thickness: .001" 167650K
Wall thickness: .100" 1676K (weight: 862 gm)
Wall thickness: .200" 838K (weight: 8620 gm)
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Might as well stick with Al and use a thicker wall.  That
puts all the pressure on the scales though, but I think that's
a much easier problem than weighing actual 455 nm light.  Just
keep pumping 5W into the cavity for 72 hours, and the hole
for introducing the 5W could be a sapphire disc with a 455 nm
transmission band, otherwise, too much energy would be lost.
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What do you think, still out of bounds?  Yeah, I think so.

Use this site instead:
https://aluminyumburada.com/en/aluminium-weight-calculation
Weight of a sheet of 100x100mmx0.1mm (Alloy 5083) is 3.0 grams.
Not bad for my estimation (2 gr) for the weight of the cavity, with
inner surface of 7,850 mm², and thickness of 0.1 mm.
And it will not melt. Not even heat enough for you to feel it.
Most of the energy will be confined in the inner volume, not in the
walls, even with 99% reflectivity.