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

On Tue, 26 Nov 2024 18:29:11 +0000, rhertz wrote:

On Tue, 26 Nov 2024 8:27:15 +0000, ProkaryoticCaspaseHomolog wrote:
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On Mon, 25 Nov 2024 21:54:34 +0000, rhertz wrote:
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This means that half of the accumulated 955 Joules remain within the
cavity.

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477.5 Joules of energy in a volume of 5.236e-4 m^3 implies a energy
energy density u (assumed to be black body) within the shell of
9.12e4 J/m^3
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u = 4 σ T^4/c
9.12e4 = 4 * 5.67e^-8 T^4/3.0e^8
T^4 = 120,634,920,634,920,634,920
T = 1.05e5 K temperature of the black body radiation
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Your numbers don't make sense.

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Your entire post don't make sense.
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1) Your u value is wrong by 10 times. It should be u = 9.12e5 J/m^3.

Yes, I dropped a digit.
So things are even WORSE for you.
The effective black body temperature corresponding to that radiation
density is 1.86e5 K

2) You applied incorrectly how the radiation is spread within the
cavity.
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P = uc/4 is the radiative FLUX of energy being radiated in 3D, traveling
at the speed of light. The factor 4 is due to the integration of energy
moving omnidirectionally and being radiated toward infinity, which IS
NOT THE CASE HERE, INSIDE THE CAVITY.
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The 477.5 Joules ARE CONFINED in the cavity, and the radiation keeps
bouncing within its volume.
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At any case, it would give a radiative flux P = 6.84E+13 J m^-2/sec.

Read that out loud and ask yourself if that is reasonable.
68.4 trillion watts per square meter.
From a 5 watts of input power.

When you modify the Stefan-Boltzmann law by writing u = 4 σ T^4/c,
you're assuming that the energy propagates spherically without
attenuation. At any case, T =  186,364 K, a value that HAS NO MEANING
AT ALL.

It has a very definite meaning.
See the discussion in "Energy density"
https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law#Energy_density

3) Using correctly the influence of energy within the cavity filled with
air (behaving as an ideal gas), you have:
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E = 3/2 nRT
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n = PV/RT , moles of air in the volume
R = 8.3145 kJ/kmol K (ideal gas)
R = 0.287 kJ/kg K (air)
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ΔT = 2E/(3 PV) = 6 K , the increase above room temperature.

What type of numerology are you using?
Where did you get a value of E to plug into that?
Check your units. PV has dimensions of energy, so 2E(3 PV) is a
dimensionless number.

As you can see, this is a much more reasonable value.
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What would you expect if you use a 5 W incandescent lamp heating the
cavity?
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Only 6"C increase in the inner temperature after a couple of minutes.

Yet you claim that that 5 watt incandescent lamp will raise the
temperature to 707 K.