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From: hertz778@gmail.com (rhertz)
Newsgroups: sci.physics.relativity
Subject: Re: Want to prove =?UTF-8?B?RT1tY8KyPyBVbml2ZXJzaXR5IGxhYnMgc2hvdWxkIHRy?=
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Date: Tue, 26 Nov 2024 18:29:11 +0000
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On Tue, 26 Nov 2024 8:27:15 +0000, ProkaryoticCaspaseHomolog wrote:

> On Mon, 25 Nov 2024 21:54:34 +0000, rhertz wrote:
>
>> This means that half of the accumulated 955 Joules remain within the
>> cavity.
>
> 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
>
> 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
>
> Your numbers don't make sense.

Your entire post don't make sense.

1) Your u value is wrong by 10 times. It should be u = 9.12e5 J/m^3.

2) You applied incorrectly how the radiation is spread within the
cavity.


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.

The 477.5 Joules ARE CONFINED in the cavity, and the radiation keeps
bouncing within its volume.

At any case, it would give a radiative flux P = 6.84E+13 J m^-2/sec.

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.

3) Using correctly the influence of energy within the cavity filled with
air (behaving as an ideal gas), you have:

E = 3/2 nRT

n = PV/RT , moles of air in the volume
R = 8.3145 kJ/kmol K (ideal gas)
R = 0.287 kJ/kg K (air)



ΔT = 2E/(3 PV) = 6 K , the increase above room temperature.

As you can see, this is a much more reasonable value.

What would you expect if you use a 5 W incandescent lamp heating the
cavity?

Only 6"C increase in the inner temperature after a couple of minutes.