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From: tomyee3@gmail.com (ProkaryoticCaspaseHomolog)
Newsgroups: sci.physics.relativity
Subject: Re: Want to prove =?UTF-8?B?RT1tY8KyPyBVbml2ZXJzaXR5IGxhYnMgc2hvdWxkIHRy?=
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Date: Tue, 26 Nov 2024 20:50:43 +0000
Organization: novaBBS
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On Tue, 26 Nov 2024 18:29:11 +0000, rhertz wrote:

> 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.

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.
>
>
> 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.

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:
>
> 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.

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.
>
> 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.

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