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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: Wed, 20 Nov 2024 18:37:40 +0000
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On Wed, 20 Nov 2024 17:44:28 +0000, ProkaryoticCaspaseHomolog wrote:

> On Wed, 20 Nov 2024 17:35:09 +0000, rhertz wrote:
>
>> This IS NOT the case of the cavity under discussion, because its
>> temperature increase with the permanent supply of energy, until it's
>> destroyed, partially or entirely.
>
> Why should the foil box not radiate away energy as its temperature
> rises?
>
> Forget the innards. I don't care whether the reflectance is 0.85 or
> 0.99999. Those numbers only dictate the power levels reached within
> the cavity.
>
> What happens to the exterior surface of the box?

Consider this, using the Stefan-Boltzmann law:

1) Reverse the problem: think that the inner area has spherical shape,
with a radius of 5 cm. It gives A = 78.54 cm² = 0.00785 m².
Now cut the cavity and spread the area in 2D. It's a planar surface.

2) Consider that such area EMITS (due to perfect reflection)
5 Watts/unit area. This would give 0.03927 W/m². You have to spread the
5W all over the surface. Dividing 5W by 0.00785 m² is INCORRECT.

3) Apply Stefan-Boltzmann law:


0.03927 W/m² =   5.67E-08 Watt/m²/K⁴ x T⁴

T⁴ = 0.03927/5.67E-08 K⁴ = 6.926E+05 K⁴

T = 28.85 K

This is cold, isn't it?

AGAIN: You have to consider that the entire surface (0.00785 m²) is
emitting, not receiving 5 Watts/unit area. In this case, 0.00785 m²
is the unit area, so such area emits 0.03927 W/m².

The calculations are based on 100% reflectivity.

Do you see any error in my interpretation? I just reversed the entry of
power, taking it as power emitted by the reflecting surface. Maybe I'm
wrong, but I don't think so.