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

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.