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

On Wed, 20 Nov 2024 19:57:18 +0000, ProkaryoticCaspaseHomolog wrote:

On Wed, 20 Nov 2024 18:37:40 +0000, rhertz wrote:
>

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.

>
1) You multiplied to get 0.03927 Wm² (funky units!) instead of
   dividing to get 637 W/m².
2) I was using a cube geometry, not a sphere. So our numbers are going
   to differ by a substantial amount.
3) You should not be starting from absolute zero, but from room
   temperature.
4) Look at my original calculation, where part of my calculation is
   T_f^4 - T_i^4 since I am not starting from absolute zero.

I told you that it was the unit area (effective). You can't use 637 W/m²
because a surface of 1 m² doesn't exist. Only a surface 127 times
smaller exist, which is the inner area of the cavity.
Stefan used his formula first to calculate the temperature of the Sun,
with a HUGE effective area. He calculated the "radiant power" by
estimating it in terms of the Sun's luminosity, decades before Planck.
In that case (even when I disagree about using Kirchoff's black body
radiation), he obtained a surface temperature that SEEMED TO LET HAPPY
most astronomer of his epoch (and even today). Even IF it's FALSE.
But, as in many other things in science, if you had NOTHING, something
seemed to be useful and was adopted by 1870. Using such law, stars
temperatures were calculated using their relative luminosity. But NOBODY
was accountable for the errors, if such equation was/is FALSE.
Go close to a star or the Sun and MEASURE its temperature. You can't,
can you? It's THE SAME CRAP as using Planck's equation to measure the
CBR by the COBE/WMAP satellites and (curiously) getting a perfect match.
I think that it's incorrect to use Stefan's formula in this case. How do
you spread 5 Watts over 0.00785 m²? Using 637 W/m², the formula gives
325.6 K = 52.45 "C. This is not reasonable, because scaling down, it
would represent a STEADY STATE of 41,350 K for such small surface.
The Stefan formula is empirical, and doesn't scale down well. It was
developed for big surfaces.
Better to use calories, converting the loss of energy in Joules. Then
using an approximation of 1 calorie --> 1"C rise/gram of water, you
could have a better result.
Now, how do you propose to use Stefan with a surface of 0.00785 m² that
radiates 5 Watts of power?