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

On Tue, 26 Nov 2024 2:25:07 +0000, ProkaryoticCaspaseHomolog wrote:

On Mon, 25 Nov 2024 21:54:34 +0000, rhertz wrote:
>

Prokaryotic, I was thinking about what you wrote on the cavity behaving
as a black body and, as I wrote before, I completely disagree to take it
as a black body radiating energy, once equilibrium has been reached.
>
My main doubt was that, once in equilibrium and having gained as heat
all the energy supplied by the 5W laser, the aluminum cavity HAD TO
radiate using the external surface AS WELL AS the internal surface. I
thought that almost HALF of the heat was going to be radiated INTO THE
CAVITY.

>
Yes, the aluminum radiates inwards as well as outwards, but the heat
radiated inwards is reabsorbed into the aluminum.
>

Injecting 5 Joules/sec makes the cavity (initially at room temperature
of 300K) to reach thermal equilibrium in a couple of minutes.
>
I used ChatGPT, which calculated the thermal equilibrium at 707 K, which
is reached in 191 seconds.

>
Let's borrow a hot plate from Paul. I would ask for a Bunsen burner,
but Bunsen burners don't work in vacuum.
>
Take a solid aluminum ball, emissivity 0.13, radius 5 cm. Heat the
ball to 707 K in a vacuum chamber whose walls are 293 K.
>
The net heat radiated by the ball is
P = ε σ A_e (T_f^4 - T_i^4)
P = 0.13 * 5.67e-8 * 0.03144 * (707^4 - 293^4)
P = 0.13 * 5.67e-8 * 0.03144 * (249,849,022,801 - 7,370,050,801)
P = 56.2 watts
>
Replace the solid ball with a hollow ball.
Are you claiming that the heat radiated by the ball depends on
whether it is hollow or solid?
>

This means that half of the accumulated 955 Joules remain within the
cavity. The extra mass added to the 2 grams cavity would be 5.306E-12
grams, adding an extra weight of 0.052 nanoNewtons.
>
Even if this is a very low weight (or mass), it's almost 10,000 times
higher than in previous (and wrong) calculations.
>
I believe that such weight can be measured by advanced technology and,
besides, it's a steady value, so measurements are not limited by time.
>
>
>
I can't transcribe the answers of ChatGPT (I failed once), so I write
here my questions so you or anybody can duplicate the chat.
>
....................
Q: I have one sphere made of aluminum (2 grams), with a tiny hole of 3
mm^2 used to inject a 5 watts (550 nm) green laser beam. The sphere has
a 5 cm radius and a thickness of 0.1 cm.. It has been calculated that
the 5 Joules per second are completely absorbed by the aluminum. Does
the sphere (a cavity) radiates the  absorbed 5 Joules per second into
the cavity, as well a part of it by its external surface? In what
proportions?

>
You're using a MUCH thicker shell than in previous thought experiments.
So the exterior surface area is 4% more than the interior surface area.
The aluminum radiates 96% as much into the cavity as it radiates
outwards into space. However, the interior surface of the shell
_absorbs_ as much energy as it radiates, so that the net power going
in and out of the cavity is zero. The power radiated out into space
will be 5 watts more than the power absorbed by the ball from the
environment.
>

A: approximately 49% of the absorbed energy radiates into the cavity,
and 51% radiates externally.
...................
>
Q: What would be the temperature of the sphere?
>
A: The equilibrium temperature of the sphere is approximately 707 K (434
°C).

>
No.
The steady state temperature of the ball does not depend on whether it
is solid or a hollow shell.
>
Assuming that the environmental temperature is 293 K and that the
experiment is conducted in vacuum,
>
P = ε σ A_e (T_f^4 - T_i^4)
5 = 0.13 * 5.67e-8 * 0.03144 * (T_f^4 - 293^4)
5 = 0.13 * 5.67e-8 * 0.03144 * (T_f^4 - 7,370,050,801)
21,575,509,276 = T_f^4 - 7,370,050,801
T_f^4 = 28,945,560,077
T_f = 412.5 K (actually, the numbers only justify 1-2 sig figures)
>

....................
>
>
Q: Exist a specific wavelength at which the heat is radiated?
>
A: Using Wien's Displacement Law, the peak energy is radiated  across a
broad spectrum, primarily in the infrared. Most radiation is in the
mid-infrared range (2 - 25 μm), with a peak at 4.1 μm.

>
Your numbers are off because your final temperature is off.
>

.................
>
Q: How many seconds would take to reach thermal equilibrium at 707 K?
>
>
A: The sphere would take approximately 191 seconds (about 3 minutes) to
reach thermal equilibrium at 707 K under the 5 W laser input.

>
The word is "steady-state", not equilibrium.
>
If the sphere were perfectly insulated, it would reach T_f in a time
which you would calculate as mass times thermal coefficient and
temperature rise divided by power. But it is NOT perfectly insulated,
so you have to integrate a constant input power minus an increasing
amount of power lost as the sphere heats up.
>
You also have to specify an endpoint. An appropriate end-point might
be when the temperature reaches to within, say, 0.01 K of its steady
state value.
>

...................
>
>
>
ChatGPT used formulae from Stefan, Wien, Planck and many others to
provide the results.

**************************************************************
I don't understand why you didn't follow the questions that I did to
ChatGPT, which were only four. I also explained that it was impossible
for me to transcribe the complex and detailed calculations that ChatGPT
did, excelling in clarity and meaning.
I wrote those four questions and summarized each answer the best I
could. My idea of posting those questions was for you TO REPRODUCE the
extensive reasoning behind the answers. Easily, exceeding three pages
full of explanations, formulae, reasoning and calculations. ChatGPT
excelled working over this, being a sign that it really learn from
previous chat with me.
BUT, you resorted to IGNORE what I wrote and kept stuck with your own
interpretation and misconceptions about this experiment, like
introducing the concept of a solid sphere. Why?????
At any case, I made my mind: You ARE WRONG, and ChatGPT is correct (this
time).
If you want to try a further refinement of the idea, I have two (at
least):
1) The cavity is suspended by a thin wire within a thick box of
REFRACTORY MATERIAL (in the same way that the BBC at the Berlin PTB
was). I also published a picture of the original BBC used from 1893 to
1901, and it was huge, with porcelain as refractory material, and heated
electrically up to 1,600°C.
2) The entire block volume is cooled on the outside at -170°C to prevent
heat escaping such volume (1 m^3 is OK for you?).
I used 0.1 cm thickness since the beginning, to obtain a spherical
cavity of 2 grams of aluminum, so check your wrong comment.
If you want to deal with this topic with fairness, I suggest you try the
FOUR QUESTIONS, in that order, so we both can have the same calculations
without the excessive burden of transcribing them.
Finally, it seems that you are renegading of your principle of steady
state equilibrium, when I wrote that half the heat remains inside the
cavity (in the IR range) and the other half stay outside, confined at
the 1 cubic meter of a thermally isolated chamber.
Don't be like Paul, repeating all the time your calculations. Show some
respect to AI, which after all is closing its interpretation of the
problem by using refinements. ChatGPT remember all my previous chats.
I'll be thinking about further enhancements to my proposals. I'm far
from being knowledgeable in thermodynamics, but I learn from succesive
refinements of my understanding on this particular problem.
HERE ARE THE FOUR QUESTIONS, EXACTLY. DISMISS THE ANSWERS THAT I
SIMPLIFIED AND GET A CHANCE TO CHATGPT. IT'S GETTING BETTER MONTH AFTER
MONTH. NOBODY CAN DENY THIS FACT, KNOWN WORLDWIDE.
//////////////////////////////////////////////////////////////////
...................
Q1: I have one sphere made of aluminum (2 grams), with a tiny hole of 3
mm^2 used to inject a 5 watts (550 nm) green laser beam. The sphere has
a 5 cm radius and a thickness of 0.1 cm.. It has been calculated that
the 5 Joules per second are completely absorbed by the aluminum. Does
the sphere (a cavity) radiates the  absorbed 5 Joules per second into
the cavity, as well a part of it by its external surface? In what
proportions?
A: approximately 49% of the absorbed energy radiates into the cavity,
and 51% radiates externally.
..................
Q2: What would be the temperature of the sphere?
A: The equilibrium temperature of the sphere is approximately 707 K (434
°C).
...................
Q3: Exist a specific wavelength at which the heat is radiated?
A: Using Wien's Displacement Law, the peak energy is radiated  across a
broad spectrum, primarily in the infrared. Most radiation is in the
mid-infrared range (2 - 25 μm), with a peak at 4.1 μm.
................
Q4: How many seconds would take to reach thermal equilibrium at 707 K?
A: The sphere would take approximately 191 seconds (about 3 minutes) to
reach thermal equilibrium at 707 K under the 5 W laser input.
..................
//////////////////////////////////////////////////////////////////