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

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.
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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.
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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.
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Even if this is a very low weight (or mass), it's almost 10,000 times
higher than in previous (and wrong) calculations.
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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.
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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.
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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.
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Q: What would be the temperature of the sphere?
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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)

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Q: Exist a specific wavelength at which the heat is radiated?
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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.

.................
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Q: How many seconds would take to reach thermal equilibrium at 707 K?
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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.

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