On Sun, 24 Nov 2024 16:23:34 +0000, rhertz wrote:
On Sun, 24 Nov 2024 10:04:13 +0000, ProkaryoticCaspaseHomolog wrote:
>
Just for fun, I posted the last part with calculations to ChatGPT. I
wrote that I disagree with the idea of treating it as a black body
radiation problem. Here is the answer:
ChatGPT apparently used markup and/or unicode that did not render
properly on NovaBBT. I don't know about other newsreaders.
However, I was able to get the gist of what might have been its
response.
Let’s go through this text step-by-step and analyze it in detail,
including the key concepts, assumptions, and the areas where you might
disagree or where clarification is needed.
>
1. Overview of the Experimental Setup
The experiment involves injecting a 5-watt green laser into a cavity
made of aluminum. The chamber is initially at room temperature (T_i =
293 K), and as the laser is introduced, the temperature of the system
increases. The system eventually reaches a steady-state temperature
(T_f), where the heat input from the laser matches the heat lost by the
system.
>
Key concepts:
>
Steady-State Temperature (T_f): The final temperature when the system
stops warming up, and the heat input (from the laser) equals the heat
output (radiation to the surroundings).
Heat Radiation: The system will have two types of radiation at
steady-state:
Coherent laser light: Reflected inside the chamber and eventually
absorbed.
Black body radiation: Emitted by the chamber walls at temperature T_f.
Heat Absorption by the Walls: The aluminum shell absorbs energy,
increasing its temperature from T_i to T_f.
2. Calculation of Steady-State Temperature (T_f)
The equation provided to calculate the steady-state temperature is:
[UNABLE TO READ MARKUP AND/OR UNICODE]
Where:
[UNABLE TO READ MARKUP AND/OR UNICODE]
P is the power input (5 watts),
[UNABLE TO READ MARKUP AND/OR UNICODE]
ϵ is the emissivity of aluminum (0.13),
[UNABLE TO READ MARKUP AND/OR UNICODE]
σ is the Stefan-Boltzmann constant,
[UNABLE TO READ MARKUP AND/OR UNICODE]
is the external area of the sphere (0.03144 m²),
[UNABLE TO READ MARKUP AND/OR UNICODE]
is the final temperature,
[UNABLE TO READ MARKUP AND/OR UNICODE]
is the initial temperature (293 K).
After rearranging and solving the equation, the result is:
[UNABLE TO READ MARKUP AND/OR UNICODE]
=412.5K
This means the temperature of the chamber reaches 412.5 K after the
system has stabilized. This is a significant rise from the initial
temperature of 293 K, indicating the laser’s energy input has caused a
substantial increase in temperature.
>
3. Black Body Radiation Energy
The energy corresponding to the black body radiation inside the chamber
is given by:
[UNABLE TO READ MARKUP AND/OR UNICODE]
U=aVT
[UNABLE TO READ MARKUP AND/OR UNICODE]
Where:
[UNABLE TO READ MARKUP AND/OR UNICODE]
U is the total black body energy,
[UNABLE TO READ MARKUP AND/OR UNICODE]
a is the radiation constant,
[UNABLE TO READ MARKUP AND/OR UNICODE]
V is the volume of the sphere,
[UNABLE TO READ MARKUP AND/OR UNICODE]
T is the temperature.
The initial and final black body energies are calculated as:
[UNABLE TO READ MARKUP AND/OR UNICODE]
The increase in black body energy is:
[UNABLE TO READ MARKUP AND/OR UNICODE]
This increase is relatively small compared to the thermal energy stored
in the aluminum shell, as the laser’s energy is primarily absorbed by
the walls and converted into thermal energy.
>
4. Thermal Energy in the Aluminum Shell
The increase in thermal energy in the aluminum shell is calculated based
on the temperature rise from T_i to T_f (119.5 K). The text states the
increase in thermal energy in the foil is 220 J. This is a much larger
value than the change in black body radiation energy, which suggests
that the majority of the energy input is being absorbed by the aluminum
shell and not the cavity’s radiation.
>
5. Laser Energy Inside the Chamber
The equation given for the laser energy inside the cavity is:
[UNABLE TO READ MARKUP AND/OR UNICODE]
Where:
[UNABLE TO READ MARKUP AND/OR UNICODE]
P is the power of the laser (5 watts),
[UNABLE TO READ MARKUP AND/OR UNICODE]
D is the average distance between bounces inside the chamber (0.07 m),
[UNABLE TO READ MARKUP AND/OR UNICODE]
c is the speed of light,
[UNABLE TO READ MARKUP AND/OR UNICODE]
R is the reflectivity of the aluminum (0.99).
The final laser energy within the chamber after reaching steady-state is
calculated as:
[UNABLE TO READ MARKUP AND/OR UNICODE]
This is a very small amount compared to the thermal energy stored in the
shell (220 J), which aligns with the idea that most of the laser’s
energy is absorbed by the shell and only a tiny fraction remains as
coherent laser light inside the cavity.
>
6. Other Considerations: Vacuum vs Air
The author mentions that the experiment is likely performed in a vacuum.
If the experiment were conducted in air, convection currents would
interfere with the measurements, and the computation of the final
temperature would become more complex. This is a valid point, as in a
vacuum there are no convective heat losses, and the temperature rise
would be more directly attributable to radiation absorption.
ChatGPT didn't really seem to get the point that the air currents
would push the sphere around by unacceptable amounts.
7. Discussion of Emissivity and Reflectivity
The author mentions that the emissivity of aluminum (0.13) plays a
significant role in the final temperature. Emissivity reflects how
effectively a material emits thermal radiation compared to a perfect
black body (which has an emissivity of 1). The relatively low emissivity
of aluminum means that much of the thermal energy is retained in the
aluminum shell, and only a fraction of it is radiated away.
>
The reflectivity of aluminum (0.99) is also important, as it indicates
how much of the laser light is reflected within the cavity. This high
reflectivity implies that the laser light will bounce around multiple
times inside the chamber before being absorbed, which is consistent with
the assumption that only a small amount of the laser energy remains as
coherent light inside the cavity.
>
8. Potential Issues in the Calculations
There are a few areas where errors could potentially arise:
>
Approximation of D: The average distance between bounces (
[UNABLE TO READ MARKUP AND/OR UNICODE]
D) is approximated as 0.07 m, which is crucial for calculating the laser
energy within the cavity. The accuracy of this assumption depends on the
exact geometry of the cavity and the path taken by the laser light. If
the cavity is not spherical or if the reflectivity varies with angle,
this could affect the calculations.
However, since the energy of the laser light bouncing around in the
cavity is many orders of magnitude less than the total energy of the
system, which is dominated by shell heating, even large errors in
estimating D make no significant difference in the final result.
Emissivity and Material Properties: The low emissivity of aluminum
(0.13) was used, but if this is an average value, variations in
emissivity across the surface could lead to slight discrepancies. If the
aluminum surface is treated or coated to increase its emissivity, this
could change the temperature calculations.
True.
Conclusion
The primary issue in the text seems to be an implicit treatment of the
aluminum as a perfect black body in some places, while in reality,
aluminum has low emissivity and high reflectivity.
ChatGPT displays its ignorance of some basic facts about black body
radiation. At equilibrium, a closed cavity will contain black body
radiation corresponding to the temperature of its surrounding walls,
regardless of the wall's composition. The wall's emissivity,
reflectivity, color, and other material-specific properties do not
affect the result.
This discrepancy
means the final temperature and energy calculations could be influenced
by material-specific properties more than the text suggests.
>
Your disagreement with black body radiation likely stems from the
realization that aluminum, as a metal, does not absorb all the laser
energy
The conditions of the experiment are that virtually all the incoming
laser energy will be absorbed, only a tiny amount managing to make it
out through the tiny entrance hole.
On the other hand, ChatGPT is correct that the ball will not radiate
like a perfect black body.
The actual
behavior of the aluminum shell—reflecting, absorbing, and emitting
energy with low emissivity—should be more accurately modeled, especially
considering that a significant amount of laser energy remains in the
form of coherent light rather than being entirely absorbed and converted
into heat.
Typical ChatGPT Bullshit. An *INSIGNIFICANT* amount light remains in
the cavity.
**********************************************
>
I didn't want to re-write the equations that appeared in the answer. You
better post yourself the entire text and read, on your computer screen,
the real answer.
>
It appears that ChatGPT agree with me that treating this as a BBR
problem is incorrect.
At some points, ChatGPT spit out correct sentences. At other points,
ChatGPT descended into total bullshit. Unfortunately, ChatGPT hardly
ever lets you know when it is bullshitting.
I described, prior to ChatGPT answer, how the experiment was set up:
data about the cavity (5x5x0.1), 550 nm 5W laser, etc.
>
Try it yourself.
>
BTW, I checked ChatGPT for a couple of other subjects, and IT'S SCARY. I
read something about AI lectured in the last summit (Musk, Microsoft,
Google, Altman, etc.). They consider that AI will hold 99% of human
intelligence before 2030, rendering human work useless.
>
Scary as hell, and I believe it's going to happen, unless developments
be stopped and/or carefully limited.
>
Musk sugested about AI being in love with humanity (human values), to
prevent that the human eco-system be destroyed due to the concentration
of greedy power in the hands of very few. Only in medical sciences, the
advances in diagnosis and R&D are astonishing TODAY.
>
Hawking was right.
Want to prove E=mc²? University labs should try this!
Re: Want to prove E=mc²? University labs should try this!