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From: tomyee3@gmail.com (ProkaryoticCaspaseHomolog)
Newsgroups: sci.physics.relativity
Subject: Re: Want to prove =?UTF-8?B?RT1tY8KyPyBVbml2ZXJzaXR5IGxhYnMgc2hvdWxkIHRy?=
 =?UTF-8?B?eSB0aGlzIQ==?=
Date: Sun, 24 Nov 2024 17:34:48 +0000
Organization: novaBBS
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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
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