Path: ...!weretis.net!feeder9.news.weretis.net!i2pn.org!i2pn2.org!.POSTED!not-for-mail
From: tomyee3@gmail.com (ProkaryoticCaspaseHomolog)
Newsgroups: sci.physics.relativity
Subject: Re: Want to prove =?UTF-8?B?RT1tY8KyPyBVbml2ZXJzaXR5IGxhYnMgc2hvdWxkIHRy?=
 =?UTF-8?B?eSB0aGlzIQ==?=
Date: Wed, 20 Nov 2024 17:09:12 +0000
Organization: novaBBS
Message-ID: <cc08e9080b69815683c3bbe9c0dd2086@www.novabbs.com>
References: <b00a0cb305a96b0e83d493ad2d2e03e8@www.novabbs.com> <9687d391072c6f5d19d3e4cad9e944ba@www.novabbs.com> <01685fa9d16c8f15a4b8fd63f5b42ed2@www.novabbs.com> <b6c06d66a1d5da3a239a49ba5f903e2e@www.novabbs.com> <3cccb55b7c7c451a385b8aad5aac6516@www.novabbs.com> <cfcd6e742c4f3c2f8a5f69d4db75206f@www.novabbs.com> <d793169808c9c1e887527df5f967c216@www.novabbs.com> <98a0a1fbdc93a5fcc108882d99718764@www.novabbs.com> <fd4937f7b180bac934eb677cca8f5c55@www.novabbs.com> <ebcad35958736e6602cf803fddfdb0fd@www.novabbs.com> <141e19a1c6acd54116739058391ca9f8@www.novabbs.com> <a4f98fa5d026bfbf5127fcbc6a585772@www.novabbs.com> <b859adf3e138697712b038bd1d73902e@www.novabbs.com> <3da6cb303f9998fa49034c557d5c314b@www.novabbs.com> <6bdb52ff942fd2465d8344d6c61488dc@www.novabbs.com> <a3c12890fdabb69ea3891b0cb506b158@www.novabbs.com>
MIME-Version: 1.0
Content-Type: text/plain; charset=utf-8; format=flowed
Content-Transfer-Encoding: 8bit
Injection-Info: i2pn2.org;
	logging-data="3377789"; mail-complaints-to="usenet@i2pn2.org";
	posting-account="Ooch2ht+q3xfrepY75FKkEEx2SPWDQTvfft66HacveI";
User-Agent: Rocksolid Light
X-Rslight-Posting-User: 504a4e36a1e6a0679da537f565a179f60d7acbd8
X-Rslight-Site: $2y$10$kE5gM9cndU68xSi1lBxZq.HgCFOiJL286QHlW5yqLGHkvdIkWYMIa
X-Spam-Checker-Version: SpamAssassin 4.0.0
Bytes: 3038
Lines: 33

On Wed, 20 Nov 2024 15:57:23 +0000, ProkaryoticCaspaseHomolog wrote:

> No, because the temperature of the foil rises until the power emitted
> equals 5 watts. Once that happens, no further increase of stored
> energy occurs.
>
> P = ε * σ * T^4
>
> Let us assume a 10x10x10 cm cube with surface area 0.06 m^3
> So 5 watts total power emitted means 83 watts/m^2
> Assume ε = 0.13
>
> Let T_i be 273 K
>
> 83 = ε * σ * (T_f^4 - T_i^4)
> 11,260,344,593 = (T_f^4 - 5,554,571,841)
> T_f = 274.8 K
>
> In other words, the temperature of the foil rises to 1.8 degrees
> above room temperature.

All this presumes that we are running the experiment in vacuum,
of course. If running in air, convection currents will carry away
heat and will make accurate weighing impossible.

If running at far higher power levels, we will have to worry about
how the radiation pressure of the emitted IR will affect the weight
measurements. A foil cube resting on a balance pan is not symmetric.
Radiation in the downwards direction will be different than in the
other five directions.

For 5 watts, the emitted radiation will be 0.83 watt in each
direction, implying a radiation pressure of about 2.8e-9 Newtons,
the weight of about 0.28 nanograms.