Path: ...!news.nobody.at!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: Tue, 26 Nov 2024 02:25:07 +0000 Organization: novaBBS Message-ID: <8f11ac85790ebb31563797cd72e917cc@www.novabbs.com> References: <b00a0cb305a96b0e83d493ad2d2e03e8@www.novabbs.com> <98a0a1fbdc93a5fcc108882d99718764@www.novabbs.com> <fd4937f7b180bac934eb677cca8f5c55@www.novabbs.com> <ebcad35958736e6602cf803fddfdb0fd@www.novabbs.com> <141e19a1c6acd54116739058391ca9f8@www.novabbs.com> <a4f98fa5d026bfbf5127fcbc6a585772@www.novabbs.com> <aGN%O.47052$2d0b.43303@fx09.ams4> <45ed9424edce8c13db24c1dbb8752c26@www.novabbs.com> <c8df6716ae871b79524720426a3f229a@www.novabbs.com> <7adfc9e5c6884729def0c6a0097c9f37@www.novabbs.com> <humdnTd1BNWduN36nZ2dnZfqn_idnZ2d@giganews.com> <092fa494db9895ba52cfac350be5e744@www.novabbs.com> <afe961104287110aab310b0cc3b5f8ef@www.novabbs.com> <98654d26cc4f5fd326f071ea7d4317b8@www.novabbs.com> <6292a6508a7a1b7e2f7d13951685410d@www.novabbs.com> <7387e2f099b81abacc7cf1184a11db86@www.novabbs.com> <c25f832f113e2f2e620db970e654daaf@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="4151917"; mail-complaints-to="usenet@i2pn2.org"; posting-account="Ooch2ht+q3xfrepY75FKkEEx2SPWDQTvfft66HacveI"; User-Agent: Rocksolid Light X-Rslight-Posting-User: 504a4e36a1e6a0679da537f565a179f60d7acbd8 X-Rslight-Site: $2y$10$tQUzNXvc4Z7S.cfVdBxjcu7h8nd0uWxG.gkr.NexkbmdjGnJUnAg. X-Spam-Checker-Version: SpamAssassin 4.0.0 Bytes: 6987 Lines: 129 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.