Guest Post by Willis Eschenbach
For those who enjoy mathematical puzzles, I’m putting this one out there for your pleasure.
Suppose we have a 1 metre by 1 metre by 1 metre concrete block floating in outer space. For the purposes of the puzzle, let’s suppose that there is no longwave background radiation at all.
The block is insulated on four sides, as shown in blue below, with the front and back of the block uninsulated. We’ll further suppose that the insulation is made of Unobtanium, which is a perfect insulator, so no heat at all is lost from the four insulated sides.
Next, let’s assume the emissivity “epsilon” of the concrete block is 0.95. [And as a commenter pointed out, let’s assume that the emissivity and absorptivity across the spectrum are both 0.95 everywhere. Yes, I know this isn’t reality, but it’s a thought experiment.] And we’ll say that the thermal conductivity “k” of the concrete is equal to 0.8 watts per metre per kelvin (0.8 W/m K^-1)
Finally, let’s assume that it gets full-time sunshine on the front side at a rate of 1360 watts per square metre (W/m2). Figure 1 shows the experimental setup.
Figure 1. Setup for the thought experiment. The concrete block (gray) is a one-metre cube. The blue insulation prevents any heat from escaping from the four sides. However, the block is free to gain heat by radiation on the front side, and to lose heat by radiation from both the front and the back sides.
Here’s the puzzle. If the concrete block starts at absolute zero, it will slowly warm up until it is at steady-state, neither warming nor cooling.
So the question is: at steady-state, what will be the temperature T_hot of the hot side and the temperature T_cold of the opposite cold side?
w.
REQUESTS: First, let me ask that when you comment, please quote the exact words you’re discussing. It avoids many problems.
Next, as my high school math teacher would say, please show your work.
Finally, please focus on the question and the answers, and leave out all ad hominems, personal comments, and insults, as well as abjuring any discussion of your opponent’s education, age and species of likely progenitors, improbable sexual habits, or overall intelligence.
Don’t have time to work this out, but as the block warms up the energy absorbed will decrease. It is proportional to the difference of the 4th powers of the emitting and receiving surfaces (i.e. sun, block) per S-B. I think we can assume a constant sun surface temperature. I suspect Willis intended his hypothetical block to have a 0.95 absorbtivity as well as a 0.95 emissivity.
Also, Dr, Spencer is correct, I think, that it requires solving two simultaneous equations.
389K and 327K, as long as there no gas inside conducting heat from the front to the back.
Oh it’s a solid block! silly me.
Once the block is in equilibrium, the hot and cold side would be the same temperature.
The perfect insulator guarantees this, so long as the conductivity is greater than zero.
Correction. I thought the cube was insulated on 5 sides.
correction to correction. Got the same answer using 4 sided insulation, by treating block as a part of much larger sphere. In the absence of a greenhouse effect, the material is predicted to be isothermal.
Fred:
Like many, you are confusing static equilibrium with dynamic steady-state conditions. This problem is clearly the second case.
The sun-facing side is (net) absorbing power, and the back side is (net) outputting power. So there will be a temperature difference between the two ends that leads to ongoing conduction.
First, kirchoffs law states that absorbtivity equals emissivity. There fore, use boltzman to calculate the absorbed surface energy.
The use boltzman to calculate the required surface temperature to emit that energy flux, assuming radiation to a 0 degree sink. That will give you Tc of the concrete.
Next apply fourier’s law of conductivity to determine the temperature gradient across the block of concrete (conduction coefficient provided by Willis). That will give you the temperature on the hot side.
This neglects surface back radiation to other cold sinks that may be available on the hot side of the block because that side will emit as well unless there is no “visible” low temperature sink.
Sorry gents, you misunderstand kirchoff’s law, maybe better is you are misapplying it. Google the sort of bible of material properaties, nasa ref pub 1121, dated april1984, “Solar absorption and thermal emittance of some common spacecraft thermal coatings.”
It is not the most up-to-date complete list, but it is a good reference to begin with.
Needless to say, we design many materials to have quite differnt a and e.
@Frenchie77
No, it appears you do not understand that Kirchoff’s Law states that absorption and emittance are equal at the same wavelength.
Pub 1121 defines ‘a’ and ‘e’ at different wavelengths (i.e. black-body temperatures): absorption => 5800K and emittance => 300K. In other words, sunlight is primarilly absorbed at wavelengths less than 1 micron, but emitted from 5 to 35 microns.
Well, do you think that the concrete is going to get as hot as the sun? Hence, why I think you are misapplying it here.
125.46 C Hot
3.27 C Cold
Is it anywhere near a black hole?
Hi Willis,
To solve this problem you need to know both the optical absorbance and the infrared emissivity.
Thanks, Steve. Another commenter pointed this out. I’d just assumed 0.95 for both, but I hadn’t specified that. I’ve modified the head post to clarify that.
w.
You can’t all be right but you CAN all be wrong.
Is this problem something to do with why no one makes concrete frying pans? Otherwise while I am sure it is fascinating as an academic exercise I will stick to my iron, non-Teflon coated, steak pan.
In other news, the Chinese lunar explorer fried an egg on the surface of the moon.
The wanted sunny side up, but it wound up scrambled.
Your cube is completely possible as a model of the Eartha atmosphere without GHG, because the ‘sides’ join each other in a sphere. You don’t need unobtainium.
And in an atmosphere without GHG (or circulation) the temperature is isothermal.
So the hot and cold side must be the same temperature. So my original answer was right, but for the wrong reason.
He could have a front and a back fixed apart with poles at the corners and with no sides at all, that saves obtaining the unobtainium.
is that true on mercury?
No it’s the same as the earths crust, is the surface the same temperature as the core?
Simple students problem from 2nd course.
Why?
Alex February 28, 2020 at 8:47 am
True. Why? Three reasons.
1) For fun, and
2) For folks who are interested in such matters but haven’t taken the “2nd course”, and
3) For Zoe Phin.
Zoe has a blog where in my opinion she’s misleading lots of folks. I couldn’t get through to her, and I hate to see folks on the skeptical side putting out incorrect information. It doesn’t do scientific scepticism any good when that happens.
So I thought that if folks here worked out and discussed the issues in public it might help her see that her theories are incorrect. In particular, she thinks that at steady-state, the energy flux passing through a solid object is different than the flux going out of the far end of the object … me, I say that’s not possible.
Best regards,
w.
“steady-state” is what’s not possible.
Kurt, I don’t understand this. After some period of equilibration, the concrete block will be neither warming nor cooling, i.e. “steady state”. Why do you say that’s not possible?
w.
The ISS and most satellites seem to managed it but apparently it is impossible because Kurt says so.
Hey Willis, there is no helping Zoe. She’s off into flat-earther mentality. The rest of the world is mad and only here little clique of fans” seem to be able “understand” this amazing discovery. This thread has its own merits but don’t imagine it will help Zoe, she’s needs to be special and ain’t going to give it up.
I explained where she mistakenly assumes that the “cold end” of her bar is cooled by emission simply because it is not specified. That is her fundamental mistake. She avoids addressing that. I explained that her idea the you can have “two different fluxes” at a surface violates conservation of energy. She can’t deal with that either.
She is convinced that a flow of 2.5W/m^2 in the bar can lead to 557W/m^2 at the other end. I suggested she patent the technique. If I can have over 1kW of space heating from 5W of electrical input, I’ll pay good money for one those concrete bars.
Greg February 28, 2020 at 10:33 am
Jim, I truly don’t know what Zoe can and can’t learn, and it is my practice to do my best to avoid hindering someone with my thoughts. Heck, in my youth I used to believe that Chairman Mao Tse-Tung was one of the good guys, and I got over that, so anything’s possible …
Next, I suspect that many readers of her blog also read WUWT. So although she may or may not change her mind, those readers might do so. In particular, she claims that a geothermal heat flux on the order of a few tenths of a watt per square metre is keeping the surface at ~ 15°C … and I hate to see people walk away foolish thinking that that is true.
Best regards,
w.
Willis, I also like to start from your generous view of someone. My current view comes from trying patiently to explain to Zoe where she went wrong and her stubborn refusal to engage in a logical argument to arrive at a commonly agreed position.
She seems very young and I expect it will take her about the same length of time to work this one out as it took you with the Mao thing.
It’s a fair point that this may be enlightening to others, that is also why I put some time into her blog trying to straighten it out. I final realised I was peeing into the wind.
Hi Willis!
It would be nice to see a new quest:
1. You have a ball shape size like earth, with no condutivity recieving an energy radiation of 1361w/m2, albedo = 0,3 – but now rotating with a speed of 360degrees/24hours. T-hot and T-cold?
2. Same as 1. but now divided into 3 zones. Zone a= +-30degrees longtude, zone b=+-60degrees logitude – zone a and zone c = +-90 degrees logitude – zone a – zone b
kind regards
SteenR
Your “perfect insulator” is just a perfect mirror.
The solution is straightforward.
The transverse size of the block does not matter.
We have incident Power flux on the front surface.
There is a temperature Tf established at this surface.
The front surface radiates a power flux according to the Stefan-boltzmann.
The back surface equilibrates at temperature Tb.
So, there is power flux inside the block due to the temperature gradient.
The back surface also radiates due to the Stefan-boltzmann.
We have two temperatures to find and two equations: total power fluxes at the both surfaces must be zero.
The system is well defined and solvable.
Will not work out numbers.
Too trivial.
Alex, certainly it may be “too trivial” for you. And I’m sure that for some folks, problems that you find very difficult are “too trivial” for them.
Me, I write for a mythical person that I call the “interested layman” … and as the number of different answers proposed here shows, the problem is not “too trivial” for many folks reading this, whether they are in that category or not.
Finally, I may be wrong myself, which is always worth writing about … but nobody so far has gotten the answer that I get. They are close to my answer, but not exactly the same, which is curious in itself.
Perhaps you could unbend a bit and give us your answer, so we can see how it should be done and whether I’m wrong or not.
Thanks,
w.
Willis,
I agree,.
Not actually solving the problem does not win the cookie.
Anyone can say “I know how to do it but it is too easy to bother with.”
Personally, I do not know the equations for solving this, and there is no point in me looking them up because other people here know much more about this than me.
But I am struck by how many people here claim to know exactly how to do this exercise and yet disagree with each other.
I do have a question…maybe I overlooked something.
Do you not need to know the specific heat of concrete to be able to say how hot it gets and thus how much it radiates from the sunny side of the block?
Or maybe that cancels out by the time thermal equilibrium is reached?
But what if it is rotating?
Nicholas
You said, “But I am struck by how many people here claim to know exactly how to do this exercise and yet disagree with each other.” Yes, I’m also struck by the fact that there is often disagreement over what seems to be basic science or engineering, between people who claim or act as though they experts in the field. I’d be happier if there was more “consensus” on basic physics calculations.
Alex February 28, 2020 at 8:59 am
The system is well defined and solvable.
There’s a very funny story about an Engineer, a Physicist, a Mathematician and a midnight fire in the bathroom wastebasket.
“A solution exists.”
w.
Alex
Yes, if the Unobtainium has an absorptivity of zero, and hence an emissivity of zero, [at all wavelengths] then you have your “perfect mirror” with a reflectivity of 100%. Your can get that with a material with a complex refractive index with a very high real index of refraction (n>>10) and an imaginary (as in sq rt of -1) extinction coefficient approaching infinity. With nothing in empty space to conduct to, all the internal energy has to be confined to the channel between the front and back, and can therefore only exit by radiation.
60.2ºC for both
Divide incoming by 2
I would post the correct answer and the associated math behind the numbers.. but I don’t want to appear superior, so I won’t. Good Luck!
1what would happen if you replaced the earths atmosphere with a perfectly transparent material 1 meter thick?
Sunlight would warm the surface of the earth, and the earth would warm the transparent material. The question now becomes, is there a difference in temperature between the inside and outside of the transparent material.
If there iis a difference how does one explain the prediction of an isothermal atmosphere. Inn the absence of GHG?
Using Stefan-Boltzmann T-hot = 398.8K
If the thermal conductivity “k” of the concrete is equal to 0.8 watts per metre per kelvin (0.8 W/m K^-1) then the T-cold side must be finally emitting 0.8×398.6 W/m2 which is 318.9W/m2.
Put that 318.9W/m2 back into the S-B equation with the same emissivity of concrete and T-cold = 277.4K
leitmotif
You appear to be touching on an essential part of the problem that is not well defined. If the rate of conduction is less than the rate of arrival, then the surface facing the source of energy will continue to heat, irregardless of the emissivity of the face radiating. This strikes me as being similar to the classic differential calculus mixing problems where one needs to know most of the variables. Conductivity seems to be a key parameter here. If the thermal conductivity is too small, it becomes a ‘bottleneck’ and the front surface will continue to heat, while the back surface reaches an equilibrium determined by the conductivity and emitting flux. I’m not sure that the “Thought Problem” is adequately posed to calculate the result.
Assume a Tcold for the unlighted side.
Calculate how much heat is radiated to space from Tcold. 0.95*5.67e-8*Tc^4
Call this Qrcold
This amount of heat had to be conducted through the concrete from Thot
So by heat conduction from hot to cold (Thot-Tcold)*0.8 is equal to Qrcold….
So your calculated Thot is Tcold+(Qrcold/.8)
Calculate how much heat is radiated to space from Thot …..0.95*5.67e-8*Th^4
Call this Qrhot
Qtotal =Qrhot+Qrcold
Select new Tcold until Qtotal is 1360
Answer…I got Tcold=222.7 Kelvin, Thot=388.3 Kelvin
This assumes the “albedo” of the concrete is 0 in keeping with Willis’s definition of the problem. Actual albedo of concrete is about 0.5, which makes about 15 degrees difference.
This problem is equivalent to the infinite sheet problem, where the sheet is 1 m thick (see Carslaw and Jaeger, or Crank). Since it’s steady state, the solution for the temperature profile has a zero time derivative. That solution will have linear dependence on T with respect to depth within the sheet. So the solution will be one that satisfies the boundary conditions. On the hot side, you have a flux in (from the sun), a flux out (S-B thermal radiation) and a flux through to the cold side. These have to sum to zero (taking direction into account). The flux through will be determined by the temperature difference between hot and cold and the thermal conductivity. On the cold side, there is flux from the hot side and flux out to space (I’m neglecting leftover radiation from the big bang, but if you like, you could factor that in). The 2 fluxes on the cold have to sum to zero. I haven’t done the calculation, but it should be straightforward.
Of course, the above assumes that the thermal conductivity is temperature independent (often not true) and it assumes that the block is totally opaque (also sometimes not true at some wavelengths), so there is the possibility of reality biting one in a “fundamental” way.
Plugging in the numbers and assuming constancy of heat conduction, no big bang, epsilon emit = epsilon absorb) etc., I get T(hot) = 383.3 K (110.1C) and T(cold) = 221.4 K (-51.7C).
Thanks, Michi. You’re the first person to get the exact answer I got (see here), so at least we’re either both right or both wrong …
w.
That makes a bunch of us including Roy if you take Glen’s comment as being correct.
As a sanity check, I set the emission/absorption at 1. The hot side was 383.7 K and the cold side was 219.4.
Indeed, Bob, many folks have gotten the correct answer.
Regards,
w.
Yes Willis. As I write this there are 453 comments on your story. The number of comments that show a more-than-rudimentary understanding of thermodynamics is truly impressive. Is there an alarmist blog that could muster such a performance?
MichiCanuck
You assumed “constancy of heat conduction.” Is that reasonable? I think that the rate of conduction would be driven by the temperature differential between the two faces. It further assumes that the material is capable of any rate of conduction.
It’s unlikely to be realistic, but it’s a necessary assumption given the problem as stated. For many materials, some of the heat conduction is actually via radiation across internal voids or because the material isn’t perfectly opaque at important wavelengths. I remember hearing an interesting talk a long time ago about how important blocking IR radiation was to making effective fabrics for cold weather survival suits. Once blackbody radiation enters the picture, non-linearity rears its ugly head. The problem’s still soluble, but the temperature profile across the block/sheet would no longer be linear. You’d also have to use numerical methods to obtain a solution to even the steady state equation.
One dramatic effect that always impresses me about a temperature dependent parameter is electrical conductivity in metals. If you’ve ever seen an RF coil used to heat up a metal bar, you’ll have seen that it heats very slowly at first. That’s because at low temperature, its conductivity is high and Ohmic heating isn’t very effective. However, as temperature climbs, resistance in the metal increases and heating speeds up. Eventually, it only reaches steady state when heat losses equal the absorbed RF energy. In air, conduction and convection can be important. In a vacuum, it’s good old black body radiation that has to do the cooling, which isn’t all that effective until things get quite toasty.
h=temp hot surface
c=temp cold
s=stefan’s constant
(1)Incident radiation=.95×1360 (some reflected)
(2)Emitted rad hot=.95sh^4
(3)Emitted rad cold=.95sc^4
(4)Conducted heat=k(h-c)/1 {temp gradient 1m}
Balance at 2 surfaces
(1)=(2)+(4) at hot
(3)=(4) at cold
Solve numerically: 2 equations in 2 variables h and c – ignore complex and negative temps 🙂
I was going to post the same thing but you got in first. Exactly right and easier to solve numerically than exactly. Guess C, use equn (3)=(4) to calculate H and then use H to to calculate (1)-(2)-(4). Adjust C until (1)-(2)-(4) equals zero. A couple of minutes in excell
Nice exercise Willis. I get Thot = 383.3 K, Tcold = 221.4 K
Work in Excel shown here in a screenshot with formulas written out. As Roy Spencer notes simulaneous equations are used.
Qh and Qc are watts out for the hot face and cold face.
Procedure: Enter trial Th, calculate Qh out via S-B, calculate initial Qc out by conservation of energy, calculate initial Tc via S-B, calculate resulting Qc out by conduction and iterate trial values of Th until the two Qc values converge.
Instead of solving it algebraically, I programmed the heat budget equations for the front and back surfaces in Excel. I whipped it up quickly, but I get 388.52K for the front surface and 222.74K for the back surface. I had to specify a heat capacity (I used granite), but the final answer does not depend on that, the heat capacity only affects the time it takes for the front and back surfaces to come into thermal equilibrium (with granite it took over 30 days to approach equilibrium).
Thanks, Dr. Roy. Kudos, by the way, to all who solved it iteratively in Excel. I use the “Solver” function in Excel for lots of these kinds of problems, it works like magic.
w.
Here’s an R solution if Word Press doesn’t chew up the code:
epsilon = 0.95
sigma = 5.670374419e-8
R = 1360
k = 0.8
err = function(T.hot){
T.cold.1 = (R / epsilon / sigma – T.hot ^ 4) ^ (1/4)
T.cold.2 = T.hot + (epsilon * sigma * T.hot ^ 4 – R) / k
(T.cold.1 – T.cold.2) ^ 2
}
T.hot = nlm(err, 389)$estimate
T.cold = T.hot + (epsilon * sigma * T.hot ^ 4 – R) / k
Joe, always good to hear from you. Thanks for the sweet R code, always more for me to learn.
w.
Roy,
Do you think Boltzmann and Planck were dummies?
Why didn’t they get this solution?
I would accept either if you use flat sheet laminar flow and just to be clear because Zoe seems messed up on this there is a gradient thru the block.
Dr. Spencer,
The temperatures you posted indicate no correction of incoming heat duty from 1360 watts to 1292 watts with 0.95 emittance. If you correct to 1292 watts, you should get the 383.3 K and 221.4 K temperatures that David posted above.
I was going by Willis’ original post, which suggested the heated side was absorbing 1360 W/m2.
I made hot side 385.5K and cold side 222K
At the hot side, at 389 K, there is 1227 W/m2 of radiation to space and 133 W/m2 of conduction to the cold side of the cube, at 223 K, from which there is 133 W/m2 of radiation to space.
Kind Regatds
Assuming there is no matter past the cold end, the final temperature of both sides will be equal.
Ah, the old “radiation needs to know where it will land before it leaves” hypothesis. You obviously are ready with an explanation of how radiation emitted 13 billion year ago “knew” it was going to find Hubble space telescope before it left home.
Please lets us know how that works.
‘the old “radiation needs to know where it will land before it leaves” hypothesis’
Yes, very one Boltzmann and Planck used to derive their laws.
Tell me the new junk science that can’t derive them.
http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html
Please tell me which scientists originated what you claim, and why did you believe him?
So how does light from 13bn years ago “know” it is going to “resonate” with HST before leaving home? You never answer a question. You always divert with a different question.
Greg,
The space between the EM tethered objects got stretched, if you believe in the big bang.
There’s no EM rope trying to find an end. All was tied up in the beginning, then got stretched.
Zoe is right.
It is a popular misconception that light leaves A at time zero and arrives at B at times 1. Even calling c the “speed” of light is an inaccurate colloquialism. Light travels at the speed of – well – light, and thus experiences no time.
Light is outside of time. An EM photon does not experience the passing of time.
Take Betelgeuse, 700 light years away. Did light leave Betelgeuse 700 years ago? No. Light took no time at all to get from there to here. But time over at Betelgeuse is 700 years different from here.
It’s hard to get your head around, but this is correct cosmology and a full understanding of what Einstein illucidated. Time and space are one. The “light year” is actually a very good term since the distance between earth and Betelgeuse can just as well be described as 700 years as a number of meters.
Sorry but it’s just WUWT old boy science to talk about light taking time to travel a distance. Light experiences no time so it takes no time for get anywhere. As Matt O’Dowd of PBS SpaceTime YouTube channel explains very nicely, the speed of light is not even about light. It’s the speed of causality.
https://youtu.be/msVuCEs8Ydo
Phil Salmon
So in Willis’ example if the universe the sun and block are in is curved in such a way that light from the sun is on one side 93,000,000 miles away and on the other side 13 billion light years away in Zoe’s system when we place the block in its position when does the far side of the cube start to cool or radiate, instantaneously or after 13 Billion years?
Phil Salmon wrote:
“Light experiences no time so it takes no time for get anywhere.”
That’s so wrong that “wrong” is the wrong word to describe how wrong it is. LOL
First, we’ll explore what a photon “is”.
A photon is the interaction of the electronic and magnetic fields, oscillating in quadrature, geometrically transformed into a spiral (because a sinusoid is a circular function and a circular function spread axially over space-time is a spiral). When we see electromagnetic energy as a sinusoid on our oscilloscopes, we’re literally looking at a shadow of reality, because we don’t have 3-D oscilloscopes, and the oscilloscope only looks at the electronic field, not the magnetic field.
The photon, being massless and circularly polarized when considered singularly, doesn’t carry its energy in its linear momentum, it carries it in its angular momentum.
http://staff.washington.edu/bradleyb/spiralsynth/fig3.1.gif
This is why, when traveling through transparent mediums of differing refractive indexes, the photon energy does not change, while the apparent photon speed does. If its energy were carried in its linear momentum, photon energy would necessarily change as it transited different mediums.
E^2 = p^2 c^2 + m^2 c^4
pc is the magnitude of the momentum vector. Since c is fixed in vacuum, p must change for the photon’s energy to change.
p=ħk
where k is the wave vector (where the wave number k = |k| = 2π/λ), ω = 2πν is the angular frequency, and ħ = h/2π is the reduced Planck constant.
The energy and momentum of a photon depend only on its frequency (ν) or inversely, its wavelength (λ):
E= ħν = hc/λ
This is why Planck’s constant has units of angular momentum (J-s), and the reduced Planck constant represents the quantum of angular momentum.
———-
The photon itself (if it is traveling at c, which it may not necessarily do… it is dependent upon the medium it is transiting. c is for vacuum.) _experiences_ no time, which is why it is persistent and doesn’t entropy into the background zero-point quantum vacuum field (at least, until after it impinges upon invariant-mass matter or experiences a sufficient gravitational field (because light doesn’t travel in a straight line, it follows the path of least time, and gravity is a manifestation of the warpage of space-time, which is why light ‘bends around’ large celestial objects, because invariant-mass matter expands the surrounding space-time and slows down time.).
But just because the photon _experiences_ no time doesn’t mean no time has _elapsed_.
Only for entities at c would the photon seem to not experience any time… for objects at anything less than c, that photon does indeed take time to traverse space. It’s called the theory of *Relativity* for a reason.
If “Light… takes no time for get anywhere.”, then causality is irreparably broken.
Light travels at 299792458 m/s in vacuum… it transits 299792458 meters for every second in our frame of reference.
Now we have to describe what we mean by ‘vacuum’…
hard vacuum (no invariant-mass matter, quantum vacuum zero point field exists)
perfect vacuum (no invariant-mass matter, no QVZP field)
In fact, it is the quantum vacuum zero point field which is the medium which limits the speed of light. In a hypothetical perfect vacuum (which cannot exist, because the metastability of invariant-mass matter is dependent upon the existence of the QVZP field, thus we could never construct a machine capable of a perfect vacuum without the atoms / molecules of that machine’s vacuum-facing components undergoing beta capture and thus transmutation [1][2][3][4]), there would be no speed limit except for the energy field the photon itself lent to the perfect vacuum in transiting that vacuum.
[1] https://journals.aps.org/prd/abstract/10.1103/PhysRevD.11.790
[2] https://web.archive.org/web/20190713220130/https://arxiv.org/ftp/quant-ph/papers/0106/0106097.pdf
[3] https://web.archive.org/web/20190713225420/https://www.researchgate.net/publication/13330878_Ground_state_of_hydrogen_as_a_zero-point-fluctuation-determined_state
“We show here that, within the stochastic electrodynamic formulation and at the level of Bohr theory, the ground state of the hydrogen atom can be precisely defined as resulting from a dynamic equilibrium between radiation emitted due to acceleration of the electron in its ground-state orbit and radiation absorbed from zero-point fluctuations of the background vacuum electromagnetic field, thereby resolving the issue of radiative collapse of the Bohr atom.”
[4] https://web.archive.org/web/20180719194558/https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20150006842.pdf
“The energy level of the electron is a function of its potential energy and kinetic energy. Does this mean that the energy of the quantum vacuum integral needs to be added to the treatment of the captured electron as another potential function, or is the energy of the quantum vacuum somehow responsible for establishing the energy level of the ‘orbiting’ electron? The only view to take that adheres to the observations would be the latter perspective, as the former perspective would make predictions that do not agree with observation.”
LOL this I have to hear.
So how does the emission leaving the cold end know ahead of time there is matter out there somewhere?
Does it send out some special faster than light communication asking for all matter to respond?
Or perhaps it just omipotent know?
The ISS is up there right now and I am seriously wondering why it isn’t all the same temperature given your statement.
Zoe: It might help to use the form of S-B:
Qrad = e A sigma (T1^4-T2^4)
(In Willis’ problem A =1 , e = 0.95 , Sigma = 5.67 E-8)
Where T1 is the warmer T2 is the colder surface temperature.
You will find that T1= 398.59 K and T2 = 0, you get 1360 watts/m2
With T1 = 398.59 and T2 = 388.5 you get 132.5 watts/m^2, the heat Flux to the hot side of the cube.
With T1 = 222.7 and T2 = 4 you get 132.5 watts/m^2, the heat Flux from the cold side of the cube to space.
And 0.8 * (388.5 – 222.7) = 132.5 watts/m^2, the heat Flux from the warm side to the cold side.
So the temperatures are all reasonable and the 3 heat flows are equal as required for a system in thermal equilibrium. I think this exercise has been a good demonstration that being an excellent computer programmer is not enough to get correct answers. You must also understand the problem and apply the correct solution method.
Rick,
You missed the part where 90% of energy from the source was rejected, while pretending absorptivity is 0.95.
You can’t solve HSR = CHF = CSR
So you solve the cold end, CSR = CHF, and ignore HSR = CHF or you reject 90% of incoming HSR to get your equality, but then your Thot is fake.
Zoe: No, you missed the basics of thermodynamics. Everything with a temperature above absolute zero (which is literally ‘everything’) will radiate energy. If you have two objects at different temperatures, there will be a NET heat flux from the warmer to the colder. The 90% of the heat you’re talking about is not rejected, it is absorbed raising the surface temperature of the block from 0 to 388 K which results in 90% 0f the heat absorbed being radiated back out to space. The net rate of heat flow will be proportional to the difference in the 4th powers of each object’s absolute temperature. Thus, a 1000 C radiant source surface can heat a 975 C surface, but not very much. But both surfaces would radiate a lot of heat to a cold object. They don’t stop radiating energy when you position them to face each other. However, the same 1000 C source will heat a 20 C surface by 100’s of degrees. This is really basic thermodynamics stuff. Take a course – there are lots of them.
Rick, you should just ignore Zoe. She/he is just prankster blogger.
Correct DMac, I thought she was serious and just being dense but yep I worked that out as well.
Unfortunately, I fear that she truly and passionately believes what she’s saying. That’s the scary part. Go to her website and read some of the many pages. It’s far too large, complex, detailed, and too much work to be just a prank.
And sadly, people believe her. I was directed to her site by a very enthusiastic friend who claimed that she was revealing amazing stuff …
In large part, that’s why I wrote this—in the hopes that fewer people would be fooled. In that, I think I’ve been fairly successful because she’s revealed her contradictions and lack of understanding in very clear terms.
Regards,
w.