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.
Please show the solution when the hot and cool sides of the block are just thin layers and have vacuum in between them, and do not forget to include back radiation from the cool layer.
Also show the solution when a VIS-transparent & IR-absorptive glass layer is put in front of the warm surface with a small vacuum gap in between, again including the effect of back radiation on the sun-ward layer from the new glass layer. In this case, it is a concrete block again.
Please show the solution.
Hey Johnny,
I loved that stunt where you rode down an escalator in a shopping cart full of bowling balls and glass bottles!
In the final stationary state T2=T1=T.
Then, the irradiated surface of size A, S1, receives G = alpha * 1360*A Watt and loses by radiative loss L1=2*pi*eps*sigma*A*T^4 Watt and – by conductive loss to the interior – the same amount, which in turn is radiated from surface S2, L2=L1. The thermal conductivity (together with the specific heat) just determines how long it takes until the stationary state is reached.
Hence, with alpha = eps (= 0.95 T) = (alpha *G / (4*pi*eps*sigma))^0.25 = 212 K.
R. F.
People have a hard time understanding what conduction is.
The conduction formula = KA(Th-Tc)/L
If A is in the (x,y) dimension
L is in the (z) dimension
K is in units W/(m*K), and m is also in the (z) dimension
In the conduction formula: K/L ~ 1/m /m = 1/m^2
The conductive m^2 is (z) dimension divided by (z) dimension = still (z) dimension
Conductive m^2 is different from Radiative m^2.
Do you not see that A is not in the same dimensions as K/L ???
Willis et al believe they can compare flux through the (x,y) dimension to the (z) dimension.
This is silly.
Boltzmann and Planck found that
CSR = εσ(Th-CHF*L/K)⁴
And CHF is free to approach ZERO. When it does,
CSR = εσ(Th)⁴
Thermal equilibrium! SB Law!
https://en.wikipedia.org/wiki/Black_body
“A black body in thermal equilibrium (that is, at a constant temperature) emits electromagnetic radiation called black-body radiation. The radiation is emitted according to Planck’s law, meaning that it has a spectrum that is determined by the temperature alone (see figure at right), not by the body’s shape or composition.”
But hey why do real physics, when self-deceptive rhetoric and ideological mathematics will do?
Willis never specified that the concrete block was equivalent to a black body in equilibrium. In fact, by stating that there was conduction through the block and an exposed, non-insulated rear surface away from the radiation input side, he gave fair warning that IN NO WAY could the concrete block be modeled as a blackbody!
Gordon,
He models a 0.95 * Blackbody. Basic laws don’t change. Sheesh, if wikipedia doesn’t mention a 0.95 * Blackbody. I must be wrong, right? How pedantic.
You posted: “He models a 0.95 * Blackbody.” No, he does NOT.
It is you who is attempting to argue a blackbody has relevance to Willis’ postulated problem. It simply does not.
And there is no purpose of me trying further to gently convince you of that. Good luck.
A blackbody will emit: sT^4
A greybody will emit: esT^4
where e = [0..1]
Are there any others laws evoked?
Are you arguing that he doesn’t use a grey body therefore the emission law is completely different, say sT^3-4T^12+PI or sT^2-4ln(T), or whatever?
What are you saying?
where does Boltzmann and Planck refer to CHF ? That is just you own BS mixed up derivation from assuming YOUR block was cooled by SB radiation when there is nothing to say that is the case.
You are still unable to address what I’ve pointed out at least six times now. You made a false assumption.
Do actual blocks of concrete in space do real physics Zoe…or do they whine about units and pretend they know what they are talking about?
It’s the same A. The magnitude of the flux is inversely proportional to the length, since the longer the bar, the greater the thermal resistance.
Wrong again. m is the dimension of A divided by the dimension of z. L^2/L in dimensional analysis.
The more you come out with the more you show your ignorance and basic lack of understanding of any of this. But don’t let me stop you, this is fun.
Greg,
“Wrong again. m is the dimension of A divided by the dimension of z. L^2/L in dimensional analysis.”
Why would you divide a face in the (x,y) by z. What does that yield? And you forgot K.
You’re forgetting that the flux equivalent is:
Q/(A*dt) = K(Th – Tc) / L
Now both sides are W per m^2
But the two m^2 are in completely different dimensions.
“Why would you divide a face in the (x,y) by z.”
You are not dividing a face by z. You are dividing the power by area to get the flux and multiplying by the length, to derive a property of the material which is not a function of an object’s shape or size.
You could say Wm/K/m^2 which is what it really is but the m on top and bottom cancel, so it becomes W/K/m. The ‘per metre’ is not a linear length it is the result of dividing an area by a length. Dimensionally it is L-1 i the units.
They’re not, since K ( usually k ) includes “per unit area” in its full form: Wm/K/m^2 . That is why you divide by length and multiply by temperature. What is left is power / area.
“You are dividing the power by area to get the flux and multiplying by the length, to derive a property of the material which is not a function of an object’s shape or size.”
(Q/(A*t)*L) = K (Th-Tc)
Now you LHS is meaningless as it divides z-dimension by (x,y).
So you’re right (LOL), you derive a meaningless nonsensible property.
“They’re not, since K ( usually k ) includes “per unit area” in its full form: Wm/K/m^2 . ”
No it doesn’t. k or K is in W/(m*K).
“In the International System of Units (SI), thermal conductivity is measured in watts per meter-kelvin (W/(m⋅K)).” — Wikipedia and every place else.
You have no idea what you’re talking about.
I’m not saying anyone every writes it that way ! I’m trying to explain the derivation of the unit and what it means, to explain to you that it is division by a length at all.
Energy flux of 1361W/sq.m through hot face. Energy is conserved so, at steady state, 1361W/sq.m must exit the cold side. That requires a cold side temperature of 399K ([1361/.95/5.67E-8]^.25). The hot side temperature must be at 2100K (1361/0.8/1*1+399) to achieve the steady state heat flow through the block of concrete.
The hot side also radiates, no?
Yes of course it radiates. But the problem shows 1361W/sq.m being absorbed by the hot face. Radiation is, by definition, a result of THE electro-magnetic field. It is singular and unidirectional at any point in space and time. A proper solution requires solving Maxwell’s equations for the electrical-magnetic field that the object exists in. That requires details on the heat source and more detail on the orientation of the concrete block. Approximate solutions can be achieved by applying view factors and using radiant heat transfer equations; again requires details on the heat source and the orientation of the block relative to the heat source..
Nonsense and gibberish.
With all due respect.
To properly account for conservation of energy (actually, in this case, power) you need to account for radiation off the front surface of the block at its equilibrium temperature.
Come on, Nicholas and Gordon! Everyone knows that a black asphalt parking lot in the tropical noon sun, receiving 1360 W/m2 of solar input doesn’t radiate away any power at all…
(Need I say /sarc?)
If the asphalt is receiving 1360W/sq.m then that is the end of the story. The radiation energy is singular. It occurs by virtue of THE electro magnetic field – singular. There is only ONE EM field no plethora of fields that are adding or subtracting.
Are you seriously arguing that a black parking lot in the hot sun does not radiate away energy while it is in the sun??? You don’t need any sophisticated sensors to know that it does — you can feel it yourself!
Of course you can add the effects of multiple sources together. If I shine two lights on you, the EM field that reaches you will be stronger than if I only shine one light on you.
This is basic, basic stuff!
I think it was sarc
Yes, and I guess all those infrared cameras actual image unicorn dust instead of RADIATED thermal energy, particularly in the absence of any incident radiation.
/sarc
The limiting cases for this problem are:
When the thermal resistance of the concrete is zero (high conductivity concrete) then both faces have the same temperature which energy balance reveals to be 330.9K.
When the thermal resistance is infinite then
and 
I will choose an intermediate value for the thermal conductivity of concrete
which is well within the various values for concrete.
Now I have to iterate between two equations, which are energy balance at the two faces:
1.
and
2.
I started with
and iterated between the two equations six times until I was bored….
Oops….
1.
and
2.
I started with
and iterated between the two equations six times until I was bored….
For some reason I cannot correct a typo in eq. 2….It is missing a factor of
in the second term.
For some reason this software won’t let me correct the error in equation 2.
1.
and
2.
I started with
and iterated between the two equations six times until I was bored….
Well, I read through a bunch of comments and did not see one person follow the instructions and give an answer as one would on a test and also show work or, where I went to school anyhow, you get a zero in this type of class.
Some people said things that were probably true but unhelpful.
Some told how to set it up but gave not one detail that would instruct anyone who does not already know how to do it.
And some said stuff that was either mystifyingly abstruse, perplexingly opaque and also likely wrong, and just plain made no sense.
Others pretended that it is impossible to have such a situation, which in the case of a perfect insulator on the sides is likely true, but there are very good insulators and anyone who knows this stuff knows how to calculate a limit or so I would think, so it should be easy to say what would happen as the insulator approached being perfect from a state of being pretty good.
We can put block of concrete with the described characteristics in orbit and measure it…using some ultralow density stuff like space shuttle tiles or aerogel or some crap like that.
It seems that is the simplest way, despite the assembled wit and wisdom of the commenters/comments I was able to read, before becoming bored to tears, anyone not already an eggspurt on this topic could arrive at an answer, given the condition that putting some concrete on the very first Virgin Galactic flight would be easier than getting the required edumacation in thermodynamics.
It is not impossible to show your work here.
Write it down, take a picture, post it on photobucket or twitter and post a link.
I have done stuff like that a thousand times.
Nicholas:
Several of us posted step-by-step analyses of the problem just as we would for a technical class, all reaching the same answer. I have taught technical subjects at the university level, and I would give full credit to the others whose analyses matched mine.
Thank you Bo…Yeah, I wrote this after reading a few dozen of so at the very beginning.
Maybe less.
I did see that several people has already done the exercise using what appears to me to be the proper logic.
I often do not read all through a thread before commenting.
Especially when the site is acting up.
(For a while there, maybe a week or more, comments were posting immediately and the page would return to the part of the thread one posted their comment in.
Now suddenly comments are posting after a delay off variable length again, and after clicking the blue button the page returns to the top and stays there.
Had to find one’s way back sometimes…)
Solving word problems, as they used to call them in grade school or high school math or algebra classes, is very difficult for a lot of people, even in advanced University science and math classes.
Willis,
You really hooked me today. I found an online solver Wolfram Alpha. Using t1 for the hot face deg K and t2 for the cold face deg K, and expressing the two simultaneous equations to solve using only the two temperatures, here is the result as a link.
https://www.wolframalpha.com/input/?i=solve%7Bt1%5E4*0.95*5.67E-8%2Bt2%5E4*0.95*5.67E-8%3D%3D1360*0.95%2C%28t1-t2%29*0.8%3D%3Dt2%5E4*0.95*5.67E-8%7D
Just like the Mathematica solver you used, it gives the desired answer at the end of the list.
I hope this works as a link, as I suppose it should.
Nice, Dave, the link worked perfectly. I have standalone Mathematic on my computer, same answer, but it doesn’t give the cool graphic showing the two real solutions.
w.
wow, I did not realise that T^4+T^4 was a closed loop though I probably should have recognised the similarity to x^2+y^2=r^2 , a circle. With the forth powers I suppose that’s like squaring the circle and it ends up making a … squared circle.
When I did my graphical solution below, just plotted the intersections and did not scan out to a scale where I would have seen the full form. Nice.
Th = temperature of hot side (uniform across the face)
Tc = temperature of cold side (uniform across the face)
Qh = heat radiated by hot side = ε σ Th^4
Qc = heat radiated by cold side = ε σ Tc^4
Qs = heat from sun = 1370
Qi = heat input to cube = ε Qs
Qt = heat conducted thru the block = k(Th-Tc)/t
t = thickness (from hot to cold side)
(All values for Q are W/m^2)
(the front side and the back side will be at constant temperatures across each surface)
There are two constraints relating these two variables.
1) the heat conducted through the block must equal the heat radiated by the cold side
Qt = Qc
k(Th-Tc)/t = ε σ Tc^4
2) the total heat in = total heat out
Qi = Qh+Qc
εQs = ε σ Th^4 +ε σ Tc^4
Note that emissivity drops out here, so
Qs = σ Th^4 +σ Tc^4
Two equations, two unknowns — so just solve.
OK, OK, that is much easier said than done! Solving either equation for Th and substituting back into the other equation gives nasty combinations of Tc, Tc^4 and/or Tc^0.25. I am not sure there is an analytic solution, nad right now I don’t have time to do a numerical approximation. >/i>
Partial credit for anyone who first defines their terms.
No more than partial credit for anyone who does not.
Both are regardless of correct answer.
(Getting the logic part correct is 75% of the possible points.)
I think the “right answers” are
383.6 °K for the lit side, and
183.7 °K for the dark side. Why?
Assuming Stefan-Boltzmann constant of 5.67×10⁻⁸, then the lit side will be reflecting 5% of 1360, leaving 1292 W absorbed. It will come to asymptotic balance emitting 95% of that, or 1227 W. This corresponds to 383.6°K or 110.4°C.
Likewise, the net transmitted thru also asymptotically approaches (absorbed – reemitted), or 64.6 W. Using σ (SB constant), this corresponds to an emission temperature of 183.7°K or –91.8°C.
The reason one doesn’t have to figure the thermodynamic conduction part is because of the unobtanium wrap. Over long enough time, thermal balance is in steady state.
⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅
GoatGuy, your solution doesn’t check out. You have the hot side radiating 941 W/m2 and the cold side radiating 61 W/m2. This only adds up to 1,002 W/m2, but the block is absorbing 1,292 W/m2.
No bueno.
Sorry,
w.
Graphical solution:
gnuplot> Th2(x)=(1360/sigma-x**4)**0.25;
gnuplot> Th1(x)=(sigma*epsilon*x**4)/k+x;
gnuplot> plot Th1(x) tit “(sigma*epsilon*x**4)/k+x”,Th2(x) tit “(1360/sigma-x**4)**0.25”
take intersection where both values are +ve : 221.4 , 383.34
There are 3 scenarios under which Tcold will not equal Thot, i.e steady-state gradient condition.
1) k increases thru the block
2) A increases thru the block
3) There’s radiative/conductive/convective leakage through the 4 walls.
Willis defines exclusion of all these things, and therefore his solution is invalid. He never researched what would create a gradient.
Of course his other comments reveal there is no gradient, only a small sliver at 383K, while the rest is 221K.
It’s fantastic pseudophysics. I’m impressed how he keeps this game going without any reference to literature or experiments.
Plug in given Th into CSR = CHF assumption to get Tc, but ignore HSR = CHF, then 90% of energy from HSR goes back to the source (making it hotter?) while you pretend aborptivity was still 0.95 and albedo is 0.05. LOL
You need to read all of Willis’ comments to understand this.
You have it exactly backwards:
“This” (meaning whatever what you wrote is purported to mean) can never be understood, at least not in this century.
Perhaps in several millennia your particular brand of delusional thinking will be understood.
But it will take at least one or two rounds of Twenty Questions.
Zoe says “There are 3 scenarios under which Tcold will not equal Thot”
Actually, there is 1 scenario under which Tcold WILL equal Thot” — the thermal conductivity, k, is infinite.
Or unobtainium is covering 5 sides
I forgot a 4th: mass variation.
To Zoe,
I posted fhis comment on your site, please please stop trying to re-invent physics and math.
READ “Transport Phenomena” by Bird Stewart and Lightfoot
If you don’t like second order differential equations don’t bother.
It’s rude to send people on wild goose chases. If that book had the solution to what we’re discussing you’d be more specific with the section, pages, etc.
Haha!
The rudest person all day gets her knickers in a twist over being asked to do some research!
touche`
Nicholas, my expensive lingerie doesn’t get in a twist.
I’m not going to read a giant textbook just because someone says so. If they find something useful and applicable I will gladly read that chapter.
You not feeling like doing something does not make it rude.
You could have just asked for a summary or a page number.
What is it called when one takes the less than polite alternative?
Nicholas,
You know nothing of our previous conversation, but you butt in with your opinion on etiquette.
Do you have any self-awareness of your own rudeness?
Yes, plenty.
I am rude to people who deserve it.
“Treat me nice, I’ll treat you better.
Treat me bad, I’ll treat you worse.”
I extend “me” to include anyone for whom I have respect from prior interactions.
If you treat them badly, I’ll treat you worse.
And BTW…you know nothing of what I do or do not know regarding previous interactions.
And that is irrelevant in any case to calling out hypocrisy.
This is a science site.
Getting things correct is a virtue unto itself, so it is not a rudeness to say that someone who is wrong, is wrong.
It is simply the truth.
* To quote the great D.B. Stealey, wherever he is.
I wish he was here right now.
“Of course his other comments reveal there is no gradient, only a small sliver at 383K, while the rest is 221K.”
Zoe, how do you think the pressure in a hose varies along its length while water is flowing. Is it all at tap pressure with just a “slither” atmospheric pressure just before the outlet, or is there a pressure drop ( a gradient ) along the hose?
Willis said that which flows thru the hose must leave the hose. But when it comes to entering the hose (HSR), for that we can just reject 90%. But it already entered!
As you do EVERY SINGLE TIME you avoid answering the question and reply with some diversion.
All the solutions here are based on flux in = flux conducted = flux out. What is your 90% “we” can reject ?
Greg, pay attention,
The flux in is spread out to both sides to get flux out.
Flux in = HSR Out + CSR Out
Does that happended with water?
First Wilis stipulates that all water entering must leave. Then it turns out that only a small fraction of it leaves, and the rest can’t go through and spills out where it should come in.
He has 1000 “W”+ of water going through but only ~130 “W” (I forget the numbers) leaving.
In fact he’s got more water coming out the way it should be going in then goes through it.
If you can’t make sense of this, it’s because Willis’ analogy is crap and he contradicts his core assumption, but I guess it’s all OK because radiation is “different” now then his water analogy.
Zoe, the water hose was an ANALOGY, and like all analogies it is not perfect. My main point was simple.
The water flowing THROUGH the body of the hose must equal the water flowing OUT of the far end of the hose. That was the only reason I brought it up. So forget anklebiting the damn analogy and let’s get back to the actual question
You keep claiming that the energy flowing THROUGH the block is NOT equal to the energy flowing OUT of the far side of the block. Nobody here, including myself, has a clue how that would be physically possible.
If more energy is flowing THROUGH the block than is radiating OUT of the far side of the block, where is the extra energy going? And if more is radiating OUT of the far side than is flowing THROUGH the block, then where is the extra energy coming from?
You make the same bizarre claim on your blog with respect to geothermal heat. You claim that somehow a geothermal energy flux of a few tenths of a W/m2 is sufficient to heat the earth surface to 15°C, which is losing energy by radiation at about 380 W/m2 or so.
Madness. I’d hoped that if I weren’t the only person pointing out this facially incorrect claim, you might notice … seems I was wrong.
Now, please don’t take this and start off in another direction. Just answer the questions. Here they are again:
If more energy is flowing THROUGH the block than is leaving OUT the far side, where is the extra energy going? And if more is coming OUT of the far side than is flowing THROUGH the block, then where is the extra energy coming from?
Answer, please. No handwaving. No misdirection. No personal attacks on me. No invective.
Answer, please.
w.
“Nobody here, including myself, has a clue how that would be physically possible.”
Of course they don’t. They distegarded all my arguments and spouted their ideological math.
The CHF in the sun’s photosphere is TINY compared to its CSR.
It doesn’t matter to me that ideological mathematicians disregard observations and claim its not possible within their ideology.
I’m interested in science, not ideology.
My solution differs slightly from papijo’s.
I used 5.670367 *10^-8 for the Stefan-Boltzmann constant.
Hot side: 383.28 K
Cold side: 221.43 K
So the flux to the cold side is 129.48 W
After some iterations the radiation of the hot side was set to 1162.5, that leaves for the cold side 129.5 W, that agrees well with its flux.
Wow!! All this is way over my head.
What I take from this exercise is the following:
This problem relates to a stationary cube.
The cube is composed of a uniform and constant substance.
The properties of this substance is known
The cube has perfectly flat and uniform surfaces.
Only two surfaces of this cube are interacting with the environment.
A constant source of external energy is applied one suface.
Sounds simple right?
I am left wondering how the same problem can be applied to a spinning globe with an uneven surface composed of wildly differing materials. This globe is then regularly subjected to variations in orbit around a variable energy source. The orbit has variations both in distance from the energy source and variations in the angle of incidence to the energy source. All the while, the globe is shrouded by a complex mix of gaseous fluids which interact both with the surface of the globe and the incoming heat source.
If I don’t believe Al Gore I am labelled a denier.
Elementary my Dear Watson.
I didn’t even try to kid myself that I might be able to solve this one – IQ limitations and all that jazz.
As an aside, the solution is the same no matter how the 1,292 W/m2 is added at one side.
It could be a star far away applying 1370 W/m^2.
It could be a 394 K blackbody surrounding one face applying 1370 W/m^2.
It could be an IR laser spread uniformly over the surface 1370 W/m^2.
It could be a 1292 W electric heater on the surface.
This realization may help squash a few red-herrings like ‘heat back to the source’. In every case, we are assuming a fixed input such that 1292 W/m^2 is absorbed. The exact nature of the source is immaterial to the problem at hand.
“As an aside, the solution is the same no matter how the 1,292 W/m2 is added at one side.”
Actually, no. One has to consider the view factor for the radiation emitted off the front face. If the view factor is ESSENTIALLY that of a hemisphere of deep space, then front face radiation is to an “environment” that is essentially at 3 K over 2*pi steradians.
If the view factor is that presented for the front face of the concrete block at, say, 100 km altitude orbiting a brown dwarf having a diameter 200,000 km and looking onto a “surface” temperature of 700 K, the front face radiation “environment” is essentially 700 K over ~2*pi steradians.
W/m² K °C
1.360,00 398,62 125,46
0,95*P 1.292,00 393,54 120,38
1.292,00
T2*Ϭ4 129,49 221,43 -51,73 = T2
T1*Ϭ4 1.162,51 383,29 110,13 = T1
(0,8) * (T1-T2) 136,30 161,86
0,95 * Pi = 1.292 W/m²
Tcold = -51,73 °C = 221,43 K
Thot = 110.13 °C = 383,29 K
I posted a video to a youtube video that debunks my critics. Not surprisingly no one responded. No references to any literature.
I very much enjoyed the insults. I do not feel that my best criticisms got an intellgient response.
The best part is that I have all the references I need from the scientific literature to back up what I said. I will unroll it all on my blog over the year, and I will continue enjoy being attacked by ideological mathematicians posing as physicists.
Youtube is the most inefficient way to communicate that sort of thing. Why not just write it down.
Oh hang on, it’s going to take you a year to do that ?
You should be able to do it in a couple of paragraphs. I know, why not make it ten years. Then you can hope that everyone will have forgotten by then and you will never have the embarrassment of having to eat crow pie in public after having insulted everyone in sight.
Watching a thermally imaged radiator in real time action debunk the theories espoused here is a bad way to communicate?
Great. Thanks for confirming your ideological math trumps experiments. Not surprised denial runs deep.
Yes, it’s a bad way to communicate because it is totally different to what is discussed here and what is in your own version. The heat loss in the radiator is predominantly due to lateral losses which were expressly set to zero in both exercises.
You can waffle for 20min on a video and not say what can be put down in a few lines written and understood in two min. Yes, it’s a bad way to communicate.
I did not mention a single bit of maths in that last comment, “ideological” or otherwise.
Can you explain what “ideological math” is ? It sounds like a new and interesting field of study that I seem to have missed.
Greg,
“The heat loss in the radiator is predominantly due to lateral losses which were expressly set to zero in both exercises.”
Exactly, the radiator had more disadvantages than this block, and it still reached thermal equilbrium (CHF=0).
You lack critical thinking skills.
Zoe, among your many wrongnesses, you seem to have a very poor understanding of language, and what certain words mean.
To name a single example, you have no idea what the word “debunk” means.
“I posted a video to a youtube video that debunks my critics. Not surprisingly no one responded. No references to any literature.”
Psst a video can not do this.
Psst references to literature cannot do this.
Simple question:
Why do my critics convert all of conduction to radiation on the cold side, but on the hot side they reject most radiation becoming conduction?
Why the hypocrisy?
Why does the cold end decide to reject most of the energy and cause the hot side to return most of the energy that could’ve went into heating it?
That energy returning from the hot side back to the source must make it hotter, no?
Not only does the source have its own energy but its getting backenergy from the block! That must make it hotter. And hotter objects emit more! Does the block ever get to see this amplified source?
How could my critics remove energy from the block back to the source and then forget about it?
I don’t get it. I really don’t understand your math outside of the ASSUMPTIONS you put in it.
I completely understand what you guys did, and it’s still smells funny. But I guess you can’t smell it, because winning an argument is more important. It all actually makes perfect sense within the framework of your idealogical mathematics, I’ll give you that.
“Why do my critics convert all of conduction to radiation on the cold side, but on the hot side they reject most radiation becoming conduction?”
Heat will move by any and all means available from warmer areas to cooler areas.
* On the hot side, heat can conduct toward the cool side and can radiate toward cool space, so both happen. Some conduction and some radiation.
* On the cool side, heat can radiate toward cooler space, but cannot conduct anywhere. So only radiation.
So the hot side can send photons right through the photons coming in. Cool.
Or the hot side can emit EM waves right through the EM waves coming in.
Brilliant!
If Planck thought that there was two-way photon travel in his cavity, his radiation formula would have an extra 2x factor, but there isn’t because Planck knew there was only one standing waving per frequency between all two opposing molecules of opposing walls.
Where is it written that Planck described his standing waves as being between two opposing molecules?
light , EM and sound waves can all flow in both directions at the same time. The fields superimpose. When you look in a mirror, does the light coming from you face block the reflection?
You should really think before you post, you are making yourself more ridiculous at each step.
Have you ever heard the expression: when in a hole stop digging?
Excellent example, Greg. When you look right at your eyes in a mirror, the light goes from some light source to your eyes, thence to the mirror and right back to your eyes.
And that obviously is what Zoe ridicules as “two way photon travel”, which despite her ridicule happens all the time.
w.
Willis,
We’re talking about normal radiation. Obviously photons can pass through angles.
What you failed to show is that radiation from your eyes comes back to your eyes [thus making it warmer due to the mirror.]
“heat can radiate toward cooler space”
Tim , what makes you limit that statement? Radiation happens equally in all directions, it does not poll what is happening on the other side of the universe before setting out.
This is a common misinterpretation of the second law of thermodynamics. In Willis’ example the sun will radiate more heat towards the object than the other way round, thus 2nd is obeyed. That is NET flow of energy, it does not forbid radiation to be emitted from the block back to the sun.
Greg,
“Radiation happens equally in all directions, it does not poll what is happening on the other side of the universe before setting out.”
Tell me the scientist that originated this idea and quote him.
Greg, I was trying to emphasize that heat = Q (the net flow of thermal energy between two regions) is always from hot to cold. Ie
* Heat always goes from hot to cold
* Radiation is one form of heat
* Therefore radiant heat is always from hot to cold.
I was trying to say the same thing you were saying. Radiation itself goes both ways; the net radiant heat is always from hot to cold.
“Why do my critics convert all of conduction to radiation on the cold side”
… because that is all there is happening on the cold side.
Why the offensive language. The fact that you can not understand the difference is your problem, it does not impinge on the integrity of others.
Pretending your own short comings are someone else’s fault is hypocrisy.
It does not “reject” it ( apart from the 5% reflected ) , it absorbs it, gets hotter and re-emits it. It is not “returned” to the sun, it is radiated over a full hemisphere. You could calculate what minute proportion of that is in the sun’s direction and how many milliwatts that will be but you can be sure it’s not something the sun is going to notice against its 63 MW/m2 output. So technically, yes, it will heat the sun but you’d need a whole lot of sig figs to write down the difference.
Because anyone with half a brain can see it’s insignificant.
Oops, projections. Hypocrisy.
Greg,
You could take an example from Tim Folkerts. He understood my point 100%.
In my original post that inspired Willis, my block was supposed to demonstrate geothermal heating the hot end. No radiation input.
Willis’ example was not supposed to be clone of what you did, it was in fact considerably harder to solve but it does show conservation of energy and constant fluxes through out the system.
So now you are dropping the whole discussion here and trying to divert back your original post on wordpress.
You do realise this is looking a lot like retreating to where you came from. Did you feel you were doing better in that discussion?
“You could take an example from Tim Folkerts. He understood my point 100%.”
No, I didn’t. Your understanding of the basic physics of this problem is wrong.
“So the hot side can send photons right through the photons coming in. Cool.”
Of course it can. Just like I can shine a flashlight at you and you can simultaneously shine a flashlight at me. Photons easily pass through each other.
“If Planck thought that there was two-way photon travel … there was only one standing waving ”
A standing wave IS two-way travel! A standing wave is two equal waves heading in opposite directions.
Thanks Tim, I was going to point out that she did not understand what a standing wave was but you have to limit the number of issues you raise at one time. It seems that a single question is already information overload for this self proclaimed successor to Einstein.
“A standing wave IS two-way travel! A standing wave is two equal waves heading in opposite directions.”
Great! You have 1361 W/m^2 of waves going from sun to block, right at the beginning. So what is this 1162 W/m^2 now going back?
Is that in addition or is it overlaid with what came in? And if it is overlaid then how is it a loss? 1361 covered it in the beginning – i.e. two-way 1361 as you say.
“i.e. two-way 1361 as you say.”
Sigh. I never said any such thing. YOU brought up standing waves in a different context (cavity radiation). I pointed out your misunderstanding in that context — that within a cavity, the standing wave is the superposition of 2 equal waves in opposite directions.
Nothing in the concrete cube scenario is radiation within a cavity. The 1360 W/m^2 of solar radiation is definitely not a 2-way standing wave.
Tim,
Planck had ONE standing wave, not 2 waves superimposed creating the illusion of one.
If what you said was true, Planck’s formula would have a 2x factor for this increased photon density you proposed.
Sigh, sophists will sophize.
“ … That energy returning from the hot side back to the source must make it hotter, no? Not only does the source have its own energy but its getting backenergy from the block! That must make it hotter. And hotter objects emit more! Does the block ever get to see this amplified source? …”
—
Incredible, you’re actually serious. The illuminated side of the block does NOT redirect its (absurdly characterized “rejected”) energy back towards its original “source”. You actually think the 0.95% of the 1,361 w/m2 light must go back to the emitter? You don’t even seem to realize that re-emission (and even scattering off a surface) can go in all available directions, into open space. At this point I’m amazed how misguided you are, and how you got such basic observational things so discombobulated. But your ongoing attempts to avoid all direct questions and requests for answers, and ignoring of all calls for a reasonable explanation, or citations. And you claim people respond with “ideological mathematics”, when you can’t even make output balance input!? However, you want people to take you ‘seriously’ and if they don’t you act as though they were wrong to reject your many false assertions?
I’m left wondering what your motive is because it sure isn’t to get the details of EM radiation and thermodynamics hammered out straight and true.
Now imagine a john Q public who doesnt have Willis’ skill.
They will happen on this discussion and have no clue who is correct; willis or Zoe?
( psst Willis is )
But imagine our John Q public. How will they judge? they can’t work the math themselves
How will they judge?
by vote?
by credentials?
by identity? willis is a nice guy, Zoe is a piece of work
or will they say ‘hey this isn’t settled?”
what would be a rational approach for someone who cannot check? who doesn’t get it?
The rational approach for people who DO GET IT is clear. watch. observe how they reason together.
what about Joe bag of donuts? how does he rationaly decide who is right
Wow, I hate to be the one who points out the *elephant in the room*, but everyone knows there are tiny bubbles of CO2 gas trapped inside all Unobtainium, and in the concrete itself. CO2 gas, as every climate scientist knows, changes the amount of heat trapped inside the concrete by a catastrophic order of magnitude. One would have to model the concrete and level of CO2 in order to determine the amount of heat as no physical process exists to explain it.
Sorry for those of you trying to take this seriously, but there you go!
Ah, that’s it !
That is how Zoe manages to get 557W or for 2.5W in, it’s CO2 amplification.
Now my critics are suggesting backradiation is OK, but backconduction is not. How silly.
“silly ” is not a scientific argument, what is you point?
Again, it’s not “backradiation” it’s radiation. All surfaces do that.
I love how Willis turned my conduction to radiation problem which acted like a euphamism of geothermal to space into a point-source radiation to conduction to radiation problem in order to show me wrong. Assuming your premises true: great job Willis! but there is no backradiation into Earth. The surface will get what it gets from geothermal, and there won’t be magical backradiation from Th in order to deprive Tc.
Or are you now going to claim backconduction from Th?
Well why not backconduction from Tc in this example?
“acted like a euphamism of geothermal to space”
That word euphamism, I don’t think it means what you think it means.
” … I love how Willis turned my conduction to radiation problem which acted like a euphamism of geothermal to space into a point-source radiation to conduction to radiation problem in order to show me wrong. …”
—
Actually your assertions of such a geothermal flux were already debunked due to being inconsistent with observations and implications for the near surface terrestrial geotherm profile, plus ocean basin thermal profiles. If you were correct about such a high geothermal flux deep mining would not be possible as we couldn’t provide viable cooling to survive a kilometer underground. And long-chain hydrocarbons would have already been thermally eliminated before humans came along. So you wouldn’t be wearing plastic knickers. And if earth’s interior were so hot how do earthquakes in a rigid solid mantle still occur down to 680 km? We know its rigid rock that’s fracturing there because measured seismic wave velocities show it is. If it was liquid there would be no quakes at all, no faults, no fracturing, no mineral bonding within liquid for storing convective cell motions as elastic energy with. No mechanism for creating the observed epic acoustic shockwaves so far down that we see occur every day.
WXCycles,
“If you were correct about such a high geothermal flux deep mining would not be possible”
You obviously lack reading comprehension.
You lack the ability to make a coherent point when challenging a factual statement made by someone else.
WX made a reasoned argument with facts and numbers and the best you can do is a vague “lack reading comprehension”. Do really you think he or anyone else will go : gee you are right, why didn’t I see how flawed his position was ?
Was that comment even meant to say he was wrong? Once again you fail to make any meaningful comment.
Greg,
“WX made a reasoned argument with facts and numbers.”
You also lack reading comprehension.
The geothermal heat flux (CHF) is 92 mW/m^2.
The geothermal emission (CSR) OUT of the ground is like 335 W/m^2.
All references to mining should use CHF, and not CSR.
WXCycles purposefully confused the two to score some cheap points.
“If you were correct about such a high geothermal flux”
OUT OF THE SURFACE, NOT IN THE GROUND.
“… OUT OF THE SURFACE, NOT IN THE GROUND.”
—
Is the wall of a mine shaft not a “surface” emitting into the same gas mixture as above? Please explain why we still need active cooling with giant fans pushing air down vents to survive the thermal emission from the walls? Specifically, why do some IR photons emit from the “surface” of a mine’s walls, but not the whole 335 W/m^2 enchilada?
Please explain in plain English Z so we can all understand your reasoning.