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.
The sun emits 63MW/m^2, and the conductive heat flux in the photosphere is TINY.
Why don’t you check your BS with observations?
What EXACTLY are you claiming is “BS” ?
What is the relevance of conductive heat flux in the photosphere? TINY compared to what ?
You are just spouting gibberish again. Try to make a coherent point for once and explain what you mean.
According to religion on this forum, CSR = CHF
The photosphere is what emits 63MW/m^2 (CSR), but the CHF in the photosphere is TINY compared to this.
I understand that the religious fanatics on this forum will never understand or acknowledge this. How could they? Their ideological mathematics is inoculated from observations.
Their only plan is to not understand what I’m saying. That’s your plan, isn’t it?
Oh no, what if Zoe is right?
Nah, Zoe is so stupid, she doesn’t understand how ideological math is real science.
At equilibrium, the flux is 1360W in, through, and out.
So working backwards, and using engineering toolbox calculator (it’s not cheating, it’s working smart):
https://www.engineeringtoolbox.com/radiation-heat-transfer-d_431.html
(I had to iterate the inputs to get the required output)
Tc = 125.6 C
Again, using the calculator for conductivity:
https://www.engineeringtoolbox.com/conductive-heat-transfer-d_428.html
Th = 1825 C
According to Stefan – Boltzmann you would be emitting 1043586.8538327338 Watts from the front face. I find that hard to believe.
I have learned some things today:
If I strap my wifi dongle to the wireless router I won’t get a signal because the distance to the aerial is shorter than the wavelength.
If I shine a torch into space I will get no reading on the ammeter and voltmeter on the battery. There will be a reading if I place my hand in front of the torch.
Zoe is very entertaining and passionate in her beliefs.
I don’t want Zoe on my side. If she were to agree with me I would have doubts in my own conclusions.
If she were to agree with me, I would doubt my own sanity !
She is displaying exactly the same kind of refusal to deal with logical scientific arguments as you get from flat-earthers. It is more important to them to cling on to their little brain-fart “discoveries” which make them a special person than it is to work from a common basis of understanding to get to the truth.
This is a sociological phenomenon, not a science question.
This whole exercise illustrates with poignant clarity why discussing science with a climate alarmist is like peeing into the wind.
If someone who considers themselves skeptical of AGW is this far disconnected from logical argument, what kind of hope do you have with someone whose entire worldview, political self-identity and peer-group social standing is grounded on screaming “DENIERZ” instead of seeking the truth.
We really have reached the end of the age of enlightenment. We will probably descend rather rapidly into a new dark age now. The next glaciation will take care of what remains.
You have to decide at what point to stop beating your head against a brick wall.
Alex,
There’s more than one wavelength in thermal interaction.
If your wifi was based on 2m signal, it would work at 2m, 4m, 6m, etc., but not 1m,3m,5m, etc. It would be terrible technology.
“If I shine a torch into space I will get no reading on the ammeter and voltmeter on the battery. ”
When WMAP satellite at L2 shined its 4K reference load at the stars, only 2.275K came out. And we called this Microwave Background Radiation.
According to theology on this forum, all 4K should have come out.
Great thought exercise by Willis.
Now for extra credit: What if we were to apply a gravitational field/attraction to the hot side of the cube. Would the cold side warm to a higher temperature? (gravitational field considered to be produced by non-emitting invisible force).
I approached it as an energy balance equation
Energy in = 1360 x 0.95 x area = 1292 W
At equilibrium, Energy In = Energy lost from the front face + Energy lost from the rear face.
So 1292 = δ x (Tfront x 0.95)^4 + δ x (Trear x 0.95)^4
Energy conducted from the front to the rear is a function of the temperature of the two faces, (and a unit distance). Energy conducted = 0.8 x (Tfront – Trear)
Simultaneous equations will get the answer, or you can load the two equations into excel and play around with the two temperatures to get an answer close to Tfront = 379K and Trear = 217K
Given your equations, yes, you could load the equations into Excel and play around . . . but you would never get the correct answers for the front and rear face temperatures . . . and your’s are not correct.
You see, in the S-B equation for radiation power exchange, the emissivity factor (0.95) is not brought together with the absolute temperature term (T) into a term that is then raised to the fourth power: as your expressed it, (Tfront x 0.95)^4 and (Trear x 0.95)^4.
It is only the absolute temperatures that are raised to the fourth power.
The cold side will emit Th-Tc 0.8 W/m2 (as L is 1m) = qc.
Th ^4 = (1360 W/m2 -qc)/0.95σ
Tc ^4 = qc/0.95σ
qc = ((1360 W/m2 -qc)/0.95σ -qc/0.95σ) 0.8 W/m2
qc = (1360 W/m2 -2qc) 0.8W/m2/0.95σ
I’ll have to call that coefficient C
1/C = 1360W/(m2.qc) -2
Soccer is on but the arithmetic from here shouldn’t be difficult like typing this on a phone.
I
The cold side will emit (Th-Tc) 0.8 W/m2 (as L is 1m) = qc.
Should have figured it out on a piece of paper first.
exchange heat with void?
i just don’t understand…
why the heck add unaobtainium?
unaobtainium , is that a material which you can only get one bit of?
“Unobtanium” is the material used to fabricate replacement aircraft parts, the originals of which were fabricated from “Breakseasium”
I hope that clears things up.
I can’t believe how many ****ing wrong answers there are here.
(unit system all mks)
Th=389K
Tc=223K
Yeah, I ran to 3 sigfigs since I don’t think you’re interested in just one
Th= hot side temperature (kelvin)
Tc = cold side temp
k = thermal conductivity = .8
Conductance through the cube
1) q=k*(Th-Tc)/l
Power radiated from cold face
2) Pc = Ec*Bk*Tc^4*A
A = area of a face = 1m
Bk=Stefan-Boltzmann constant = 5.67 x 10 -8
Ec = emissivity of concrete face = .95
Steady state, power radiated from cold side equals q through cube
3) q=Pc
Power radiated from col and hot must equal power incident on hot side
4) Pi=Pc+Ph
Pi = power incident = 1360
Need to substitute Tc for Th:
q = Pc
k(Th-Tc) = Ec*Bk*Tc^4
Th=Tc + (Ec*Bk/k)*Tc^4
Subbing in for Th:
Pi = Ec*Bk*Tc^4 + (Tc + (Ec*Bk/k)*Tc^4)^4
If you want to solver for Tc, be my guest. I used a matlab program to settle on the values and verify eq 4)
Matlab
—————————–
Pi=1360; %incident power, W/m2
k=.8; %conductance W/mK
Ec=.95; %emissivity of the concrete
Cc=.75; %specific heat of concrete kJ/kG*K
Bk=5.67e-8; %boltzmann constant
Th=0; %temperature of the hot side
Tc=0; %temperature of the cold side
minErr=1e-4;
cmpVal=Pi/(Ec*Bk);
estimateArr=zeros(1,1000);
tst=0;
while (tst<=1000)
tst=tst+1;
tstVal=(Tc^4)+(Tc+(1/k)*Ec*Bk*Tc^4)^4;
diffVal=nthroot(cmpVal,4)-nthroot(tstVal,4);
Tc=Tc+.1*diffVal;
estimateArr(tst)=diffVal;
if (abs(diffVal)<minErr)
break
end
end
estimateArr(tst+1:end)=[];
plot(estimateArr);
Th=Tc+(1/k)*Ec*Bk*Tc^4;
Ec*Bk*(Th^4 + Tc^4)
title(['Th= ' num2str(Th) ', Tc= ' num2str(Tc)]);
PseudoNhymm says:
“Power radiated from col and hot must equal power incident on hot side
4) Pi=Pc+Ph
Pi = power incident = 1360”
No. Power radiated from cold and hot must equal power ABSORBED on hot side.
Pa = Ec Pi = 1292
Since you used the full Pi, your numbers are both a little high.
I can’t believe that after saying that you added a wrong ****ing answer. LOL.
4) Pi=Pc+Ph
Pi = power incident = 1360 *0.95
If the cube is perpendicular to the radiation from the Sun, then there are 5 cold faces of 1meter squared, each radiating. So, the power radiated from the |FIVE cold faces is 5*Pc.
In the steady state, Pc=1348/5=5.67 x 10 -8*0.95*Tc^^4
Thus Tc = 262 Kelvin
The exercise stipulates that four of the faces are covered with a perfect insulator and are not radiating, and the insulation absorbs no energy from the cube either.
Correct, which is why this is mathematically equivalent to an infinite sheet problem. There is heat flux only through the top and bottom of the sheet. There is none out the edges, which are at infinity. It’s this sort of trick that allows one to convert sheet, cylinder and sphere 3-d diffusion problems into 1-d problems. In this case, we have steady state and the time derivative is zero. The solution for T to the 1-d differential equation is a linear function of position from a surface of the sheet. So the problem boils down to figuring out the values of T at the boundaries, subject to flux (proportional to dT/dx) values at the surfaces, which have to account for incoming energy and blackbody radiation out. The cool thing is that heat capacity is not important here. Different heat capacities merely affect the time it takes to reach steady state, not the final temperature profile.
I remember solving this sort of problem decades ago to show that continuous lasers could be used to uniformly heat small silicate mineral grains. Good time, good times.
Willis: This is the epitome of a “well-posed problem.” If I may, I’d like to add one more problem to the mix, one that’s equally well-posed, yet whose solution may elude everyone – or at least astonish everyone.
First, let’s establish sign conventions:
If energy enters the system, its sign is positive.
If energy leaves the system, its sign is negative.
If work is done on the system, its sign is positive.
If work is done by the system, its sign is negative.
Now, given these sign conventions, what is the time rate of change of entropy of the system you have described? There are three choices: zero, some number greater than zero, or some number less than zero. Other than zero, a numerical answer is not required…
If you are including the energy source ( sun ) that is losing energy which is being dispersed in all directions, positive. The bit of energy in beam hitting the concrete is then also dispersed in all directions at each end. Both those processes are an increase in entropy.
Thermal energy at the hot end is flowing to the cold end. Since this is descibed as steady state the temperatures are not coming closer together but you need to consider the loss of energy at the cold end flowing out into emtpy space in all directions. Entropy ++ .
You’ve taken a well-posed problem and turned it into a complete mess. Who said I was “including the energy source (sun)”? Just look at the system Willis has defined.
Willis, scrolling through the comments this morning, I see why you did this. Good for you.
For others reading this now or in the future, please consider that Willis gets this right, because the fundamentals from which he constructs his problem and answer are plain for all to learn and verify for themselves.
When I first saw the head post of the problem posed, I thought, “I got this.” But it took me a while to knock off the rust and get the problem framed into the necessary equations. So I learned again that humility and persistence pay off. And I learned again there may be alternate paths to the same correct answer.
David,
I imagine many of us can relate to your comment.
Now Zoe is hilarious
https://phzoe.wordpress.com/2020/02/12/average-moon-temperature/
I do give her credit for sharing her code. it lets folks find her error immediately
The error is not in the code. It is in the false idea that you can average temperatures !
The false idea is that you should convert temperatures to radiation, average radiation, and then convert back to temperatures. That is so insane.
Let’s say you have 2 objects with equal mass and heat capacity. One at 0C and the other 50C.
You bring them together. What will their equilbrium temperature be?
Will it be 25C?
or 27.94C? (The Willis Method)
If the two objects have the same heat capacity, it’s legit since temp is then ( and only then ) proportional to the energy content. But that’s a big if. That is exactly the reason that you should average energy content not temperature. ( see intensive vs extensive properties ) You can impose special conditions under which temp is a valid proxy for heat content. In that case it works.
If you are trying to calculate the instantaneous SB thermal emissions you want to be using the “Willis method”.
As you can see in the graph in this article land temps change about twice as quick as SST. That is because land ( wet rock ) has about half the specific heat capacity as sea water.
https://judithcurry.com/2016/02/10/are-land-sea-temperature-averages-meaningful/
Averaging the two ( where land represents only about 30% of the earth’s surface ) biases the result high. That in essence is why it is done.
In that article I suggested a weighting of land temps by SHC and area. The traditional 70/30 weighting becomes 85/15.
Of course, all that assumes that you have not spent the last 40 years rigging the various datasets to support you agenda in the first place !
As you see if you take the time to read it, he was not trying to average the thermal energy of the surface by averages T^4, he was trying to get the emitted thermal energy to do energy budget calculations. That seems perfectly sensible to me.
He is also very open about the limitations and the context of what he is doing, in fact he does both with the aim of comparing. So saying “Willis’ method” as though he only sees one way of doing it and applies to all is disingenuous.
“he was not trying to average the thermal energy of the surface by averages T^4”
In your quote that’s exactly what he says he did.
read the quote again.
That why I never average temperatures. Only Tony heller does that
HadCRUT, GISS LOTI, BEST all average temperatures. The whole climate discussion is based around average temperatures. Worse they are land + sea averages. Different media.
400 comments in barely 24 hours. Nice! 🙂
I think half of them were mine 😀
Having trouble getting through, though.
The solution is straightforward. Two equations and two unknown. 1) T1^4 + T2^4 = 1360/sigma, 2) T1-T2 = sigma • T2^4/k, T1 is the front side temp and T2 is the back side temp
Unfortunately, the equations are non linear so a graphical solution is the simplest way. We know that T1 > T2 for equilibrium heat flow, we know that T1max = 393.5 K thermal conductivity = 0, T2 < 330 K (the point that radiation from both sides is equal to the input flux. In equilibrium the radiated heat from the back = the heat flow through the concrete.
We need only plot radiation from back side sigma T2^4 and heat flow Q = k (T1-T2) delta T (T1-T2) over the range of T1 = 330 K and 393.5 K and look where the curves cross T1 = 383.727 K and T2 = 218.99 K.
Albedo 1 above. The 0.95 albedo only is a 1.3% correction and only shows up in eq. 2. T1= 383.28 K, T2 = 221.32 K
I posted that yesterday and again about 4 comments above. Essentially you use the same method.
Zooming in a little closer to gain accuracy I get :
221.42 and 383.28 K
Why do you think your cold end is significantly different. ?
Greg,
I’m actually impressed with your gnuplot solution. I learned something new.
Thanks for the positive comment. I’m glad you found it useful.
Which scientist first made the claim that hot and cold will send photons to each other? That there will be two-way heat flow, but only one NET heat flow?
Who started it?
Well, the idea of photons was developed by Einstein. So he would be the first to make such claims.
But sure, feel free to correct Einstein.
So obviously finding such a claim by Einstein should be easy.
I don’t see how calling waves photons and returning back to the corpuscular theory of light was progress, when dozens of scientists explained photoelectric effect using waves.
I’ve mentioned this several time but it seems convenient for you to ignore it. The current concept is NOT a return to Newton’s corpuscular light, it is wave-particle duality. A recognition that light is neither billiard ball particles nor purely wavelike but displays both kinds of properties. It depends on the situation as to which model is most useful in describing observations.
Remember, all the “laws of physics” are not discoveries, they are human inventions. Attempts to capture the key aspects of what happens and model them in a way our brain can deal with in an attempt to have enough understanding to predict future behaviour.
Neither does a wave have to a continuous stream or two way standing wave. Consider the concentric rings spreading out when you throw a stone into a pond. Now imagine the cross-section of the wave, there’s a main peak with smaller ripples ahead and behind which is propagating forwards. This is similar to the idea of a photon as a wave packet travelling through space.
Einstein in his 1917 paper has this to say about the heating of a gas by photons:
http://inspirehep.net/record/858448/files/eng.pdf
During absorption and emission of radiation there is also present a transfer of momentum to the molecules. This means that just the interaction of radiation and molecules leads to a velocity distribution of the latter. This must surely be the same as the velocity distribution which molecules acquire as the result of their mutual interaction by collisions, that is, it must coincide with the Maxwell distribution. We must require that the mean kinetic energy which a molecule per degree of freedom acquires in a Plank radiation field of temperature T be
kT / 2
this must be valid regardless of the nature of the molecules and independent of frequencies which the molecules absorb and emit.
Einstein understood like no other before or since the unity of mass, light and time. The equation e=mc^2 encapsulates this. Here he also sees that heating of matter by photons isn’t that different from heating of matter by matter. And the whole IR absorption blanket idea is false.
From your various comments, I think the bottom line question you are seeking is: what is the justification for a body emitting into a void , without another body, another “receiving ” molecule or a standing wave between two bodies?
Off the top of my head , I can’t tell you who is responsible for that crazy idea. 😉
You seem to refuse the concept of a photon entirely and are not happy with the concept of wave particle duality.
Just as an aside, as an undergrad, I recall being given the exercise to calculate the diffraction of a 20 tonne truck, driving at 60mph between two narrowly spaced buildings, as one would do for a photon going through a narrow slit.
You can do it, but the answer is some insanely small deflection that would never be measurable. The point being that you can even model a truck as wave.
https://en.wikipedia.org/wiki/Photon#Historical_development
Waving a rope can be used to create a whipping force.
If EM waves travel through aether they too can have a momentum just like the rope.
Science works by flasification, not advocacy of one group’s POV.
Why don’t you mention those that debunked the corpuscular view of light and unified everything under wave theory, rather than those who failed to do so and left an unnecessary duality.
You should watch some of the PBS space-time YouTube videos by Matt O’Dowd. This one explains that matter and radiation aren’t that different from each other (e.g. the photon box thought experiment at 1 min 29 sec:
https://youtu.be/gSKzgpt4HBU
So much effort to solve a “thought exercise” that bears no relation to the real physical world. The gray surface model stipulated by Willis reduces the exercise to a simple mathematical problem that is irrelevant to the real world heat transfer problem.The student should rather read the engineering literature concerning topics such as spacecraft thermal control ,solar collectors etc.,to understand the critical effect of non gray behavior on heat transfer involving solar radiation.
Until you are able to deal with such a simple idealised exercise you are unlikely to get too far with a more complex , realistic analysis.
AM, you missed the important revelations of this “thought exercise” completely.
They are:
(1) the percentage of presumably-educated individuals that make logic mistakes in considering the solutions to the postulated problem, which in reality is not all that hard to comprehend and solve.
(2) the obstinate, condescending attitude some people take when they are shown to be poorly informed or flat out wrong, and
(3) how easy it is for humans to make mistakes in performing what are really simple algebraic calculations.
Now with this as a basis, how do you think humans will fare in the nearly intractable issue of “fighting climate change” . . . however you want to interpret that phrase. Feel comfortable spending $50-100 TRILLION trying to do so, given the results seen here of Willis’ “thought exercise”?
Yes I have, it’s my job. Remember this is an exercise intended to provoke thought and interest. The most critical aspect oversimplified is the absorptivity and emissivity are different at different wavelengths, henc the almost ubiquitous use of white paint.
Exactly. A waste of time with no insight how the world works. I am still confused be Willis statemt that the Tao buoy readings show back radiation. I’m not saying they don’t, bit it’s far from clear to me. For the life of me I can’t understand the focus on radiation when convective forces dominate in the troposphere. It’s a little like asking the drunk who lost his keys why he’s lóoking for them under the street light.
“when convective forces dominate in the troposphere.”
Yes. But radiation dominates from the top of the troposphere to space. Neither one by itself gives the whole picture. Both are needed to understand climate and temperatures in the atmosphere.
Nelson February 29, 2020 at 3:24 pm
What’s not clear? As I said, downwelling longwave radiation from the atmosphere has been measured all over the planet for decades, both by scientists and by automatic recording stations like the TAO buoys and the SURFRAD stations. That’s just a fact. Google it if you doubt me. Heck, go here, you can plot the SURFRAD downwelling infrared (longwave) for yourself from any of the SURFRAD stations.
As a global 24/7 average there’s about 160 W/m2 of shortwave solar energy absorbed by the surface, and about 340 W/m2 of downwelling longwave absorbed by the surface. So longwave radiation is a huge part of the surface energy budget, more than twice that of solar energy, and as such it deservedly gets lots of focus.
Regards,
w.
The solution tom 2 decimal places is:
T1= 383.28 K
T2= 221.43 K
These assume sigma is 5.67E-8
Part of the exercise was to “show your work”. So, apart from copying the consensus result from several earlier contributors an asserting it like you are a sage to whom this is abundantly obvious, what is your method for arriving at that result?
In the postulated problem:
— the concrete block dimensions are given to one significant figure,
— the block’s emissivity is given to two significant figures,
— the block’s thermal conductivity is given to one significant figure.
Given the great importance of the block’s frontal area and emissivity in deriving the correct face temperatures, it is improper to give temperature solutions to five significant figures.
“It is the mark of an educated mind to rest satisfied with the degree of precision which the nature of the subject admits and not to seek exactness where only an approximation is possible.” — attributed to Aristotle
This Cambridge University research reported at Science Daily, analysed the atmosphere on a planet 125 light years away with a mass twice that of earth. It’s in the habitable zone with the possibility of water at the surface.
https://www.sciencedaily.com/releases/2020/02/200226212008.htm
Here’s the approach they used to analyse the atmosphere:
Given the large size of K2-18b, it has been suggested that it would be more like a smaller version of Neptune than a larger version of Earth. A ‘mini-Neptune’ is expected to have a significant hydrogen ‘envelope’ surrounding a layer of high-pressure water, with an inner core of rock and iron. If the hydrogen envelope is too thick, the temperature and pressure at the surface of the water layer beneath would be far too great to support life.
This conveys the suspicious impression that knowing the mass density and (gravitational) pressure only, together with distance to the star, tells you the temperature.
Oddly enough, I missed any mention of CO2. Shouldn’t atmospheres freeze to absolute zero without at least one molecule of CO2 present?
This rings a bell – didn’t two scientists propose a similar idea recently – that mass, pressure and distance to the sun is sufficient to know ground level atmospheric temperature? With composition irrelevant? Nikolov and Zeller I believe – it was that Volokin and Rellez?
N&Z got their causality backwards.
Turn off geothermal and solar, and the entire atmosphere solidifies and falls to the ground. This solid “gas” mass still exerts a pressure, but there is no atmosphere. Obviously pressure doesn’t make it hotter.
There is only one sound theory (mine):
https://phzoe.wordpress.com/2020/02/13/measuring-geothermal-a-revolutionary-hypothesis/
https://phzoe.wordpress.com/2020/02/25/deducing-geothermal/
“(mine)”
Now that was funny!
After all of this you finally said something amusing.
I have always believed that you can get a measure of a person by how often they something intelligent, interesting, or funny.
One word out of a few tens of thousands does not cut it.
Still, at least for a moment, you said something that was hysterically funny.
I actually did LOL.
Most satellites in Earth orbit run at roughly 250-350K. The big problem on telescopes is keeping the sensors cool and the structure at a constant temperature so there is no thermally induced distortion of the optics.
And did I mention electronics?
Another interesting thought experiment is the “photon box” that Matt O’Dowd discusses starting about 1 min 20 seconds into this:
https://youtu.be/gSKzgpt4HBU
It turns out that matter and radiation are not fundamentally different from each other.
Check out photonic matter. We’ve been able to convert invariant mass matter to energy for quite a long time, but to convert energy to invariant mass matter is a relatively new discovery. I consider it the first tiny step toward a Star Trek-type replicator… a long ways off still, to be sure, but we’ve at least taken that first step.
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.
It 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)
which is where the energy of a photon is carried.
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.
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.
Apparently, converting photon angular momentum into rest mass is the key.
If anyone is still reading, a better problem is what is the relative mean temperatures of a sphere that is illuminated from a point source and such extremely high thermal conductivity that the whole surface has a constant temperature, compared with a sphere of insulating material so that each section of the surface behaves like an isolated black body.
I don’t have it on me but I did a back of envelope (still required a few sheets of paper) and got about a factor of 2.5. The main simplification was to have linear relation of T with angle of incidence on the illuminated side from 0 to Tmax in the latter.
1360 W * 0.95 = 1292 W
783.28 W + 508.72 W = 1292 W
T_hot 70 C (783.28 W)
T_cold 35 C (508.72 W)
70 C = 343.2 K
343.2 K * 0.8 W/K = 274.56 W
783.28 W – 274,56 W = 508.72 W
(Relationship between degrees Celsius and Blackbody temperature according to Wolfram Alpha.)
A simple problem:
Place a frying pan on a gas stove. Turn it on. We know the bottom will get 400F. After a long while, what will the top of the frying pan emit?
Assume emissivity = 1.
Next question: what will the conductive heat flux be?
Zoe Phin February 29, 2020 at 4:41 pm
said
How on earth do you know that? You really really do not know that. The temperature that the pan reaches depends on the size of the pan and the power(wattage) of the flame.
You are completely out of your depth.
This is not a difficult problem.
https://youtu.be/Var3o_eko9U
Anyone doubt that the CHF is nearly zero?
Anyone still doubt that CSR != CHF?
Come on sophists, do your sophistry …
Everyone who thinks this problem is unrealistic or that the two sides are the same temp: Go outside and find a large flat rock. Shaped like a lens would be best, but not required.
Hold the rock up to the Sun.
Wait.
Feel the side in the Sun after a nice long while.
Feel the other side.
Are they the same?
Why not?
Now the Sun goes down.
What happens?
How does the hot side feel after a few hours, compared to when it was in the Sun for several hours straight?
How about the other side? How does it feel, compared to when the Sun was shining?
How long does it take the hot side to cool down?
What happens to the energy that had made it warm?
On a cold clear night with low humidity, place a thermometer under your car sitting in the driveway.
Place another above the car, but not lying on it, maybe on a wooden post right next to the car. Make sure it is not pointing at the sky…put a piece of cardboard overhead of it. As an experiment, place another one face up on top of the cardboard. Cheap alcohol in glass thermometers can be had for a small price at any home store of big box market or on Amazon.
In the middle of the night, place your hand on the metal of the roof of the car…or the top of the hood or the top of the trunk if your car has a fabric roof.
Now place your hand on the metal of the bottom of the car.
The thermometers will tell you that the air temp is about the same under the car as above it, although very close to a concrete surface that was in the Sun all day, it will stay quite a bit warmer.
The thermometer on top of the cardboard facing the sky will show a much lower reading that the one facing sideways under the cardboard.
The car roof will be very cold to the touch, under the car far warmer.
Try it with a car on grass another time.
Another way to understand radiation and heat conduction is to spend a lot of time outside all night when the temp is cold enough and the sky clear enough, winds light enough, and air dry enough, for frost to form.
Frost will form on car roofs first, at 38°F…far above the melting point of ice.
Next will be grassy surfaces and blades of grass themselves.
If the day time temp is warm, you will never get frost on concrete.
You will never get frost under a tree with leaves on it.
Here in Florida, there are a lot of nights in the central part of the state when it is warm in the day, and chilly or very cold at night. In agricultural areas, it can be nearly a hundred nights a year or more, between October and April.
How can frost form when the air temp is above the melting point of ice?
How can frost form on grass when the ground two inches under the surface is tens of degrees higher than freezing?
Why does concrete protect plants from frost, while a few feet away on grass, the same type of plant will be damaged from ice crystals?
How does spraying water on plants protect them on cold nights, when if a person gets sprayed with water on the same night in the same place and the same time, you will be suddenly freezing cold…far colder than if you do not get wet?
Materials all have their own capacity for storing and transmitting thermal energy, both from the surfaces and through the material.
Dirt has one, stone and concrete have their own, sheet metal, especially thin sheet metal, has another.
Why is it as much as ten degrees cooler under a live oak tree during the day, but that night when it is cold, under the same tree is five or ten degrees warmer than a foot away from the tree canopy?
The sky is very cold when the air is dry at all levels.
Dry air has a far lower heat content than air of the same temp that is moist.
Radiation from a surface is far faster at dissipating energy than the rate of heat conduction through many materials, such as stone, dirt, or concrete.
But heat conducts very rapidly through metals, and so a layer of sheet metal can get very cold if exposed to a cold sky.
Tree leaves are the same as the air temp at night, or maybe ever a little colder, but still far warmer than the sky, so they are radiating down to the air and ground under the tree, preventing frost and keeping it warmer than where the ground and air is exposed to cold sky.
Now for a surprise, and this I have verified by many nights outside protecting a plant nursery full or tropical plants from cold weather:
On a night with a cold clear sky, dry air, and temperature falling rapidly, it happens occasionally that a thin layer of cirrus streaks overhead from the southwest.
What happens then is immediate and dramatic.
Within minutes the air temperature at eyeball height will not just stop dropping, it will jump up, by several degrees.
Nothing else has changed…the dew point is the same, still no wind, no mixing of air layers…
Just a thin layer of ice crystals many miles up in the air.
Raises.
The.
Temperature.
Of.
The.
Air.
Near.
The.
Ground.
And.
Keeps.
Plants.
Alive.
I do not care who does or does not believe me, and I have seen this not once or a few times but dozens and dozens of times.
Entire orange and strawberry crops have been saved from killing frost or hard freeze by a wisp of ice crystal clouds many miles up where the temperature is far far below freezing.
Anyone who knows me knows I am no warmista, and am not a member of any sect withing the skeptical community.
I made these observations over many years, decades in fact, while I was building and running a tropical plant nursery in Central Florida, and for the first five or more years of such experience, there was no such thing as global warming alarmism…it was the early and mid eighties.
And at the beginning of that time I was in college, studying physical geography, chemistry, geology, math, physics, astronomy, biology, meteorology, climatology, botany, and every other topic in science offered by Florida Universities.
I have chronic insomnia…I only sleep a few hours a night.
I used to sit outside all night back in those days, sky watching, observing the weather and the wildlife (our farm is in the middle of nowhere if ever such a place exists)…and I still do, although not as much as back when there was no internet, no weather channel, no social media…nothing much to do but read and make observations.
I have no idea why anyone thinks photons will not leave a material in the direction of a warmer object, let alone why anyone thinks they will not leave a surface of any object unless they can see some matter for them to impinge upon.
Do stars radiate in all directions all the time or not?
Do stars in orbit around each other radiate differently than if they were in isolation?
Does a slightly cooler star in orbit around a slightly warmer star, close enough that the warmer star occupies a large part of the “sky” of the cooler star…stop radiating on that part of that side?
Our Sun sends out photons in all directions, even though it is a mathematical certainty at least some of those photons will never strike anything but will keep going forever.
Anyone who proves different has a big fat Nobel Prize and lifelong fame and an immortalized name waiting for them
Interestingly and coincidently, tonight is a cold and clear and dry air night here in Florida, and there may be some cirrus clouds coming in from the Pacific ocean across Central America, if not tonight then perhaps tomorrow night.
Dew point where I live is 30°F at the moment.
Forecast low is 41° although it may be a little colder and maybe we will get frost.
I am gonna see if the proper conditions exist to demonstrate when I was just talking about.
I have some old smartphones and a tripod and one of those phones will take time lapse video in HD.
I think one problem we have these days, and I know it is only one of many, is that there are a lot of people who have never made a single scientific observation in their life, never done a lab experiment, never spent whole days and nights outside in remote locations, never sat alone and just observed things for any length of time at all.
The little girl from Sweden (or wherever she s from…it hardly matter) just travelled around the world and took photos of herself in hundreds of places and posted them on social media.
Every one of them was some scenic locale of big city, and in most it was remarkable for the scenic splendor and the places.
But after months of such travel…she told us all again about how the planet is dying and in a crisis.
She has never showed one place with any crisis, or any place that is “dying”.
We can learn a lot from books, but not everything.
We can learn a lot from school, but the most important thing to learn is HOW TO THINK CRITICALLY AND FOR OURSELVES.
When what one thinks or believes does not match up with what can be seen, that person needs to change their mind, or they will just be wrong for the rest of their life.
Greetings Willis,
I did this last night on a napkin and got 383 Kelvin and 221 Kelvin.
But you asked us to show our work, so this had to wait until this morning. So here it is, including the units analysis. Thot = Temperature hot (front), and Tcold = Temperature cold (rear)
At equilibrium, energy in = energy out. So, Insolation Front = Radiation Front + Radiation Rear
At equilibrium, Heat Flow through the block = Radiation Rear
Professor Lehman taught us in Thermodynamics that if you can cross out all of the matching units on both sides of the equation and you have none left over, then you will not get the wrong answer. So, crossing out all of the units, striking the terms that appear on both sides of the equations, we can reduce the above two equations to just…
Two equations with the same two unknowns, Thot and Tcold.
So, we solve the second equation for the hot temperature (Thot) …
Then insert the Thot result back into the first equation to find the cold temperature (Tcold), and we have …
I cannot factor 4th power terms to even attempt to isolate Tcold. But I can guess, repeatedly, starting with say 200 Kelvin, and work my way up to find an answer that results in 1360 watts.
We use the Define Names part of the Formulas tab in Excel to name the cells that contain our constants and our results so that we can use them in our equations …
sigma 5.670374419E-08
epsilon 0.95
k 0.8
Answer 1360.0000000000
Tc 221.4224502148
Th 383.2796281619
The formula for the Answer cell (showing 1360.0000…) is “=sigma*power((Tc+(sigma*epsilon*power(Tc,4))/k ),4)+power(Tc,4))”
The formula for the Th cell is “=Tc+(sigma*epsilon*power(Tc,4))/k”
The Tc cell contains plugged values and is changed repeatedly until 1360 pops out as the answer.
Reporting the results with the same unreasonable precision that we see in orthodox climate science as practiced by the high priests of the Church of Anthropomorphic Global Warming, we have …
Temperature of the front of the block (hot): 383.2796281619 Kelvin
Temperature of the rear of the block (cold): 221.4224502148 Kelvin
I actually solved a similar real-world problem from INSIDE the block some years back.
I had a Digital Equipment Corporation VAX Cluster that we used to monitor the temperatures of IV bags during sterilization in a set of autoclaves at the pharmaceutical company I work for. The VAXen were bullet proof and ran continuously for decades without complaint or failure, dutifully recording data from dozens of temperature probes that were placed amongst the IV bags to endure that the product achieved lethal temperatures for the required time, and was therefore safe to inject directly into patient – which was me once during a hospital stay, so I was very happy to see our logo on the saline bag.
One day the electricians came to me and said they had to shut down the Liebert CRAC units (Computer Room Air Conditioners) for four hours to complete required maintenance to the feeder circuits, and they wanted me to turn off the VAXen. I didn’t want to do it because I owned the campus-wide continuous running computer uptime record, exceeding 7 years, and I wasn’t going to let that record go easily – it’s a geek thing – my computer is better than yours, etc. So, we clamped an amp meter onto the power cables and determined that the VAXen were soaking up just shy of four thousand watts. A tape measure gave us the dimensions of the room. We looked up the R values of the box’s materials, which were concrete floors, insulated steel panel walls, glass windows, and a foam tile ceiling, so we could calculate your equivalent “k” values. We wheeled in an industrial fan to keep the air moving, so conduction and convection ruled, not radiation. The problem was more complicated than your steady state problem, as the (Th-Tc) quantity that drives heat into the walls was changing over time as the room warmed up.
The math said the room would be 85F at the end of four hours.
When the electricians were done at the end of four hours, the room was 85F.
I enjoyed the problem.
Thanks Willis.
My pleasure, Thomas, and thanks for walking through the exercise including the units. Cancelling out the units is something I do on a regular basis in my research, my thanks to Mrs. Henniger, my high school Chemistry teacher for insisting that we do that.
w.
Willis,
To be fair, you should reask this problem, but with a conductive heat source, and a radiative output – just like my original.
Zoe, assuming you’re talking about this problem, I fear your problem is ill-posed. Why? Because it’s not at steady-state.
If the hot end is fixed at 75°C as you show, then it would need to have energy being constantly added to it … but you haven’t included that energy in your problem
You also haven’t included the radiation coming off of the hot end at 75°C. I understand that you don’t think that something can both absorb energy and radiate it at the same time, but if you think about a hot rock in the sun you’ll have to admit that yes, it’s not only possible, it happens all the time. And assuming an emissivity of say 0.95, that end is radiating 791 W/m2 … which doesn’t appear anywhere in your problem.
Without the incoming energy necessary to keep the hot end at 75°C, then the block is not at steady state, and the problem cannot be solved.
Which is why I put up the problem in the head post, which actually is well-posed and solvable, as proven by the large numbers of people who got the same answer that I got, 383K for the hot end and 221K from the cold end.
Until you understand that and can do the calculations for it yourself and get that correct answer, I fear you won’t understand why your problem is ill-posed.
Best regards,
w.
Willis,
I presented a very simple problem you can find in a high school textbook. In fact it’s usually the first problem your presented with when you learn about heat.
Have you used this calculator?
https://www.engineeringtoolbox.com/amp/conductive-heat-transfer-d_428.html
I didn’t specify steady state.
A steady state wouldn’t bring the cold end down from my 50C anyway.
My point was to show how CHF and CSR are related, and no, CSR != CHF at steady state.
I’m glad your guests were able to solve the problem AS IF the sun was a POINT source, rather than the actual Sun square shining on your block.
Do you even understand view factor?
All your guests and you solved the problem wrong. But it makes for good ideological math.
Zoe Phin March 1, 2020 at 12:56 am
Me, I’ve NEVER seen a high school textbook that said one word about thermodynamics, but it sounds like your high school was different.
Perhaps so … so you can either give us a link to said mythical “high school textbook”, or admit you’re just making things up with no evidence.
Your choice.
w.
Willis you are correct about her making that claim about text books ( high school or otherwise ), I challenged her to provide such an example and, guess what, she ignored it.
There is no reason to have the hot end radiate, the temperature and heat flux are given quantities. Nothing is specified about how that outcome is arranged.
The problem is the cold end.
You may find some such idealised thermal conduction example a text book but it will NOT have the little “E=?” tagged on the end. That is her false claim. That does not come from a text book , she added that bit in. SHE MADE IT UP.
That is the fundamental error which leads to her “magic” energy creating block of concrete.
In spuriously assuming that the given heat flow and temperatures are consistent with the cold end evacuating all that flux by thermal emission, she has created a contradictory set of conditions. This leads to the violation of conservation of energy at the boundary and the magical creation of energy.
If she’d been a little more low key about this until she checked it out , she could have just said : oops I made a mistake.
Sadly, after bad mouthing everyone in sight ( and even those she can’t see ! ) and becoming more and more belligerent, she’s painted herself into a corner where she’d need to eat enough humble pie to feed the 3rd world for several years to come.
I’d erroneously assumed she was about 18, since this is the kind of age when people are often this foolish. They have the excuse of youthful inexperience of life. Sadly, she claims to be 34, an age where one would generally say : she should know better.
Oh well, it’s never too late to learn.
Have you? It clearly shows the heat flux going into , through and out of the block. I guess they must be using that damned “ideological maths” too.
BTW , nowhere on that page does it show the cold end being cooled by SB radiation. I wonder why not ? Standard text book stuff apparently. They must have missed school that day.
Thomas Edwardson,
Answers for temperatures carried out to 13 significant figures? Really?
Gordon, it was a joke. He said:
w.
Sorry, I get it . . . but even the Church of AGW and its offshoot, the Church of CAGW, do not provide values—when they provide values at all—to that extreme. 😉
With that result the average of both temperatures is 302 K which would correlate with 471 W/m2 radiated away on average.
This is nowhere close to what is received, which underlines that averaging temperatures in order to determine radiation power is a wrong approach.
If one was to calculate the average temperature needed to radiate away all incoming solar power, the answer would be 335K.
335K is not 302K. Interestingly it is exactly 33 K which seems to be the GHE.
Seems the GHE is just an issue of wrong physics.
Well, we’re rounding up on 500 comments on this thread, far more than I thought it would garner. What have I learned?
• I’ve learned that there are a lot of folks out there who understand this stuff backwards and forwards, in some cases because this is what they do for a living.
• I’ve reaffirmed that there are a number of people out there who don’t understand this stuff at all and are totally immune to facts. For example, despite having likely seen two people with flashlights shine their lights at each other at the same time, they cannot be convinced you can have two opposing streams of photons occupying the same space …
• I’ve seen that there are still folks out there who think that a colder object cannot leave a warmer object even warmer than it would be if the colder object wasn’t there. For them, I can only recommend my post entitled “Can A Cold Object Warm A Hot Object“.
• I’ve been sadly unsurprised by Zoe’s comments. Ah, well, as the old saying goes, “You can lead a horse to water, but teaching it to do the backstroke might be difficult” … or something like that. However, my intention was not to convince her to change her beliefs. It was to dissuade others from adopting her beliefs … and her angry, aggro, evasive comments have certainly been helpful in that regard.
• I’ve been amused by the number of wildly different answers to the question from folks who are totally convinced that they are 100% right.
• And finally, I’ve been happy to see that my original answers to the question and my interpretation of how to solve it agree with the answers of those who know this stuff the best.
Let me say that I don’t expect this summation to end the commenting on the post, nor is it intended to. I just wanted to nail this up as a précis of where we have gotten to so far.
My thanks to everyone who has commented, including Zoe. Even though she’s been disagreed with by those who know this subject the best, she has not turned and run. And while I don’t like that she appears to be totally immune to facts, at least she’s willing to attempt to defend her misunderstandings. Can’t ask for more than that. Well, I guess I could ask for more than that, but it doesn’t appear I’d get it, and I’m a man who tends to focus on wanting what I have and to not worry too much about having what I want.
Regards to all, lurkers, commenters, professionals, amateurs, and casual readers alike, and as always, my thanks to Anthony Watts, to Charles The Moderator, and to the many volunteer moderators for providing us with this marvelous space for public peer review.
w.
Willis,
You’re a hoot. I agree that I’m wrong from the point of view of ideological mathematicians.
If you noticed, I posted videos debunking the claims made here and all I got were crickets.
There are other points I had that went ignored, misunderstood, or ridiculed.
There is ZERO reference to any experiments or scientific literature.
What am I to make of this? Let’s give it another shot.
To be fair, you should re-ask this problem, but with a conductive heat source, and a radiative output – just like my original.
I’m still waiting for an introductory link this amazing field of study which no one seems to have heard of yet apparently we are all practicing without knowing it.
Google drew a blank on that one , can you help?
Maybe it’s a branch of post-modernist philosophy which I need to know about.
In fact , your won refusal to engage in reasoned logical discussion seems decidedly most-modernist. Maybe that’s where we are all going wrong.
“her angry, aggro, evasive comments have certainly been helpful in that regard”
You have to be angry and aggro when dealing with people in fantasyland telling you you’re wrong because their ideological math tells them so.
I didn’t get ahead on walls street by being quiet and listening to bad suggestions.
https://youtu.be/Var3o_eko9U
Anyone doubt that the CHF thru the pan is nearly zero?
Anyone still doubt that CSR != CHF?
Interested mathematicians greatly desire to know what IS the factorial of CSR (CSR!). Please expand on this.
!= mean not equal to 😉
OK, if you say so. But in my mathematical universe the common symbol meaning “does not equal” is “≠”.
Gordon, you are right about the symbol ≠. However, computer programming is done with the regular keyboard. As a result, the “!” has been chosen to stand for negation.
So for example, in the computer language called R, we have a function called “is.finite()”, which tests to see if a given value is finite or is some kind of error.
The negation of that, which tests if a given value is NOT finite, is written as:
!is.finite().
Regards,
w.
Pass. I’ve made my arguments, posed my puzzle, got the same answer as a bunch of actually sane people practicing that evil “ideological math” that’s taught in every thermodynamics class in the world, and made my point.
You have studiously ignored it all, been ugly, abusive, and abrasive, and generally done everything possible to further my goal of convincing people not to listen to your nonsense.
As a result, I fear I have nothing left to say to you. You’ve proven my case that you’re not worth listening to very neatly, my thanks for that, but my work is done here.
So hele on! I wish you well, but I’m not going to continue the discussion. As a number of people have pointed out, you never answer questions, you just take off on some other unrelated track. That’s fine, but you’ll have to go down that track without me.
My best to you,
w.
Willis,
“Pass. I’ve made my arguments, posed my puzzle, got the same answer as a bunch of actually sane people”
That’s nice, and the empirical evidence proves you and them wrong. Sorry. Empiricism wins over consensus.
“You have studiously ignored it all, been ugly, abusive, and abrasive, and generally done everything possible to further my goal of convincing people not to listen to your nonsense.”
I implored people to look at empirical evidence. If their fine with armchair philosophy, then clearly I’m wrong.
But hey, you know your tribe. If a video demonstrating that CSR != CHF can’t convince them of my arguments, then you deserve them. Please take your ideologues with you, and I’ll keep the empiricists. Deal?
But you know what? Maybe I was wrong the whole time, and you have some empirical evidence to show me that CSR = CHF….
I would love to be shown this, and I graciously await your return so I can formally apologize.
Pass, Zoe. I’m taking the advice of the man who said, “Never wrestle with a pig. They enjoy it, and you just get dirty”.
w.
I pointed out straight away your false assumption in your own concrete bar puzzle: you spuriously assumed that the arbitrary 2W flux was consistent with the cold end being cooled by SB radiation. There is not reason for that to be the case and with the values you chose, it is not the case. End of.
You have studiously avoided even commenting on that and realising that your amazing energy multiplying gadget is a simple conceptual error. ( Sadly we are still going to have to pay for every watt we need ).
Clearly you have no intention of admitting you are wrong or even responding to explanations of where you went wrong. That is not the behaviour of someone who is likely to formally apologise about anything.
It is the behavious of someone who is a little too sure of themselves and their level of knowledge. Like a freshman undergrad who has learnt a few basics and imagines he knows it all.
Questioning accepted ideas is good attitude but you have to follow through and back it up. Just coming up with a crazy statement and ignoring counter arguments will go down fine at a flat earther meeting but not with trained scientists and engineers who require answers.
I gained a degree in applied physics , so I’m fine with empirical evidence.
All I see so far is your faked text book example and spurious assumptions. Now where is your empirical evidence?
Are you actually heating your home using one of these concrete blocks with a duracell battery providing the 2W input energy?
Zoe, That cooking video is supposed to be your empirical evidence? Looking at the cold side of the veggie burger?
1) The video shows radiation via a FLIR camera, but it shows nothing about conduction. How are you ‘measuring CHF’ to know how it compares to ‘CSR’ when the only data you have is ‘CSR’ & ‘HSR’ ?
2) The veggie burger never comes to a steadystate condition, so no one would expect ‘CSR’ = ‘CHF’ .
3) There is air in that kitchen. EVEN IF the burger came to a steadystate condition, there would be conductive and convective loses from the top of the burger, meaning
CHF = CSR + CSConvection + CSConduction
Again, no one would expect CHF = CSR.
You don’t seem to even understand what your own evidence is showing.
OMG , that is what she was calling empirical evidence?
I refuse to get into U-tube videos as part of a scientific discussion but thanks for watching it.
So much for the idea of designing a controlled scientific experiment. What a joke.
Tim,
You appear to be not quite right in the head. Did the stove appear to make the top and bottom of the pan the same temperature? Then CHF = 0, and CSR is the hot yellow you see.
The burger was not mentioned. Can’t you read?
“Anyone doubt that the CHF thru the pan is nearly zero?”
Zoe, that is no better!
1) Anyone with a shred of scientific knowledge would indeed doubt that “the CHF thru the pan is nearly zero”. If that were the case, pans could not cook food or boil water. Pans work precisely because they conduct heat very well.
2) The fact that the FLIR sees intense IR radiation is proof that there is large “cold side radiation” — lots of energy leaving the from the top of the pan. This means there must also be lots of energy being supplied to the top of the pan = large “CHF” ( conductive heat flow) resupplying the energy being lost via radiation.
Tim,
You appear to be not quite right in the head. Did the stove appear to make the top and bottom of the pan the same temperature? Then CHF = 0, and CSR is the hot yellow you see.
The burger was not mention. Can’t you read?
“Anyone doubt that the CHF thru the pan is nearly zero?”
What is the resolution of the measurement that allows you to state that the top and bottom temps are the same?
Pans of different materials have completely different thermal characteristics.
The amount of heat conducted through and aluminum pan is an order of magnitude higher than that through a stainless steel pan under the same conditions.
Can you give a coherent explanation of how heat arrives on the top surface of the pan if it is not being conducted through the pan?
I seriously doubt it.
Prove me wrong.
All you ever do is say stuff.
Wrong stuff.
You remind me of the guy who used to come around arguing that there was no such thing as convection, and that meteorology was a fake science, water vapor was heavier than dry air, everything was caused by “vortexes” that only he know anything about.
He was similarly immune to facts, insulting, and by all indications 100% uneducated in anything resembling any actual science.
BTW…here is the source for aluminum vs stainless conduction rate, at bottom of the page:
“Conductive Heat Transfer through an Aluminum Pot Wall with thickness 2 mm – temperature difference 80oC
Thermal conductivity for aluminum is 215 W/(m K) (from the table above). Conductive heat transfer per unit area can be calculated as
q / A = [(215 W/(m K)) / (2 10-3 m)] (80 oC)
= 8600000 (W/m2)
= 8600 (kW/m2)”
“Conductive Heat Transfer through a Stainless Steel Pot Wall with thickness 2 mm – temperature difference 80oC
Thermal conductivity for stainless steel is 17 W/(m K) (from the table above). Conductive heat transfer per unit area can be calculated as
q / A = [(17 W/(m K)) / (2 10-3 m)] (80 oC)
= 680000 (W/m2)
= 680 (kW/m2)”
https://www.engineeringtoolbox.com/thermal-conductivity-d_429.html
Any explanation yet, why the theoretical average temperature and its associated Emission Power is 33 k apart from the real physical value?