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.
OK. At this point, I’ll give my answer, and indeed, I may be wrong.
Also, it seems some folks are using slightly different assumptions than I used, my bad, my lack of clarity in setting out the puzzle. So I’ll try to cover all of those.
We have two conditions that must be met at steady-state. First, the amount of energy entering the block must be equal to the amount of energy leaving the block. The amount entering is equal to 1360 * epsilon, which I’ve said is the emissivity (and thus the absorptivity) at all frequencies. The amount leaving the block is equal to sigma epsilon (T_hot^4 +T_cold^4). So the first equation is:
sigma epsilon (T_hot^4 + T_cold^4) == 1360 epsilon [eqn1]
The second condition at steady-state is that the flow through the block has to be equal to the flow out of the cold side. The flow through the block is k (T_hot – T_cold), and the cold side radiation is sigma epsilon T_cold ^4, so the second equation is:
sigma epsilon T_cold^4 == k (T_hot – T_cold) [eqn2]
Me, I use Mathematica for this kind of problem. This avoids me making stupid math errors. I’m good at abstract math, but I’ve been known to not be able to remember that 8 * 7 = 54 … just kidding, put down the mouse …
I’ve used “Th” and “Tc” for the hot and cold side temperatures. Here’s the Mathematica solution, with “In[N] being inputs and “Out[N] being the Mathematica outputs:
In[43]:= Clear["Global`*"] In[44]:= eqn1 = sigma epsilon (Th^4 + Tc^4) == 1360 epsilon Out[44]= epsilon sigma (Tc^4 + Th^4) == 1360 epsilon In[45]:= eqn2 = sigma epsilon Tc^4 == k (Th - Tc) Out[45]= epsilon sigma Tc^4 == k (-Tc + Th) In[46]:= sigma = 5.67 * 10^-8 k = .8 epsilon = .95 Out[46]= 5.67*10^-8 Out[47]= 0.8 Out[48]= 0.95 In[49]:= Last[Solve[{eqn1, eqn2}, {Tc, Th}]] Out[49]= {Th -> 383., Tc -> 221.}So, 383 K for the hot side, and 221 K for the cold side. Note that because of the fourth power in the terms, there are a number of either physically impossible (negative temperatures) or imaginary number solutions to the problem. Here is the full list of all the solutions:
{{Th -> -428. - 12.2 I, Tc -> 201. + 245. I}, {Th -> -428. + 12.2 I, Tc -> 201. - 245. I}, {Th -> -406. - 8.55 I, Tc -> -201. + 145. I}, {Th -> -406. + 8.55 I, Tc -> -201. - 145. I}, {Th -> 10.7 - 389. I, Tc -> -65. - 220. I}, {Th -> 10.7 + 389. I, Tc -> -65. + 220. I}, {Th -> 14.7 - 407. I, Tc -> 226. - 142. I}, {Th -> 14.7 + 407. I, Tc -> 226. + 142. I}, {Th -> 31.4 - 376. I, Tc -> -293. + 70.3 I}, {Th -> 31.4 + 376. I, Tc -> -293. - 70.3 I}, {Th -> 39.3 - 405. I, Tc -> 132. + 291. I}, {Th -> 39.3 + 405. I, Tc -> 132. - 291. I}, {Th -> 345., Tc -> -315.}, {Th -> 373. - 18.5 I, Tc -> 46.7 - 277. I}, {Th -> 373. + 18.5 I, Tc -> 46.7 + 277. I}, {Th -> 383., Tc -> 221.}}Mathematica puts the positive real-number solution last, so I’ve just shown that one.
If it is a perfect blackbody, on the other hand, then epsilon = 1 at all frequencies. In that case we get:
In[50]:= Clear["Global`*"] In[51]:= eqn1 = sigma epsilon (Th^4 + Tc^4) == 1360 epsilon Out[51]= epsilon sigma (Tc^4 + Th^4) == 1360 epsilon In[52]:= eqn2 = sigma epsilon Tc^4 == k (Th - Tc) Out[52]= epsilon sigma Tc^4 == k (-Tc + Th) In[53]:= sigma = 5.67 * 10^-8 k = .8 epsilon = 1 Out[53]= 5.67*10^-8 Out[54]= 0.8 Out[55]= 1 In[56]:= Last[Solve[{eqn1, eqn2}, {Tc, Th}]] Out[56]= {Th -> 384., Tc -> 219.}Finally, if the longwave emissivity is 0.95 but all of the incoming 1360 W/m2 is absorbed by the hot side, we have:
In[57]:= Clear["Global`*"] In[58]:= eqn1 = sigma epsilon (Th^4 + Tc^4) == 1360 Out[58]= epsilon sigma (Tc^4 + Th^4) == 1360 In[59]:= eqn2 = sigma epsilon Tc^4 == k (Th - Tc) Out[59]= epsilon sigma Tc^4 == k (-Tc + Th) In[60]:= sigma = 5.67 * 10^-8 k = .8 epsilon = .95 Out[60]= 5.67*10^-8 Out[61]= 0.8 Out[62]= 0.95 In[63]:= Last[Solve[{eqn1, eqn2}, {Tc, Th}]] Out[63]= {Th -> 389., Tc -> 223.}All comments gladly accepted.
w.
How nice, you were writing your comment at exactly the same time as me!
“The second condition at steady-state is that the flow through the block has to be equal to the flow out of the cold side. The flow through the block is k (T_hot – T_cold), and the cold side radiation is sigma epsilon T_cold ^4, so the second equation is:
sigma epsilon T_cold^4 == k (T_hot – T_cold) [eqn2]”
Uhuh, just as I predicted:
“This is so obviously silly, so he will have to pretend the solution is CHF = CSR (Forgetting HSR), he will get an answer that befuddles HSR and makes it physicially unreal.”
Do you see the problem?
Here is what you did with ”
that the flow through the block has to be equal to the flow out of the cold side ”
Model:
|[XXX]
The “|” represents a infinitesimally small sliver of molecules that are supposedly at 383K while the rest of the of the block “[XXX” is at 221K !!!
This is unphysical. Why can’t the 383K sliver pass this energy onto the next sliver, and next sliver, etc until the whole block is resonating at 383K. What happend to the 162K’s worth of energy? Did the block reject it? Did it go back to the source? If it did then the absorptivity is not 0.95 but MUCH less.
Sorry Willis, but math is not physics.
Conductive Heat Flux is not a conserved value, but goes to ZERO.
I dont get what you are saying Zoe?
The maths says there is a gradient all the way thru the block so it looks like this
383K 987654321 221K
Now that assumes flat sheet laminar flow and we have a slightly unrealistic cube setup.
Are you really saying you don’t think there will be a gradient all the way thru the block?
Why can’t the 9 heat the 8?
Why can’t the 6 heat the 5?
Where is the energy going?
Wllis believes in backradiation. He believes HSR will emit back to its heat source to make it warmer because the block prevented its cooling. This is incompatible with a gradient.
Remember, Willis only makes ideological sense when the pieces of his puzzle are in complete isolation.
He can’t handle all of them at once, because then the solution is all zeroes.
Notice he can’t do HSR = CHF at the same time as CSR = CHF. Even though this breaks his silly water through a hose analogy.
This is not “back radiation”, it’s radiation. It happens from all bodies in all directions, no matter who is looking.
Because each step is radiating in every direction.
“Willis believes in backradiation.”
He does Zoe. After all, it is the basis of the GHE hypothesis. Of course, there is no evidence to support such a belief but it suits alarmists and lukewarmers equally.
Sometimes I don’t know if the difference between alarmists and lukewarmers is only a matter of degree of commitment to a belief system.
Are they sort of like high and low church, Anglo-Catholic v Protestant proponents?
Or are they like those who believe that there are lots of unicorns as opposed to those who believe that unicorns are quite rare?
Sort of floating AGW voters?
leitmotif February 28, 2020 at 3:37 pm
No evidence? Say what? We have a couple of completely independent and very strong lines of evidence.
First, downwelling radiation from the atmosphere has been MEASURED thousands and thousands of times around the world. It’s measured directly by scientists. It’s measured automatically by things like the TAO buoys and the SURFRAD stations. It’s not some imaginary quantity as some people claim.
Second, we know that the surface of the planet is at something around 15°C. So it’s radiating at something like 380 W/m2 or so. However, we also know from both theory and from satellite measurements that only about 240 W/m2 of that escapes the atmosphere.
So the rest, perforce, must be absorbed by the atmosphere. Leaves the ground, doesn’t make it out to space, has to be absorbed by the atmosphere. Good so far?
Now take the final step. If that energy just stayed in the atmosphere, the atmospheric temperature would soon be thousands of degrees.
So that energy is not staying in the atmosphere, and it’s not going out the top of the atmosphere … you do the math. I and most everyone else says that it goes back down to the surface. It’s usually called “Downwelling longwave radiation” (DWLR), or “Downwelling infrared” (DWIR), or occasionally, “backradiation”. However, the last term is confusing, so most folks use another term.
Best regards,
w.
“First, downwelling radiation from the atmosphere has been MEASURED thousands and thousands of times around the world. It’s measured directly by scientists. It’s measured automatically by things like the TAO buoys and the SURFRAD stations. It’s not some imaginary quantity as some people claim.”
So what part of DWLR is back radiation i.e. radiation that emanated from the planet surface that is returned to the planet surface by GHGs? What part of that back radiation is anthropogenic? Feldman et al (2015)? I don’t think so. Not heard of transfer of energy by conduction by non-GHG molecules to GHG molecules. How many collisions between air molecules before a GHG molecule emits a photon? Tens of millions? Hundreds of millions? What?
“Second, we know that the surface of the planet is at something around 15°C. So it’s radiating at something like 380 W/m2 or so. However, we also know from both theory and from satellite measurements that only about 240 W/m2 of that escapes the atmosphere.So the rest, perforce, must be absorbed by the atmosphere. Leaves the ground, doesn’t make it out to space, has to be absorbed by the atmosphere. Good so far?
Now take the final step. If that energy just stayed in the atmosphere, the atmospheric temperature would soon be thousands of degrees.”
Wow! Total Bernie Sanders! You are actually averaging temperatures and fluxes? Is this your Roy Spencer moment?
“So that energy is not staying in the atmosphere, and it’s not going out the top of the atmosphere … you do the math. I and most everyone else says that it goes back down to the surface. It’s usually called “Downwelling longwave radiation” (DWLR), or “Downwelling infrared” (DWIR), or occasionally, “backradiation”. However, the last term is confusing, so most folks use another term.”
The cooler atmosphere heats the warmer planet surface? Can’t argue with that empirical evidence, Willis. Enjoy your fame. 🙂
I stumbled through almost a whole afternoon trying to solve simultaneous fourth-order polynomial equations mathematically, and of course I couldn’t. I looked up stuff on the internet and found mathematical methods that I’d never even heard of; understanding them was way beyond my simple abilities. So I did the “educated trial-and-error” method (we could all it iteration to make it sound respectable) and it only took five steps to get Th = 383.7°K and Tc = 221.4°K. Thanks for the post, Willis, it got my brain going on an otherwise lazy day.
I’m sorry Zoe, it looks as though you just don’t get it. Everything that has a temperature above 0°K and non-zero emissivity/absorbivity is both emitting and absorbing radiation all the time. The hot side is not radiating back only at the sun; it’s radiating in all directions so the fraction received by the sun is beyond infinitesimally small. The omnidirectional nature of black body radiation means that the equations we used in this case are only really accurate if they are applied to an infinite plane rather than a 1 m² concrete block, but the exercise was valuable and entertaining.
Zoe, you seem to be in thrall of the “a colder body cannot heat a warmer body” axiom that we keep seeing from less informed sceptics who want to deny the existence of the greenhouse effect. That is only true if you’re talking about black body radiation. In the atmosphere, radiation (photons) generated by molecules of water (and CO2 to a lesser extent) when they drop from an excited (higher energy) state to a relaxed (lower energy) state is not black body radiation. At least, that’s how my simple geology brain understands it. I could be wrong; I’ve been wrong before (although that may come as a surprise to those who know me).
“Willis believes in backradiation” OK here’s a simple example of what you might call “back radiation” that you can feel yourself. You have to be indoors, in winter, in a house with moderately well insulated walls but single-glazed, clear glass windows, i.e. an old house like mine. Stand in front of a window. The exposed skin of your face, normally at 37°C, is radiating outwards towards the window (and through the window assuming it’s transparent to LWIR). The window at maybe 5°C, and the snowy landscape outside, at maybe -5°C, are radiating back towards you, but at much less intensity than you are radiating at them, so there’s a net outward heat flux from your face. Your normally 37°C skin is now at (say) 30°C, and you feel a bit chilly. Now stand in front of a wall. The wall, at about 20°C, is radiating more heat towards you than the window/landscape did, so the net outward heat flux from your face is less, and your skin is now at (say) 35°C, and you feel warmer. Has the 20°C wall heated your face from 30° to 35°C? Not exactly; but it has lowered the net heat flux from your face, which has caused your skin temperature to rise. “Slows the rate of cooling” is another way of putting it.
You can feel the same effect if you compare how temperature drops after the sun goes down in a desert with how it cools in a more vegetated region where the air is humid. It cools down really fast in the desert compared with elsewhere. In a humid region, there is “back” radiation from water vapour in the atmosphere, and in the dry desert there is much less. This is the greenhouse effect, and the fact that it varies so much between deserts and jungles is clear proof that most of the greenhouse effect is due to water vapour. CO2 concentration is much the same everywhere (“a well mixed gas”) so the GHE should be almost the same in the desert and the jungle if CO2 was responsible. But it’s not.
The idea that everything radiates and absorbs all the time is a bit counter-intuitive, and it’s hard to grasp without taking the time and effort to learn how it works. It’s easy to observe radiation from hot bodies like “radiators” (if your house has hot-water heating) but the fourth-power relationship with temperature means that you have to think hard about it to observe radiative heating and cooling in everyday objects at ambient temperatures. It’s not as difficult as the clock paradox in relativity, but from the comments we see at WUWT, it’s clear that not everyone “gets it”.
LDB
The maths says there is a gradient all the way thru the block so it looks like this 383K 987654321 221K
Zoe
Why can’t the 9 heat the 8? … Why can’t the 6 heat the 5? … Where is the energy going?
Now Moi (Robert K)
I think that 9 DID heat 8, but 8 cooled to heat 7, which cooled to heat 6, which cooled to heat 5, and so on. Now 9 can no longer keep up with 8’s cooling rate to heat 7, 6, 5, etc.
9 has done all the heating of 8 it can do, and likewise on down the gradient. The energy is “going” towards the end, cooling from one place to heat another, via the successive increments of resistance that it must traverse to get through the entire LENGTH of the concrete block.
Somebody (preferably an experienced engineer), please fix me, if I’ve flubbed this. Thanks.
Smart Rock
“I’m sorry Zoe, it looks as though you just don’t get it. Everything that has a temperature above 0°K and non-zero emissivity/absorbivity is both emitting and absorbing radiation all the time. The hot side is not radiating back only at the sun; it’s radiating in all directions so the fraction received by the sun is beyond infinitesimally small. ”
OMG! Do you have any evidence of this or even a mathematical proof? The earth heats the sun? What would the earth do if the sun was not there? Heat the next nearest sun?
“Zoe, you seem to be in thrall of the “a colder body cannot heat a warmer body” axiom that we keep seeing from less informed sceptics who want to deny the existence of the greenhouse effect. That is only true if you’re talking about black body radiation. In the atmosphere, radiation (photons) generated by molecules of water (and CO2 to a lesser extent) when they drop from an excited (higher energy) state to a relaxed (lower energy) state is not black body radiation. At least, that’s how my simple geology brain understands it. I could be wrong; I’ve been wrong before (although that may come as a surprise to those who know me).”
What are “less informed sceptics who want to deny the existence of the greenhouse effect”? Do you have a scale of informed sceptics?
““Willis believes in backradiation” OK here’s a simple example of what you might call “back radiation” that you can feel yourself. You have to be indoors, in winter, in a house with moderately well insulated walls but single-glazed, clear glass windows, i.e. an old house like mine. Stand in front of a window. The exposed skin of your face, normally at 37°C, is radiating outwards towards the window (and through the window assuming it’s transparent to LWIR). The window at maybe 5°C, and the snowy landscape outside, at maybe -5°C, are radiating back towards you, but at much less intensity than you are radiating at them, so there’s a net outward heat flux from your face. Your normally 37°C skin is now at (say) 30°C, and you feel a bit chilly. Now stand in front of a wall. The wall, at about 20°C, is radiating more heat towards you than the window/landscape did, so the net outward heat flux from your face is less, and your skin is now at (say) 35°C, and you feel warmer. Has the 20°C wall heated your face from 30° to 35°C? Not exactly; but it has lowered the net heat flux from your face, which has caused your skin temperature to rise. “Slows the rate of cooling” is another way of putting it.”
Or it could be you reach thermal equilibrium quicker with a warm wall than a cold window? Hmmmm?
If it a still cold night, open the window and see how much back radiation warms you up.
Commonsense law: That which heats cannot be heated by that which it heats. This has to be a fundamental law of heat transfer or we are in a runaway heating scenario that has never happened in the existence or the knowledge of the human race. Heat, the transfer of internal energy, does not flow spontaneously from a cooler object to a warmer object.
Put two lamps 100W and 40W facing each other. Do they both get brighter or hotter because of back radiation?
Shine a light onto a white surface and then reflect the surface light spot onto itself using a mirror. Does the surface light spot get brighter?
Stop talking Bernie Sanders.
“Shine a light onto a white surface and then reflect the surface light spot onto itself using a mirror. Does the surface light spot get brighter?”
Are your eyes equipped with a light meter?
Smart Rock wrote:
“The idea that everything radiates and absorbs all the time is a bit counter-intuitive, and it’s hard to grasp without taking the time and effort to learn how it works.”
Your wording could be less ambiguous. It leads neophytes to the conclusion that objects emit as though they’re in a 0 K ambient, and absorb as though they’re in an ∞ K ambient.
In reality, the underlying physical mechanism which regulates emission and absorption of photons is radiation pressure. If the object has a higher potential than the ambient radiation pressure, it can emit photons into that ambient. If an object has a lower potential than the ambient radiation pressure, it can absorb photons from that ambient.
Here’s some web pages addressing the topic:
https://objectivistindividualist.blogspot.com/2018/08/the-nested-black-body-shells-model-and.html
https://objectivistindividualist.blogspot.com/2015/05/the-greenhouse-gas-hypothesis-and.html
LOL@Klimate Katastrophe Kooks March 1, 2020 at 8:31 pm Edit
According to LOL, nothing can radiate if it is in an environment with higher “ambient radiation pressure”.
You know, like a candle can’t radiate if it is out in the bright sunlight … oh … wait …
Or how if you shine a really bright flashlight at a dim flashlight, the dim flashlight immediately goes out … oh … wait …
Sometimes, I’m simply stunned by the level of nonsense that people believe.
Think about a standoff infrared thermometer. It works by measuring the thermal infrared (longwave) radiation from objects.
Now, according to LOL (an appropriate name given the laughable nature of his claims), the IR thermometer could only measure the temperature of objects that are warmer than the ambient temperature. His claim is that if objects are colder than the ambient they don’t radiate at all because of some mysterious “ambient radiation pressure”.
But of course, that’s obviously not true—IR thermometers can measure the temperature of any object regardless of the ambient temperature.
His whole claim about “ambient radiation pressure” makes no sense at all. The reality is that all solid objects above absolute zero radiate all the time, no matter what the ambient temperature might be.
Best regards to all,
w.
Willis wrote, “Sometimes, I’m simply stunned by the level of nonsense that people believe. … [LOL’s] whole claim about “ambient radiation pressure” makes no sense at all…”
Amen! Thank you for that.
I wondered where on earth LOL “learned” such nonsense, and of course I immediately suspected PSI. So I did a google site search of their site for “radiation pressure,” and found I two people saying such things in the comments. The second went by the handle of “Jonas,” just a few days ago. The first, on November 18, 2019, was (take a guess!)…
.
.
.
…yep, as you probably guessed, it was none other than Zoe Phin. She wrote, “If rafiation pressure from H (hot) is greater than radiation pressure from C (cold), then the photons from H “beats back” photons from C, until C doesn’t emit any photons in the direction of H. A stronger force (pressure/area) beats a weaker force and drives it back home … essentially the weaker force never leaves the house.”
So LOL is probably an alias of Zoe Phin.
I think that’s good news. It means the evident insanity has not spread to so many people, after all.
LOL@Klimate Katastrophe Kooks March 5, 2020 at 12:46 pm
Huh? You said that “radiation pressure” keeps objects from radiating. I asked, if that is the case then how can we see a candle in the huge “radiation pressure” from sunlight?
And how can a strong flashlight and a weak flashlight shine right at each other? Why doesn’t the “radiation pressure” prevent the weak one from radiating?
Your pathetic response is to wave your hands and say the magic word “piffle”? That’s it?? That’s your idea of a scientific argument?
Next, a LINK to whatever “Charles R. Anderson, Ph.D.” uses as “standard equations” would make your response at least understandable … you continue:
Glad to. Here you go:
Net flows and individual flows. The individual flows are from me to you, $100, and from you to me, $75. The net flow is from me to you, $25. Both are equally valid ways of discussing the real flows of money.
What does this have to do with cold and warm objects? It points out a very important distinction, that of the difference between individual flows of energy and the net flow of energy, and it relates to the definition of heat.
Looking at Figure 1, instead of exchanging dollars, think of it as two bodies exchanging energy by means of radiation. This is what happens in the world around us all the time. Every solid object gives off its own individual flow of thermal radiation, just as in the upper half of Figure 1. We constantly radiate energy that is then being absorbed by everything around us, and in turn, we constantly absorb energy that is being radiated by the individual objects around us.
“Heat”, on the other hand, is not those individual flows of energy. Heat is the net flow of energy, as represented in the bottom half of Figure 1. Specifically, a heat flux is the net flow of energy that occurs spontaneously as a result of temperature differences.
Now, the Second Law of Thermodynamics is only about net flows. It states that the net flow of thermal energy, which we call “heat”, goes from hot to cold each and every time without exception. However, the Second Law says nothing about the individual flows of energy, only the net flow. Heat can’t flow from cold to hot, but radiated energy absolutely can.
Do you ignore polite requests? Because that’s what you’re doing. From above in the head post:
Thanks,
w.
LOL@…, Willis has already handled the scientific arguments very well, so this comment is just about your personal remarks.
LOL@… wrote, “Dave Burton… and I have been in a months-long dispute on CFACT, wherein he’s claimed that there are no restrictions on 2LoT violations at the quantum level…”
I have no idea what you’re talking about. I do not recall ever encountering you before this month, neither on CFACT, nor anywhere else. What alias did you use?
Perhaps you’ve been arguing with a different Dave Burton. Will you please post a link to the argument?
LOL@… wrote, “…Dave Burton had his deluded arse drop-kicked by physicist Dr. Charles R. Anderson, PhD on this very topic, which is why Dave so hates the good Dr. that he’s attempted doxing him in the past.”
I don’t know who Dr. Charles R. Anderson is, and I certainly don’t hate him. To the best of my knowledge, I’ve never had any interaction with him. Will you please post links to the alleged arse-kicking and alleged attempted doxing?
To Dave Burton:
I wouldn’t put it past you to claim you are not who you are, Dave… I’d be embarrassed if I were you, too… but your phraseology, interests and denials of reality coincide too closely to the “other” Dave Burton”:
https://disqus.com/home/discussion/cfact/un_climate_roadshow_opening_in_nyc/
Have you experienced a dissociation, Dave? LOL
To Willis Eschenbach:
CLAES JOHNSONprofessor of applied mathematics
http://www.csc.kth.se/~cgjoh/ambsblack.pdf
Notice the requirement in (14.3) that T2 > T1. In the literature one finds the
law without this requirement in the form
Q_12 = σT4_2 − σT4_1, Q_21 = σT4_1 − σT4_2 = −Q_12 (14.5)
where Q_21 is the heat transfer from B1 to B2 as the negative of Q_12.
This form has led to a misinterpretation of Stefan-Boltzmann’s Law as
expressing heat transfer from B2 to B1 of size σT4_2 balanced by a transfer
−σT2_1 from B1 to B2, as if two opposing transfers of heat energy is taking
place between the two bodies with their difference determining the net flow.
Such a misinterpretation was anticipated and countered in Stefan’s original article [42] from 1879:
• The absolute value the heat energy emission from a radiating body cannot be determined by experiment. An experiment can only determine the surplus of emission over absorption, with the absorption determined by the emission from the environment of the body.
• However, if one has a formula for the emission as a function of temperature (like Stefan-Bolzmann’s Law), then the absolute value of the emission can be determined, but such a formula has only a hypothetical meaning.
Stefan-Boltzmann’s Law (14.3) thus requires T2 > T1 and does not contain
two-way opposing heat transfer, only one-way heat transfer from warm to
cold. Unfortunately the misinterpretation has led to a fictitious non-physical
”backradiation” underlying CO2 global warming alarmism.
Ready to give up and acknowledge reality yet, Willis? LOL
LOL@Klimate Katastrophe Kooks March 5, 2020 at 7:48 pm
Pass, thanks. Now that you’ve linked to it, I can see that you’re just pushing the bog-standard Slaying the Sky Dragon nonsense, and my life is far too short to engage in disputing childish misunderstandings of that nature.
w.
To Willis Eschenbach:
From my prior comment:
Such a misinterpretation was anticipated and countered in Stefan’s original article [42] from 1879:
I suggest you educate yourself, Willis. Now you’re implying that Stefan was a “sky dragon slayer”? LOL
LOL@… wrote, “To Dave Burton: I wouldn’t put it past you to claim you are not who you are… https://disqus.com/home/discussion/cfact/un_climate_roadshow_opening_in_nyc/ “
That’s not me. I have not participated in that discussion, at all. That’s “DaveBurton72,” who I suspect might be the same person who impersonated me on WUWT.
I had a conversation with that “DaveBurton72” in a different thread; here’s a screenshot:
DaveBurton72 denied that he was intentionally impersonating me. He claimed it was just a coincidence of names.
LOL@… also previously wrote, “…Dave Burton had his deluded arse drop-kicked by physicist Dr. Charles R. Anderson, PhD on this very topic, which is why Dave so hates the good Dr. that he’s attempted doxing him in the past.”
As I wrote before, I don’t know who Dr. Charles R. Anderson is, and I certainly don’t hate him. To the best of my knowledge, I’ve never had any interaction with him. My guess is that it was “DaveBurton72” again, but I’d like to know. So, will you please post links to the alleged arse-kicking and alleged attempted doxing?
LOL@Klimate Katastrophe Kooks March 5, 2020 at 9:01 pm Edit
I, on the other hand, suggest that you are far too arrogant and self-important to follow my polite request that you quote whatever it is that you are babbling about. In fact, I was reacting to the statement in your link that his work was instrumental in the Sky Dragon fantasies …
Hard pass. I have ZERO interest in your Sky Dragon madness.
w.
To Willis Eschenbach:
You left out a bit on your image… I’ve helpfully included the requisite information. You’re welcome. LOL
Mheh, it’s late. Need to sleep.
To Willis Eschenbach:
You left out a bit on your image. I’ve helpfully included the requisite information. You’re welcome. LOL
To Dave Burton:
Yes, that screen shot certainly sounds exactly like the Dave Burton I’ve been drop-kicking over on CFACT… he’s easy to spot, his hubristic attitude combined with his confused take on scientific topics is such that even when he was socked up as ‘Anonymous’ on physicist Dr. Charles R. Anderson’s website, he was easy to spot. LOL
Apologies if I’ve offended. Clearly he’s been ‘impersonating-not-impersonating’ you… he’s very passive-aggressive that way.
There you have it that is the bit why Zoe doesn’t understand 🙂
So lets do this the the layman way.
Zoe can I ask why can’t one bit be 100 then?
Why can’t one tiny bit of the metal go to many thousand of whatever unit we are using and explode into a ball of plasma ?
Why have we never seen this on the ISS and satellites?
Have you ever seen it or heard of it?
You see the problem you are setting up and just by thinking about it as a thought experiment it is obvious your answer is wrong.
“Why have we never seen this on the ISS and satellites?
Have you ever seen it or heard of it?”
Show me the telemetry data. Don’t quote me the max temp on hot side vs. min temp on cold side. We’re not told if this is concurrent. We’re not told what both sides are facing and their angles and view factors.
There’s not enough information to reach a conclusion, and I could explain it with that information.
Thanks leitmotif !
As usual Willis ignores that satellites can’t detect IR moving away from it, and all surface based instruments are measuring Upwelling-from-the-instrument-IR (which came from surface upwelling IR) which he believes is DWLWR.
He’s not shown what the instruments emit because it’s the same as DWLWR.
” … Upwelling [IR PHOTONS] -from-the-instrument-IR (which came from surface upwelling IR) which he believes is DWLWR.
—
How can an IR sensor facing the zenith detect it’s own self-emission of IR photons that are traveling away from the sensor? That’s absurd nonsense Zoey.
The photon has to be directed at the sensor, not away from it, for the sensor to detect such photons. They are only detected if coming toward the sensor’s field of view. What you are claiming would not and could not be detected!
But detectable photons do come from above, and are being measured.
And as ‘smart rock’ just explained, namely, everything absorbs and emits all of the time in every direction, at all temperatures. That is a fairly basic EM concept.
Zhoe
If what you say is true the earths crust would be the same temperature as the core.
“How can an IR sensor facing the zenith detect it’s own self-emission of IR photons that are traveling away from the sensor? That’s absurd nonsense Zoey.”
Hello, there’s a voltage gain/loss by instrument according to how much the instrument heats the environment or cooled by environment.
If the instrument is warmer than the atmosphere, the instrument LOSES temeprature, and warms the atmosphere.
A photon sensor at 10C in an environment at 10C would detect NOTHING, if it didn’t have a local thermometer.
Zoe Zhoe Zoey
They can’t even get your name right and it’s only 3 letters! :-DDD
“How can an IR sensor facing the zenith detect it’s own self-emission of IR photons that are traveling away from the sensor? That’s absurd nonsense Zoey.”
Hello, there’s a voltage gain/loss by instrument according to how much the instrument heats the environment or cooled by environment. If the instrument is warmer than the atmosphere, the instrument LOSES temeprature, and warms the atmosphere. A photon sensor at 10C in an environment at 10C would detect NOTHING, if it didn’t have a local thermometer.
—
Hi.
Which answers nothing about how it senses a whole spectrum of photons moving away from its sensor.
You suppose the designers and testers of such IR sensors didn’t think of self-noise and how to deal with it? The IR sensor will have a temperature sensor integrated at the chip level to correct for and filter self-noise. Then integrated cooling to minimize such self-noise altogether and get the sensor and apparatus below the temperature of the targeted gases in the atmosphere.
But according to your interpretation of physics, there are no thermal photons from the atmosphere coming downward toward a ground based IR sensor pointing at the zenith anyway unless the IR emitting object or molecule above is marginally warmer than the Earth.
All objects above absolute zero can and do emit photons in any direction irrespective of the temp of an object within the (cosmic) ray-path of an emitted photon. How would or does the emitting molecule, atom or particle, know the thermal geography of the entire probably infinite cosmos, and even takes into account gravitational-lensing effects on ray path, and the locations of all rarefied ‘hot’ particles, in the periphery of galactic clusters, and can plot their future location for when the photon finally reached them, BEFORE the emitting molecule chooses an emission direction which can only ever go towards a colder emission cosmic pathway?
Seems complicated.
You make this ‘inescapable’ interpretation of the physics that atmospheric gases do not emit photons in all the available directions when they have enough energy to emit one, unless the direction of the emission is at a lower temperature than the emitting object … even at cosmic scale radii!
Thus such ground-based sensors must (according to you) necessarily be presumed to not be detecting IR at all from the atmosphere’s emission, no matter what the temperature of the cooled IR sensor and its apparatus is. As it could only ever be self-noise masquerading as an atmospheric IR spectral signature, due to photons moving away from the sensor as no photons could ever emit towards a warmer atmosphere and earth below the emission altitude.
But what if all the molecules in the cosmos instantly learned that we cunningly chilled the IR sensor molecules to below the temperature of the atmosphere, and thus the atmosphere does emit toward the chilled sensor … just to mess with us?
I have to go now.
“This is unphysical. Why can’t the 383K sliver pass this energy onto the next sliver, and next sliver, etc until the whole block is resonating at 383K. What happend to the 162K’s worth of energy? Did the block reject it? Did it go back to the source? If it did then the absorptivity is not 0.95 but MUCH less.”
Beginning to get the gist of this Zoe. Say, on an abandoned city street, 383K beating down. I can’t imagine that a metre below the surface it would be 100 to 150 degrees Celsius lower no matter what insulation was used.
More sh1te physics from warmists and lukewarmists.
Hi Willis. This would make a good problem for the Mechanical Engineering PE exam. I got 388.3K hot surface and 223.6K cold surface (115.1 and -49.5 C). Quite close to Roy Spencer’s answers. I did my calculation in Inch-Pound units (BTU, feet, degrees Rankin. etc.) so some rounding and conversion differences. Also I used S-B to calculate an apparent source surface temperature for the 1360 watt Flux to an absolute 0 surface as specified in the problem. I also assumed a 4 C temperature for empty space that the cold surface would radiate to. The net heat fluxes in all three areas – sun to hot surface, hot surface to cold surface, cold surface to space – work out to about 134 watts.
Part of the challenge was to explain how you got the answer.
Yes Rick, fellow PE.,our first year mechanical engineering courses have not let us down. I got the same as you somewhere in the above blogroll. But Willis is assuming 0.95 x 1360 for heat input giving him a few degrees different temperatures. Other minor numerical differences between replied answers are a result of using 5.67037e-8 for the SB constant, or 5.6704e-8 or 5.67e-8 which also result in a degree of disparity.
Correct.
One needs two equations on the two temperatures.
Either using power power balance separately at the both sides, or select one side (you choose the cold side) and the total energy balance for the block.
The solution is naturally the same.
Yours is the same method I used except I used a simple binary goal seeking process in Excel yet got a result slightly different but not insignificantly different to yours. See 3 posts above. I did use a finer value for sigma (=5.670367*(10^(-8)) which I hope accounts for the slight difference.
Now with that answer we can calculate a average temperature of 302 K.
This is 471 W/2. So over the surface the calculated emission would be less than what is incoming.
We can calculate the temperature for which the incoming solar radiation is valid for the average and it is 335K.
The interesting part is that it calculates a 33K difference. Does this sound familiar?
So climate physics and real physics differ by 33K.
Not sure what you make of this. Let us know.
Willis obviously took my example from:
https://phzoe.wordpress.com/2019/12/04/the-case-of-two-different-fluxes/
Willis believes that there must be conservation of heat flow.
He believes that
Hot Side Radiation = Conductive Heat Flux = Cold Side Radiation.
Forget his parameters, and simplify:
HSR => [ CHF ] => CSR
Conduction Formula: q = KA(Th-Tc)/L
Radiation Formula: q = ɛσT⁴
Set Absorptivity = Emissivity = 1
Set K = L = A = 1
HSR = σ(Th)⁴
CSR = σ(Tc)⁴
CHF = Th-Tc
Assuming HSR = CHF = CSR:
σ(Th)⁴ = Th-Tc = σ(Tc)⁴
There is only one solution!
Th = Tc = 0
HSR = CHF = CSR = 0
This is so obviously silly, so he will have to pretend the solution is CHF = CSR (Forgetting HSR), he will get an answer that befuddles HSR and makes it physicially unreal.
Thanks, Zoe. I think our last two comments crossed in the ether. I actually linked to that post of yours in a comment upthread.
Best regards,
w.
PS—I don’t “pretend” anything. I tell the truth as best I know how. I asked politely, with you in mind, to lay off the ad hominem attacks. Please do so.
Willis
I wonder, based on style and muddy thinking, wether Zoe is actually Sou of Hotwhopper losing traffic.
You seem to have ignored the input radiation on the hot side.
Zoe, what is physically unreal is you example where you choose an arbitrary heat flow of 2W/m^2 and then , for no reason, assume that this is compatible with the “cold end” being cooled by SB radiation. That is not the case, it is not compatible with that heat flow.
You then use both the thermal conduction equation and the SB equation as though they represent the same physical system and come to the ridiculous conclusion that you can radiate 557W from the “cold end” of a bar which is receiving on 2W.
The fact that this is a flagrant violation of one of the foundational axioms of physics should have been a clue that you screwed up somewhere.
Greg
”
the “cold end” being cooled by SB radiation.
”
What object is the cool end heating? None
You’re confusing a radiating potential with heat transfer.
Waving EM waves in space is not a cooling mechanism. Get another piece of matter and then you have cooling.
Good Lord. For the second time today.
I just finished whacking an idiot “economist” for his belief that throwing a wad of money at people will cause them to ignore their (unfounded, but very human, especially considering the political hacks like Nasty Nancy and Micro Mike fanning the flames) fear of contracting the coronavirus, and rush into the retail stores to buy stuff and avert a recession.
Is there such a thing as Keynesian Physics? If not, I take credit for properly naming this field of foolishness.
Energy is energy. Whether it is in the physical motion of a molecule (kinetic) or the “waving” of a photon (radiation), it is the same thing.
Energy out = energy in – energy retained*. At steady state, energy out = energy in – otherwise, an object is either gaining kinetic energy (heating) or losing it (cooling). If the energy in is radiation, the energy out is also radiation.
*Before a pedant gets around to me – yes, I know that it is actually mass/energy out = mass/energy in – mass/energy retained. But under any but extreme conditions, we can happily ignore the Book of Albert.
Zoe, a potential is not a flux. The object radiating does not care where the energy is going.
Cool. You debunked Quantum Mechanics. All photons are possible, then.
Funny, Boltzmann and Planck thought that if the separation distance was 1 meter, no 2m wavelength photon will form. But I guess that assumption was wrong and so B & P didn’t derive their correct formulas.
Remember, all photons are possible and they don’t care where their going ! LOL
B & P were wrong !!! LOL
what is a “radiating potential” anyway?
Boltzmann and Plank did not use that, they did not say you needed to have another body waiting to be warmed before black body radiation would occur. The idealised cavity is not the only thing which emits thermally generated photons.
EM Radiation is a form of energy, so transformation of thermal kinetic energy into radiation is not a “potential” it actually happens. When a photon is emitted the thermal energy ( temperature ) of the object is reduced.
“Boltzmann and Plank did not use that, they did not say you needed to have another body waiting to be warmed before black body radiation would occur.”
So what how did they measure the radiation?
Was it not with another object?
“it actually happens”
Yes, to another object, and no one had observed otherwise!
I’m a photon sitting at the cool end, waiting for something to appear so that I can leave. I think I can see something a few light years away, but I can’t risk leaving in case it’s no longer there when I get there. It might not even be there now (whatever “now” means). I’m starting to get the feeling that I’m never going to leave.
Good one.
Zoe
“You’re confusing a radiating potential with heat transfer.”
Isn’t this the fundamental misunderstanding of the whole AGW hypohesis?
Some radiation falls on stony ground.
The thing is bash-awk-wards.
σ(Th)⁴ = Th-Tc = σ(Tc)⁴
This is just wrong, because the first term is of opposite direction than the two others. You need to put in power P = σ(Th)⁴ + σ(Tc)⁴ to get the ends meet. Zoe appears to think 1360 W equals σ(Th)⁴. We didn’t say that. Conservation of energy is what the calculation is based on.
‘He believes that
Hot Side Radiation = Conductive Heat Flux = Cold Side Radiation.’
Wrong conclusion. power out = HSR+ CSR = power in = ɛ 1360W.
PS I constructed the equations but didn’t bother to solve 16th grade polynomial on pen and paper. Cool.
And where does CHF fit in?
Zoe posted: “Assuming HSR = CHF = CSR:”
In Wilis’ problem statement, there is no reason to make this assumption, and in fact it is wrong. The hot side surface receives incoming radiation, re-radiates a large fraction of that energy ON THE HOT SIDE and transmits (via conduction) a relatively small amount of that received radiation to the cold side, where is is radiated to deep space.
Uhuh,
So only CSR = CHF, HSR can remain completely disconnected?
Gordon,
“The hot side surface receives incoming radiation, re-radiates a large fraction of that energy ON THE HOT SIDE and transmits (via conduction) a relatively small amount of that received radiation to the cold side, where is is radiated to deep space.”
Then absorptivity on the hot side is NOT 0.95 or the albedo is insane.
But even if that’s not the case,
Willis has a fake 1-molecule thick sliver at 383K.
Why an object would refuse energy is a mystery!
To first-order, albedo = (1- absorptivity). Therefore, a true blackbody has a absorptivity of 1.0 and an albedo of zero. Likewise, for Earth (with clouds) the average albedo can be 0.30, meaning that 70% of incoming solar radiation is absorbed in the clear atmosphere, the clouds or at Earth’s water and land surfaces. And the Moon, having no atmosphere, has an albedo of 0.12, meaning that 88% of the incoming solar energy is absorbed directly at its surface. There is nothing “insane” about any of these albedos.
Furthermore, albedo has little to do, per se, with the emissivity of a given object’s surface . . . albedo involves INCOMING radiation from objects across any range of temperatures, whereas emissivity involves OUTGOING radiation at the specific (or relatively narrow range) of absolute temperature of the emitter.
Willis’ postulated insulated concrete block is not “refusing” energy . . . it is absorbing the 1360 watts (or 1360*0.95 watts if you want to account for, and say, absorptivity ~ emissivity, and it is thus naturally reflecting back to space 1360*.05 = 68 watts in this case).
There is no mystery here.
I don’t think this is so complicated, the total out going from BOTH sides of the cube has to equal the in coming. So there can and will be a difference in temperature on either side.
Language drift is an interesting phenomenon. I’ve always been fascinated at how the Romance languages sprang up and matured so quickly after the fall of Rome.
Is it acceptable now, in American English, to use “metre” instead of “meter”?
Really just curious.
Matthew, “metre” is definitely not the American usage. I use it, quite idiosyncratically, because otherwise it gets confused in my own mind with a meter meaning a device used to measure something. I will say that it drives my spelling checker nuts.
I believe someone once commented that the US and the UK are two countries separated by a common language …
Regards,
w.
English metre comes from french metre, which ironically can also mean a length or a measuring device. The british English use of both has the advantage of removing the ambiguity, as W. says.
Thanks for setting us such an intriguing problem.
Maybe that is why in the USA you drive on the right, and in the UK we dont…
There is a more logical answer but that can wait for another time.
There is a more logical answer but that can wait for another time.
Cutting down your ‘opponent’ with the sword in your right hand. Left hand controlling the Horse.
Contrast that to the french habit of passing on the right. Similarly derived from the days of horses and swords, their convention was to pass in a way which impeded an attack by either side, since the sword hand was not on the side of he who was passing.
British English ‘Meter’ is the measuring device (Water, Gas, Electricity quantity)
British English ‘Metre’ is a distance measurement (Km, Cm or even (converted) Miles)
To help non-native English speakers, context (and inflection) is all.
(many words sound the same … to, too, two but have very different meanings and only context separates them in speech)
3×2
Such as “micrometer” — a distance — and a “micrometer” for measuring distance. The meaning depends on where the emphasis is placed (mi and e) in the first instance, or (cro) in the second. I’ve never understood why Brits place the emphasis on “lo” in kilometer. Bad habit, I suppose.
I measured the width (dimension) to the nearest micrometer (millionth of an SI metre) using my micrometer (instrument used to measure said dimension) …
English. Madness converted to a language.
But compare “nanometre” and “manometer”.
First, It is impossible to answer the question precisely without knowing what percent of the visual hemisphere of the front side of the block is occupied by the Sun that is providing the 1360 W/m^2 of incident radiation (sunshine, as specified).
However, if we ASSUME the radiating object is far enough away that it subtends of angle like that of the Sun as seen from Earth (0.5 degrees), equivalent to 6.8×10^-5 steradians, or .001% of a hemisphere, we can basically neglect this small area and then assume the front face radiates, as does the back face, to a full hemisphere of deep space that has a uniform “background” temperature of 3 K.
Thus we have three equations to solve simultaneously in order to derived the EQUILIBRIUM temperature of the concrete block:
Letting Tf = the front side of the concrete block and Tr = the rear side of the block,
>Front face net power: net radiation power received equals power radiated back to space plus power conducted to the back face, or [(1360 W/m^2) * A] = [A*e*sigma*(Tf^4-3^4)] + Pk
>Power conducted thru block (Pk): Pk = k*A*(Tf-Tr)
>Back face total power radiated: must equal power conducted thru block at equilibrium temperatures), or [A*e*sigma*(Tr^4-3^4)] = Pk,
where:
A is the cross-sectional area of the concrete block (= 1 m^2),
e is the emissivity of the block at its radiating temperature (= 0.95)
sigma is the Stefan-Boltzman constant (= 5.7*10^-8 W*m^-2*K^-4),
k is the block’s thermal conductivity from front face to back face (= 0.8 W/m/K),and
the block surfaces are assumed to behave as ideal radiators across the spectrum (i.e., are “grey bodies”) and all angles relative to the surface.
Above equations reduce to (in units of watts):
1360 = [0.95*sigma*(Tf^4-3^4)+0.8*(Tf-Tr)]
and
[0.95*sigma*(Tr^4-3^4)] = [0.8*(Tf-Tr)]
or to the single equation: (1360/(0.95*sigma)] = [(Tf^4-3^4)+(Tr^4-3^4)].
Setting up a spreadsheet to calculate the solution of this equation across a range of plausible Tf’s, we arrive at the solution of Tr = 222.5 K when Tf = 388 K, with the conducted and rear face radiation powers equalling about 132.5 W each.
Gordon,
The total emission of the two faces at the temperatures you have calculated is: 1285 (front) + 139 (back) = 1424 watts. You have an error in your calculation.
No, using the values and precision of the units that I gave (and specifying that the heat input to the concrete block’s front face was 1360 watts, not 1360*0.95 watts to account for postulated absorptivity . . . there is some confusion on this), I get 1227.2 watts radiated off the front face, 132.5 watts conducted = 132.5 watts radiated from back side and the total of these two radiations is 1360 watts, which is the net input (that I started with). It all balances.
Please recheck YOUR calculations to make sure we are using the same constants.
Gordon, I don’t see the relevance of the steradians. It is given that there is a heat flux of 1360 W/m^2. The front face will radiate to the full hemisphere whatever the size of the source. You seem to be thinking that it is not allowed to radiate back to the sun because it’s hotter. People have been banned from this site for such silliness ( when carried to momentous extremes ).
You’re making it way too difficult. While Willis did say “…sunshine…” the remainder of the text, and the sketch, both indicate uniform radiation to me. And therefore, regardless of any “back radiation” (a mythical concept, if you have taken a heat transfer class) the net result is 1,360 W/m² incoming radiation.
The only reason “back radiation” is mythical in this thought-experiment is that there are no radiatively active gases (in fact, no gases at all) surrounding the mini Borg Cube.
On the Earth, back-radiation is real and measurable (albeit with some challenges).
Red94ViperRT10, you posted “. . . regardless of any ‘back radiation’ (a mythical concept, if you have taken a heat transfer class) . . .”
OK, what then in the full S-B equation for radiation power transfer FROM flat plate 1 of area A TO parallel flat plate 2 also of area A , P=A*e*sigma*(T1^4-T2^4), would you deign to call the power quantity associated with A*e*sigma*T2^4?
Gordon
He specified that 1361W/sq.m was entering the surface. He does not need to specify anything else about the source.
He would need to frame the problem differently if required the view of the heat source to be considered.
Sorry, RickWill, but this issue in not the energy input, which as you say Willis specified as 1360 W/m^2 (although there is some confusion if this was meant to be the total input going directly into the block’s from face, or if it the energy into the block’s face needed to be knocked down to 1360*0.95 W/m^2 to account for the block surface absorptivity).
On the other hand, one does have to account for the integrated (T^4)*dA of the space that energy is being radiated toward FROM the front face of the block for accounting. For example, if the postulated concrete block was, say, orbiting a 100,000 km brown dwarf at 100 km altitude, the back-radiation would basically toward the surface temperature of the brown dwarf (say 5,000 K), not deep space at 3 K, even though the net incoming power flux could be identical at 1360 W/m^2.
Radiant energy flux is unidirectional. It exists by virtue of THE electro-magnetic field – singular. There is no two way energy. A proper solution requires solving Maxwell’s equations for the geometry but you can get close using radiant heat transfer equations setting up the view factors then simplifying assumptions as you have done.
Sorry RickWill, here’s the thought experiment for you: there is a heat lamp shining normally on a mirror, which is 90% reflective, located five feet in front of the lamp. You stand directly behind the heat lamp (receiving no radiation directly from it), facing the mirror, and notice that your face is getting warm quite rapidly. How is this possible if radiant energy flux is unidirectional and exists solely within a singular EM field.
Also, since photons are emitted off any infinitesimally small hot surface in a random directions (at least within a hemisphere), how can “radiant energy flux” be unidirectional? Wouldn’t that violate the law that light intensity falls off as 1/(distance^2) . . . as is commonly known to be the case for sunlight?
The mirror example does not make your case. What I stated is that EMR is unidirectional at any point in space and time. You would need to look at the field that is created by the mirror in combination with the lamp and the face experiencing the EMR. The presence of the mirror simply alters the field.
Please provide experimental evidence that “photons” are emitted in random directions. EMR can only emit toward regions of lower electrical potential. That is the nature of electricity.
No such thing as EMR travelling against the electric field – always down potential. No going against potential. It is not a balance of energy flows; it is a singular energy flow from region of high potential to region of lower potential.
RickWill posted: “Please provide experimental evidence that ‘photons’ are emitted in random directions.”
There is no need to go to the lab when the very basics of the underlying physics suffices:
“A blackbody is a diffuse emitter which means it emits radiation uniformly in all
directions. . .”, and
“The emissivity of a surface is defined as the ratio of the radiation emitted by the
surface to the radiation emitted by a blackbody at the same temperature. Thus,
0 ≤ ε ≤ 1
Emissivity is a measure of how closely a surface approximates a blackbody . . .
The emissivity of a surface is not a constant; it is a function of temperature of the
surface and wavelength and the direction of the emitted radiation, ε = ε (T, λ, θ)
where θ is the angle between the direction and the normal of the surface.
The total emissivity of a surface is the average emissivity of a surface over all
directions and wavelengths . . .”
—source of above-quoted text: http://www.mhtl.uwaterloo.ca/courses/ece309_mechatronics/lectures/pdffiles/summary_ch12.pdf
Of course, you could author a ground breaking paper showing that this is all wrong and all diffuse surfaces must emit unidirectional radiation per Maxwell’s equations.
Unless you can prove it in a lab it is speculation. You cannot emit EMR against the potential field even if it is an energy quantum.
This lecture might give you more insight regarding EMR:
I look forward to seeing your peer-reviewed paper of your claims enjoy all the success that it deserves.
For those that interpret Willis’s problem setup to mean that 0.95*1360 W/m^2 is the energy absorbed on the cube’s front (“hot”) face: the answers I gave above I get changed slightly to Tr = 220 K with Tf = 383 K, with the conducted and rear face radiation powers equalling about 127 W each.
Willis,
Roy is correct about the formulation, it is two equations and two unknowns.
The first equation is the overall energy balance:
0.95 * K * Tf^4 +0.95 * K * Tb^4 = 1360 *0.95 The 0.95’s cancel, so:
K * Tf^4 + K * Tb^4 = 1360 (Eq 1)
where Tf – front face temperature, Tb is back face temperature, and K is the Stefan-Boltzman constant (5.67 *10^-8)
This allows expression of the Tb value in terms of the Tf value:
Tb = (1360/K -Tf^4)^0.25 (Eq 2)
The second equation comes from the requirement that the difference in thermal emission between the two plates must be equal to the rate of heat flux from front to back:
0.95 * K * Tf^4 – 0.95 * K * Tb^4 = 0.8 * (Tf – Tb) (Eq 3)
You substitute for Tb using equation 2 above, and get this rather messy equality expressed in Tf alone:
2*(0.95/0.8) * K * Tf ^4 – (0.95/0.8) * 1360 = Tf – (1360/K – Tf^4)^0.25
Then through the magic of Excel, you can calculate the left and right sides of this equation in two adjacent cells, and try assumed front temperatures until the value of the two columns matches. The result is (to an accuracy of 0.1 degree):
Tf = 347.5 kelvin
Tb = 312.1 kelvin
Your Eq. 3 is wrong
Sorry, typo:
2*(0.95/0.8) * K * Tf – (0.95/0.8) * 1360 = Tf – (1360/K – Tf^4)^0.25
should be
2*(0.95/0.8) * K * Tf ^4 – (0.95/0.8) * 1360 = Tf – (1360/K – Tf^4)^0.25
Fixed, Steve. I hate typos, and since WordPress doesn’t offer an option to edit a post, by default I’m the WordPress Editor.
w.
The correct values are ~383.7 hot side 219.7 cold side.
I used to know a Steve Fitzpatrick. From Pearland, TX. But seems to me someone told me he’s dead, so probably not the same guy?
Or could alter it, by putting block with not insulated side up on lunar surface at the equator and other not insulated side rests on lunar surface.
The block would warm up to about 120 C during lunar noon, and would cool down during lunar night. And evenually it would warm the lunar surface around it, as that surface is almost a perfect insulation.
And block is getting less than 1/2 solar energy {on average roughly* } than a block in space.
* actually, quite a bit less than 1/2.
That would be a completely different problem. As Willis stated the problem, cold side radiates to a perfect (which I suppose I would have to state in my assumptions, some have assumed radiating to some temperature above absolute zero) black body, while the same block on the lunar surface is either radiating to the gray body lunar surface, or even conducting to the lunar surface. And either way we could assume a lunar surface with homogeneous properties for that m² but we can be virtually certain that it is not. Let’s stick to the problem that Willis laid out, it’s simpler.
Speaking of the problem that Willis laid out, rather than the insulation of imaginary Unobtainium we could have featured an equally mythical infinite sheet of 1 m thick concrete, receiving a uniform 1,360 W/m² incoming radiant energy and we are calculating the steady-state temperatures of the center (how does one measure the center of an infinite anything, because by the definition of infinite, I could move over a meter or two or ten or even a million and still have the identical infinite expanse on all sides… but I digress…) m², the calculation is the same, the point is we don’t have heat wandering off to the left or right or wherever, there’s only one direction it can go, which makes all the heat flux exactly parallel to any other heat flux.
But even more importantly, Willis you have specified “…steady state…” but if there is a flux, something is moving, so that’s not steady state, is it? Well, in Fluids we assume steady-state flow all the time, which typically just means that it has settled down to either laminar flow or turbulent, and even if it’s turbulent we still call it steady state, even though in turbulent flow the direction(s) of the fluid is(are) constantly changing. So I could support your answer. Now I really need to dig out my Heat Transfer book!
If then block is perfectly insulated on all sides (as the problem states) Then all radiation ingoing and outgoing is on the front face. There is no radiation loss on any of the other faces. That means the front and back side of the block are at the same temperature!
Jim, it’s insulated on four sides, with the front and back uninsulated. Sorry for the lack of clarity.
w.
excuse me, I’m not a scientist and I have never met Fourier or Stefan-Boltzmann or taken second course. But, if emissivity and absorptivity are equal, then hot and cold side temps are equal. After reaching whatever equilibrium temp is absorbed by 1360 w/m3, the back side emits the same. If the emissivity were less, then heat would reflect back out the hot side. If emissivity were greater, there would be a temperature difference, and a flow from high to low.
thot = tcold
Bonus points for extra credit, temps at the sides under the unobtanium are equal to thot and tcold
Nope.
You are leaving out what happens to the hot side when it heats up.
It emits radiation, as all objects do that are not at 0°K
I protest, it is far too difficult:-)
I start with two simplified models.
If the backside had also been insulated, it would have been easy:
Energy in = 1360 * 0.95 = 1292 W.
By Stefan Boltzman we get T = 4th root(1292*SBconstant*0.95) = 394 Kelvin
And without insulation on the backside, but if the concrete had been superconductive, it would also be quite easy. The temperature would then be equal on both sides, and the emitted energy on each side would have been 1292 /2 = 646 W.
i.e T = 4th root(646*SBconstant*0.95)= 331 Kelvin
You have a problem with a concrete block with final conductivity. That is mean. 🙂
But now we have the extremes, so we know that the answer must be somewhere between there
From the above we know that the hot side has to be colder than 394 Kelvin since some of the energy goes through the concrete. And it has to be hotter than 331 Kelvin since the finite conductivity in the block now hold some energy back.
We also know that the cold side has to be cooler than 331 Kelvin, since it will receive less energy than from a superconductive cube.
I leave the rest to others of my fellow nerds out there. 🙂
/Jan
Some quick thoughts:
Let Tw be the temperature of the warm side, Tc that of the cold side. For the warm side, heat flux in from the Sun is e * Qs, where e=0.95 is the absorptance, same as emissivity in this case. Radiated flux out is e*sigma*Tw^4. In addition, there is a conducted heat flux of k(Tw-Tc)/L, where L is the depth of the cube (1 meter). Equilibrium at the warm side demands:
e*(Qs-sigma*Tw^4) – k(Tw-Tc)/L = 0.
For cold side, the radiative heat flux in is e*sigma*TCMB^4, where TCMB is the temperature of the cosmic microwave background. The flux is ~3*10^-6 W/m^2 and could be disregarded. Equilibrium at the cold side demands:
e*(QCMB-sigma*Tc^4) + k(Tw-Tc)/L = 0.
This is a pair of 4th order polynomials, and a numerical solution is easiest. Using e.g. Mathematica it is easy to see that there is one real solution with both Tw > 0 and Tc > 0. With numerical values substituted, we get Tw ~ 383 K and Tc ~ 221 K, assuming I didn’t typo anything. The temperatures seem high, but not altogether unreasonable. Increasing k yields Tw ~ Tc ~331 K as it should, since now the two sides must be at equal temperatures. Decreasing it gives asymptotically Tw ~ 393 K, Tc ~ 2.7 K. In this limit, both sides are completely uncoupled, and at least the cold side temperature is reasonable, since it is just the CMB temperature we put in originally.
Here is my solution:
In the state of equilibrium, the heat flow through the cold side is the same as through the warm side. The temperature Tspace of the “outer space” is assumed to be 0 Kelvin. The calculation of the equilibrium temperature of the cold side of the cube is then as follows:
HeatFlux = epsilon * sigman * Area * (Tcold ^ 4 – Tspace ^ 4)
1360 = 0.95 * 5.67E-8 * 1 * (T_cold ^ 4 – 0)
T_cold = (1360 / (0.95 * 5.67E-8)) ^ (1/4)
T_cold = 398.62 K
The temperature difference from T_cold to T_hot:
HeatFlux = k * Area * (Thot – Tcold) / d
1360 = 0.8 * 1 * (Thot -Tcold) / 1
1360 = 0.8 * (Thot -Tcold)
Tdelta = Thot -Tcold = 1360 / 0.8 = 1700K
Thot = T_cold + Tdelta
Thot = 398.62 + 1700
Thot = 2098.62K
But there is a problem:
The heat flow of 1360W requires a certain temperature on the warm side. The warm side should be in radiation equilibrium with the sun. If the surface temperature of the warm side deviates from the equilibrium temperature, the heat flow changes, which is necessary for the radiation balance.
For the solution to be correct, the temperature, the size or the distance of the sun would have to adapt to the problem!
You can’t just set a certain heat flow. The heat flow results from the interaction:
Sun heat radiation Thot heat conduction Tcold heat radiation Tspace
I forgot the outgoing radiation from the warm side.
I thought the 1360W is the heat flow into warm side.
Using a solver and the Stefan – Boltzmann constant = 5.67037E-8
I get 383.2797K for Thot and
Tcold = 221.4225K
The way the problem is framed shows 1361W/sq.m entering the hot face. Hence your first answer is correct.
To arrive at any other answer requires many assumptions or more information. For example, the surface temperature of the heat source; the emissivity of the heat source; the orientation of the block relative to the heat source; the size of the heat source; the distance of the heat source to the block of concrete.
The concrete cube has a radiating surface area of 2 square meters.
One square meter at front and back.
The cube will reach thermodynamic equilibrium eventually, with 680 watts per square meter IR being emitted from both front and back.
The cube could be reduced in thickness with same result, so for simplicity you could say there is a thin slab of concrete with same face areas but with thickness resulting in a mass of 1 kilogram
Using https://www.omnicalculator.com/physics/stefan-boltzmann-law
and the concrete emissivity of 0.91 the two square meters of concrete will have a temp of 338.7 Kelvins
(That’s 65.7 degrees celsius)
If the slab starts at 100 Kelvins, with 1000 J/kg-k specific heat capacity, it would take about 238700 joules to heat the slab to equilibrium temperature. 238700 divided by 1360 J/sec equals 175 seconds.
Roughly 3 minutes.
There are about 2400 kg slabs in the cubic meter, so the energy needed to heat the cube from 100K to 338.7K is about 2400 time 3 or 7200 minutes.
A real world heating time is much longer, due to IR loss from the front face almost immediately after heating begins.
bwegher:
Your error here is that you assume infinite thermal conductivity (zero conductive resistance in the cube, where the problem states a finite (and rather low) conductivity value.
Thermal conductivity is not involved.
The entire mass of concrete is at thermal equilibrium with the surrounding space.
Both sides have the same temperature.
There is no “cold” side, the IR radiates isotropically, 680 watts from both front and back.
Go to the SB calculator and enter 2 square meters, concrete (emiss 0.91) and 1360 watts
Calculated temperature is 338.8 Kelvins.
If you calculate each side at 680 watts per square meter, you get the same result.
For example, with the concrete reduced to 0.1 millimeters thickness (paper thin)
the volume of the concrete is now 0.0001 cubic meters, with a mass of 240 grams.
To raise the temperature of concrete from 100 K to 338.8 K with 8.80 Joules per gram per kelvin.
240 grams times 8.8 is 2100 Joules.
2100 divided by 1360 Joules per second equals 1.5 seconds.
The face of the concrete will take 1.5 seconds to reach equilibrium temp.
The next slice will take longer since only 680 Watts are available in that direction, the face of the first slice is emitting 680 watts of IR back to the source.
So 3 seconds times 9999 takes total of 29997 seconds to raise the 2400 kilograms from 100 K to 338.8 K. Rounding off to 30000 seconds that converts to 8.33 hours.
That’s just a first approximation
After at least 8.33 hours the cube will be emitting 680 Watts of infrared energy from both faces.
In a day, the full cube will have a steady temperature of nearly 66 degrees celsius.
Of course thermal conductivity is involved!!!
Like so many others, you confuse static equilibrium (no power transfers) with dynamic steady-state conditions (constant power transfer). This problem is clearly of the second type.
In this dynamic steady-state condition, the cube is gaining energy from one end, and losing energy from the other end. Of course these ends will be at different temperatures, and there will be conduction between the ends.
If you were asked about the steady-state conditions of a long steel bar with one end in boiling water (100C) and the other end in ice water (0C), would you conclude that the bar would be isothermal in the steady state? That is in effect what you are arguing!
This is a perfect example of what I just said about translating text and using logic to set up what must be solved for.
Clearly Willis stated the conductivity of the block.
Clearly this is a limiting factor…a bottleneck.
And this has everyday real world analogous situations.
If one is in a metal box, one will roast to death in the Sun.
But inside a concrete walled box a meter thick, it will be no hotter in the day than at night, and at a certain thickness, the temp will not even vary much over a year…like being in a cave.
Stone castles are a more even temp than metal sweatboxes.
No wonder the world is in trouble.
Erratum. Concrete specific heat capacity is 0.88 Joules per gram K
.88 times 240 grams is 211 Joules K
339-100=239 Kelvins times 211 is 50429 Joules
50429 divided by 1360 J/s equals 37 seconds to reach equilibrium
74 seconds for the next 9999 slices is about 740000 seconds.
That’s 205 hours or 8.6 days for the full 2.4 tonnes concrete to reach equilibrium
(By the time I was ready to post this, Willis and Roy had done basically the same thing. But hopefully my step-by-step explanation will still be useful.)
Ahhh, a classic introductory thermal sciences problem! Let’s go through it step by step.
In steady-state conditions, Qin = Qout.
Qin = 0.95 * 1360 = 1292 W
Qout only comes from radiation from the sun-facing surface (Q1) and the back surface (Q2).
Qout = A * e * sigma * T1^4 + A * e * sigma * T2^4 = A * e * sigma * (T1^4 +T2^4)
A = 1 m^2, e = 0.95, sigma = 5.67E-8 W/m^2/K^4
(I’m using T1 for Willis’ T_hot, and T2 for T_cold.)
Qout = Qin = 1292 W = 1 * 0.95 * 5.67E-8 * (T1^4 + T2^4)
Now for the conduction. In steady-state conditions, the Q2 output must be exactly balanced by the Qcond coming through the cube.
Qcond = A * k * (deltaT/deltaX) = 1 m^2 * 0.8 W/m/K * (T1 – T2) K / 1 m
So Qcond = Q2, and 0.8 * (T1 – T2) = 0.95 * 5.67E-8 * T2^4
I’m a bit rusty on solving general 4th-order equations in closed form, so I iterated and got:
T1 (T_hot) = 383.29 K (= 110.14C)
T2 (T_cold) = 221.42 K (= -51.73C)
To check, I get:
Q1 = 1162.5 W
Q2 = 129.5 W
Qcond = 129.5W
So Q1 + Q2 = 1292 W, matching the incoming solar power.
And Qcond = Q2, balancing the cold-side power.
Using T of deep space = 3K and solving heat transfer for the two faces per Dr. Spencer:
T(hot): 383.6K
T(cold): 221.5K
First, define some terms:
ε = emissivity = 0.95 (constant for all frequencies)
I = incoming absorbed power = 0.95×1360 W = 1292 W
T = temperature (all temperatures in Kelvin)
E = radiative emissions (in Watts)
T_hot = temperature of front (hot) surface
E_hot = emissions from front
T_cold = temperature of back (cold) surface
E_cold = emissions from back
Note: due to the unobtanium sides, there’ll be no net lateral flow of heat across the front or back surface, so the front and back surfaces will both be at uniform temperatures, though not at the same uniform temperature.
Stefan-Boltzman relation:
E = ε × σ × T⁴
σ = Stefan-Boltzman constant = 5.670374419E−8 W/m²T⁴
Simplifying:
s = ε × σ = 0.95 × 5.670374419E−8 = 5.38685570E−8 W/m²K⁴ for our grey-body.
Simplified Stefan-Boltzman relation for our grey-body:
E = s × T⁴
That’s true at both the front and the back surfaces, so:
E_hot = s × T_hot⁴
E_cold = s × T_cold⁴
At equilibrium:
I = E_hot + E_cold = 1292 W
Thermal conductivity “k” of the concrete is equal to 0.8 W / m / degree K, so energy flow from front (hot side) to back (cold side) is:
0.8 × (T_hot – T_cold) W
But at equilbrium, that must be the same as the radiant energy leaving at the rear, so we don’t need a new variable for it, it’s just E_cold:
E_cold = 0.8 × (T_hot – T_cold) W
At equilibrium the power entering and leaving the front are the same, and the total energy radiated and absorbed by the cube are the same, so:
E_hot + E_cold = 1292 W
E_hot = s × T_hot⁴ = 5.38685570E−8 × T_hot⁴
E_cold = s × T_cold⁴ = 5.38685570E−8 × T_cold⁴
We now have a set of four simultaneous equations to solve, for four variables:
E_hot + E_cold = 1292
E_cold = 0.8 × (T_hot – T_cold)
E_hot = 5.38685570E−8 × T_hot⁴
E_cold = 5.38685570E−8 × T_cold⁴
Renaming, for convenience of an online simultaneous equation solver:
u = E_hot
v = E_cold
x = T_hot
y = T_cold
The equations are:
u + v = 1292
v = 0.8 × (x – y)
u = 5.38685570E−8 × (x^4)
v = 5.38685570E−8 × (y^4)
Most online equation solvers like their equations separated by commas, so:
u + v = 1292, v = 0.8 × (x – y), u = 5.38685570E−8 × (x^4), v = 5.38685570E−8 × (y^4)
But this stumped:
https://www.emathhelp.net/calculators/algebra-2/simultaneous-equations-solver/
https://www.symbolab.com/solver/system-of-equations-calculator/
https://quickmath.com/webMathematica3/quickmath/equations/solve/advanced.jsp
Rats. Okay, before I write code to solve this numerically, perhaps someone here knows of a web site which can do it.
You can do it in Excel using the solver. Just tell it to minimise the sum of the squares of the two surface flux imbalances by varying the two surface temperatures. It’s probably also good to add the constraints that the absolute temperatures must be non-negative.
I’ve never learned to use Solver in Excel, but I put the equations into Excel, and defined two error terms, and then manually tweaked the two temperatures until the error terms were very small.
Here’s the spreadsheet, exported as a web page:
https://sealevel.info/hot_and_cold_of_space.html
That should load directly into Excel, but in case it doesn’t, here’s the .xlsx file:
https://sealevel.info/hot_and_cold_of_space.xlsx
Call this the result of a “manual heuristic search”:
T_hot ≈ 383.28 K = 110.13 C
T_cold ≈ 221.42 K = -51.73 C
…and I see that Ed Bo & Rick C PE both beat me to it!
Willis got slightly higher numbers, perhaps because he assumed that all 1360 W were absorbed, instead of 5% being reflected away.
Those are the same numbers that I got, Dave.
Regarding Solver, definitely worth learning. First you have to go to “Add-ins” on the “Tools” menu. Add “Solver” to your menus.
Then you need to set a cell to measure the error, usually the sum of the squares of the distance between the desired and the calculated answers. For more than one variable, you can use the Excel function =SUMXMY2().
Then select your error cell, and choose “Solver” from the Tools menu. Tell it the variables to change, and that you want to minimize the error term. Tell it to calculate, and you’re off. Hang on … OK, I just set up a demo spreadsheet, it’s available here. The variables to change are in yellow, and the error term is in green. Start by changing the yellow cells (which currently have the correct answer) and then run Solver to solve the problem.
Best regards,
w.
Oops, that was Roy, not you. (BTW, both of you should consider my confusion a compliment!)
Hey, thank you very much for the quick Solver lesson! That’s very slick, and very useful, and your instructions were perfectly clear and concise.
Dave Burton February 28, 2020 at 2:42 pm
You’re more than welcome. I see part of what I’m here to do on this planet is to give away whatever knowledge I have … here’s a post on that subject from my own blog, where there’s lots of interesting lunacy, including the story of a most curious oceanic voyage …
w.
If today weren’t a day of mourning, I’d have written that I’m having fun with this. But the news from Princeton puts a damper on it.
Even so, I played with Solver in my antique MS Excel 2003 for a while, and it works very nicely, but I couldn’t figure out how to coax it into running a few more iterations to get maximum precision. Unsettlingly, using Solver on your spreadsheet yielded slightly different results than running it on mine.
So I tried LibreOffice 6.2 Calc, and I’m quite impressed. You have to go into options and select the “LibreOffice Swarm Non-Linear Solver (experimental)” solver engine, but after you do that it apparently runs until it has squeezed out the last drop of precision. It is noticeably slower than Microsoft’s Solver, but its answer is much more exact.
Here’s your spreadsheet, optimized by LibreOffice Calc’s Non-Linear Solver (and I added some more digits of precision to the S-B constant):
https://sealevel.info/WUWT_Puzzle2.html
https://sealevel.info/WUWT_Puzzle2.ods
https://sealevel.info/WUWT_Puzzle2.xlsx
Here’s mine:
https://sealevel.info/hot_and_cold_of_space4.html
https://sealevel.info/hot_and_cold_of_space4.ods
https://sealevel.info/hot_and_cold_of_space4.xlsx
Reassuringly, using LibreOffice Calc’s Solver, the two spreadsheets now yield results (temperatures) which are identical out to a ridiculous nine digits of precision:
T_hot = 383.279628 K
T_cold = 221.422450 K
Try Wolfram Alpha. It will solve your problem!
x = 383.28
y = 221.42
Sure ‘nuf:
https://www.wolframalpha.com/input/?i=u+%2B+v+%3D+1292%2C+v+%3D+0.8+%C3%97+%28x+%E2%80%93+y%29%2C+u+%3D+5.38685570E%E2%88%928+%C3%97+%28x%5E4%29%2C+v+%3D+5.38685570E%E2%88%928+%C3%97+%28y%5E4%29
Or, adding the “>0” constraints:
https://www.wolframalpha.com/input/?i=u+%2B+v+%3D+1292%2C+v+%3D+0.8+%C3%97+%28x+%E2%80%93+y%29%2C+u+%3D+5.38685570E%E2%88%928+%C3%97+%28x%5E4%29%2C+v+%3D+5.38685570E%E2%88%928+%C3%97+%28y%5E4%29%2C+u%3E0%2C+v%3E0%2C+x%3E0%2C+y%3E0
Thanks!
In the past I’ve given thought to similar problems, along the lines of what would a mercury thermometer read in such a situation? And would it be different if I painted the bulb with a different color paint?
But that’s why I refuse to get involved in some of the discussions/arguments about the global warming thing here at WUWT and elsewhere. I can see that the climate models simply don’t work. That is the best starting point of attack. I think even Mosher said something similar a few years ago.
Willis ignores all arguments and just reasserts his faith that CHF = CSR.
His HSR is fake and only used as a meaningless variable to derive his CHF = CSR.
Come one Willis, Surely HSR MUST = CHF too.
WHY NOT???
If you can’t answer this, then concede, if you have integrity.
Zoe, as I said before, please STOP YOUR ASININE PERSONAL ATTACKS. I tell the truth as best I know it. I say I’m wrong when I’m shown to be wrong. Claiming I have no integrity if I don’t do exactly what you want is just you being ugly.
I do say that the heat flow THROUGH the concrete from the hot side to the cold side has to equal the heat flow OUT of the cold side. It’s like a garden hose. The flow rate out of the end of the hose has to be exactly the same as the flow rate through the inside of the hose. This is not contested by anyone I know of … except you.
My mom used to say:
Here’s the rude truth. The fact that I or anyone else disagrees with you doesn’t mean that we’re operating on faith or that we lack integrity. If you look through this thread, you’ll find that not one person agrees with you. Do they all lack integrity?
Nope. THEY JUST DISAGREE. It’s what people do, it’s what scientists do. We disagree. Doesn’t mean we’re bad people. Please stop trying to make it personal.
w.
Stop playing the victim, Willis.
“I do say that the heat flow THROUGH the concrete from the hot side to the cold side has to equal the heat flow OUT of the cold side.”
Yes, I see that you set flow thru = flow out
And what about the flow in = flow thru
“It’s like a garden hose. The flow rate out of the end of the hose has to be exactly the same as the flow rate through the inside of the hose. ”
And what about the flow INTO the hose being equal to what’s flowing through?
You’re not going to escape this and you’re not going to play the victim.
“Play the victim”? STOP WITH THE AD HOMINEMS! Man, you got a nasty tongue on you.
Next, the flow into the hose is NOT the same as the flow through the hose, and the same is true about the concrete block.
Why? Because some is energy is radiated from the hot end of the block, and thus it doesn’t pass through the block. 1,292 W/m2 is absorbed by the hot end. 1,162 W/m2 is radiated from the hot end. The remainder, 130 W/m2, passes through the block and is radiated out the cold end. Steady-state.
To continue the hose metaphor, it’s as though the hose is not screwed tight to the faucet, so some leaks out. So flow in minus leakage at the faucet is indeed equal to flow through the hose.
As to me going to “escape this” … get real. We’re not playing that game. It’s a discussion. Stop the personal attacks. You’re not doing your reputation any good.
w.
Willis,
“Why? Because some is energy is radiated from the hot end of the block, and thus it doesn’t pass through the block. 1,292 W/m2 is absorbed by the hot end. 1,162 W/m2 is radiated from the hot end. The remainder, 130 W/m2, passes through the block and is radiated out the cold end. Steady-state.”
Yup, just as I predicted.
Your hot side is fake hot side that just constitutes a tiny sliver. That tiny sliver for some can’t transfer more heat to the colder end, but is required to send most of its heat back to the source. Thus making the source even hotter. Right, Willis?
A real object at 383K can transfer 383K worth of energy all round, but not your stupid tiny sliver.
“the flow into the hose is NOT the same as the flow through the hose”
Cool self-serving assertion. Why not? Are you saying the water will flow back to its source. Funny, then you only ever had the NET of water coming in. Cool, then you don’t really have 383K, you only have the NET 130 W/m2 = 221K.
Zoe, does it even occur to you to consider the actual properties of a block of stone?
We all have seen, touched, and experienced stone and it’s various properties.
If what you think, and propound in your blog example that started all of this, the inside of a stone castle sitting in the Sun could be used as an oven.
Two watts propagating through the thick stone walls would lead to a vast outpouring of energy inside the walls.
Stone exists in real life…it does not behave as you insist it does in your made up nonsense world.
Zoe Phin wrote, “Willis ignores all arguments and just reasserts his faith that CHF = CSR.”
Zoe, you are as confused as you are rude.
At equilibrium, hot-to-Cold side Heat Flux CHF (that is, heat conducted to the cold side from the hot side) must equal Cold Side Radiation CSR, because if it didn’t then the cold side temperature would be changing.
Zoe Phin wrote, “His HSR is fake…”
No, it is the radiation emitted from the hot side of the tiny Borg Cube (not counting the 5% of incoming radiation which is reflected).
You do understand that a hot grey-body in space will emit radiation, don’t you? And you understand that the hotter it is the faster it’ll emit radiation, right?
Zoe Phin wrote, “Surely HSR MUST = CHF too. WHY NOT???”
No, they are not the same.
The Hot-Side Radiation HSR is governed entirely by the hot side temperature (given that the emissivity is stipulated to be a fixed 0.95 across the entire ER spectrum).
The hot-to-Cold side Heat Flux CHF is governed by the thermal conductivity (a property of the substance from which the cube is made, stipulated to be 0.8 W / m / degree K), and by the temperature difference between the two sides of the Cube.
The only way that CHF could be as high as HSR is if the thermal conductivity of the mini Borg Cube were infinite, instead of 0.8 W / m / degree K.
Dave,
Willis already admitted:
“Why? Because some is energy is radiated from the hot end of the block, and thus it doesn’t pass through the block. 1,292 W/m2 is absorbed by the hot end. 1,162 W/m2 is radiated from the hot end. The remainder, 130 W/m2, passes through the block and is radiated out the cold end. Steady-state.”
An object that rejects 1162 of 1292 W/m2 is not an object with an absorptivity of 0.95. But one with 0.1.
Willis wants to eat his cake and have it too.
Zoe, you are conflating reflection and radiation.
A solid object with an absorptivity of 0.95 REFLECTS 5% of the incoming energy. However, other than that they both involve emissivity/absorptivity, that has nothing to do with how much it radiates. That’s determined solely by
sigma epsilon T^4
w.
Willis,
I thought you’d play this game.
An object that absorbs 1,292 W/m^2 is an object that has molecules moving at 383K equivalent. These molecules WILL bump into neighboring molecules, and so on. But you reject this, and force evacuation of 1,162 W/m2. Your molecules can only bump their neighbors at 130 W/m2 / 221K. Your Hot Side Tempetature is a fake and fraud, and is NO DIFFERENT than had it rejected 90% of incoming energy.
I went overboard and gave you credit for having a 1-molecule thick sliver at 383K, but the truth is that even they don’t exist. They can’t bump their neighbors at 383K, because they must shed 90% of their energy back to the source.
They only have 221K of bumping power, so are they are 383K or 221K ? Come on, Willis, engage !
I need to go buy popcorn before reading any more of Zoe’s dumb azz insulting full-retard nonsense.
Her words seem to suck information out of the Universe, like a text version of a CNN broadcast.
I suspect “Zoe” is some kind of AI robot, pulling technical sounding terms from WikiPedia and constructing sentences using them in a semi-random way. A bit like the way they write fake scientific papers to spook publishers.
That may explain why every reply she give is totally unrelated to the previous question and just starts another unrelated discussion.
Zoe, why are stone houses more comfortable on a hot sunny day that inside a metal walled shed?
What do you feel if you stand in front of a stone or concrete block wall that has been in the Sun all day?
This is not some impossible hypothetical. In fact it is closely analogous to what happens all ove rthe planet and the solar system alll the time, wit the exception of the insulated sides, but there is nothing technically challenging about that. Roughly the same results could be had if instead space shuttle thermal tiles were glued to four sides of the stone cube.
You know those tiles? The ones that get heated to thousands of degrees from the friction of re-entry, and conduct so slowly that the bottom side of the times is not even very hot.
All of the Apollo missions used a capsule with a heat shield as well, but one shaped like a saucer.
Those heat shields get hot and radiate away the heat as they are heating, and then until they have cooled to ambient temp.
You are so focused on believing that photons cannot radiate towards a source of heat has your mind in loopy hallucinations of cognitive dissonance.
I can tell you it is something to behold.
A person incapable of basic logic and reasoning because one preconceived notion must be protected at the expense of rationality if that is what it takes.
” … because they must shed 90% of their energy back to the source.”
—
Oh my goodness, you keep saying stuff like that! No, the energy does not go “back to the source”, it radiates in all available directions, no matter the re-emission temperature. Plus at the shaded end of this block you act like emission doesn’t or can’t occur into space alone, in all available directions, in thermal IR. No, it does not need a colder target object to emit toward, just like the illuminated side does not send radiation back toward its illumination source.
Sunlight striking the surface rocks of the moon and reflecting as ‘Moonlight’ toward the earth obviously did not go, “back to its emitter”. Nor if the sunlight was absorbed by surface rocks and re-emitted later, as IR, for the Moon has an IR output, detectable on earth or in orbit, which IR re-emission did not go, “back to its emitter”.
So drop this nonsense about:
” … must shed 90% of their energy back to the source.”
That does not happen.
Note how Zoe ignored all my questions related to a stone wall sitting in the direct sunshine.
The outside of the wall towards to the Sun gets hot.
The side of a 1 meter thick stone wall that is in the Sun all day will never get hot, even if it is in the Arctic in Summer and the Sun is shining on it all damn day long fro six months in a row.
I feel like going to the Arctic, or Antarctic, and building a one meter think stone wall just to shut her up!
The side of the stone that is one meter from the sunny side will never get hot because heat conducts very slowly through stone.
And the sunny side does not keep getting hotter until it melts because it radiates heat away far too fast for 1200-1300 watts to ever get it to it’s melting point.
In fact, anyone can do this on a bench top using a source of thermal radiation and a block of stone with insulation on four sides.
While the source of radiation is still turned on, anyone can place their hand a few inches from the heated side and feel the heat radiating off the surface.
A FLIR camera can see it.
Nine million FLIR cameras at ever possible angle can see it.
An infrared thermometer can measure it.
And after a year the back side will still only be as warm as thermal conductivity allows for energy to travel through the block of stone.
Notice how Zoe ignores all of my questions except the one in which she could point out she wears expensive undergarments.
Prove it, Zoe.
Zoe, the cube does not “reject” (whatever that means) 1162 of 1292 W incoming radiation.
1360 W of incoming sunlight are incident upon the cube. Of that 1360 W, 1292 are absorbed, and the remaining 68 W are reflected.
Once equilibrium is reached, the block also radiates 129.48574 W from the cold side, and 1162.51426 W from the hot side. But that does not represent incoming radiation “rejected” by the cube. The two sides of the cube have different emission spectra, and both of their emission spectra are very different from the spectrum of the incoming solar radiation.
Dave,
That’s not what Willis said AND
Now you have 1162 – 129 =
1033W flowing “through”
1291W flowing “in”
128 W flowing “out” @ur momisugly CSR
1162W flowing “out” @ur momisugly HSR
But Willis said that what flows through must be what flows out.
1033W != 129W
His water analogy is crap?
Zoe, what Willis meant is that 129.48574 W is thermally conducted from the hot side to the cold side, and at equilibrium the same amount of power, 129.48574 W, is radiated from the cold side to space.
The remaining 1162.51426 W (of the 1292 absorbed W) is radiated from the hot side to space.
That’s because at equilibrium:
1. the rate at which energy arrives at the cold face (thermally conducted through the Cube from the hot face) MUST be exactly equal to the rate at which energy leaves the cold face (radiated to space). And,
2. the rate at which energy leaves the entire cube (from the cold face + from the hot face) MUST be exactly equal to the rate at which energy is absorbed by the cube (1292 W). And,
3. the rate at which energy is absorbed at the hot side (1292 W) MUST be equal to the sum of the rates of the two ways in which energy leaves the hot side: 1162.51426 W radiated to space from the hot side + 129.48574 W thermally conducted away from the hot side.
Do you understand, now?
Dave,
“Zoe, what Willis meant is that 129.48574 W is thermally conducted from the hot side to the cold side, and at equilibrium the same amount of power, 129.48574 W, is radiated from the cold side to space.”
But what is conducted to the middle of the block?
I just want acknowledgement that his water analogy is crap.
Zoe asked, “what is conducted to the middle of the block?”
129.48574 W.
129.48574 W is also conducted from the middle of the block.
(It doesn’t matter where along a pipe you measure the flow, the answer is always the same.)
Also, BTW, though you didn’t ask, the temperature at the center of the block will be 302.351039 K = 29.201039 C.
As I pointed out on your discussion thread, that is simply the conservation of energy. Willis has “faith” in the conservation of energy. That seems a pretty good basis for his “belief”.
You waffled about “subtypes” but where not able to say what other energy was being ignored if the conducted flux is not equal to radiated flux as you try to insist.
I think I asked you that 4 times. You just keep ignoring the question.
In Willis’ world thermal equilibrium is impossible. Seriously!
Can he suppose one problem where thermal equilibrium is achievable outside of 0 kelvin?
No he can not.
I feel like I’m in an episode of Black Mirror.
Zoe Phin February 28, 2020 at 12:23 pm
Zoe, I have no idea why you’d think that. This whole puzzle is about what happens at steady-state, after the temperature of the block has completly equilibrated with its surroundings.
So far you’ve seen more than a dozen folks agree that the answer is that steady-state occurs when the hot side is around 383K and the cold side is around 221K, depending on your exact assumptions about emissivity.
w.
Willis,
You’ve redefined thermal equilibrium to mean preserved heat flow, rather than delta T accross an object = 0.
Congratulations! Maybe you also think there are 70 genders? Why not? Redefining words and noun phrases is OK, right?
Actually, Zoe, I never used the term “thermal equilibrium”, so I couldn’t possibly have “redefined” it.
I was very clear that what I was describing is called a “steady-state” condition. This is a condition where after some period of equilibration with surroundings, temperatures no longer change.
Finally, accusing me of thinking there are 70 genders? IT’S A DISCUSSION ABOUT PHYSICS! Stop with the personal accusations.
w.
“after the temperature of the block has completly equilibrated with its surroundings.”
The cold end is “surrounded” by the hot end, but it refuses to accept energy from it because it has to send HEAT to space, a substance that doesn’t accept HEAT in order to cool the hot end.
Remember Newton, Willis?
He said an object in motion stays in motion unless acted upon by an external force.
Can you name the scientist that first invalidated this and confirmed that molecules in motion reduces their motion by exerting their own force (EM radiation)?
Zoe Phin February 28, 2020 at 2:08 pm
I can’t. My head explodes. I just can’t. Do I remember Newton? Can’t be Isaac, wrong field of physics. Huey Newton? Fig Newton? Gravitational Constant of 6.67430×10−11 Newton⋅m2/kg2?
Pass … I can’t.
w.
Zoe what the hell are you talking about?
Every comment you make is incoherent.
Maybe if you stopped making stuff up that no one said and inserting those words into other people’s mouths, you could focus on the logic and the scientific principles involved in the discussion.
Why are the people who act like you always the ones who do not seem to have any idea what they are talking about, but act like they are the only ones who know anything?
I should just ignore you like most others have…but I am not able…you are incredible rude and irritatingly obtuse.
Just give up and ignore her Willis you can’t win this, it is because classical physics breaks down there is no actual classical physics that heat should flow. It’s an observed effect from a physics outside classical physics. Temperature is a made up observational statistic there isn’t anything fundemental about it.
It’s the same problem with trying to argue with someone that there can’t be a point on earth that gravity doesn’t exist. All the classical gravitational laws are observational there is no way to actually prove it.
Any physics uni student who knows the basic background can troll indefinitely because they know there is no actual proof you can use, they just have to keep shifting the background which is what she is doing.
Initially I thought she was just being dense but you look when I asked her about instantly exploding points on the block. She realizes she is in trouble via observation because the law of averages says if things aren’t forbidden they will occur and so we should have instances of instantly exploding pieces of the block where heat decided to pool. She would then be in the situation of actually having to explain why something doesn’t happen rather than just deflect it.
She is a troll it is best to just ignore her and there is no “proof” you can use it’s just classical physics at it’s finest.
LdB February 28, 2020 at 11:33 pm
LdB, I’m not writing to try to convince or educate her. I write for the lurkers. For every commenter I figure there are ten or more lurkers who read but rarely write. They are not set in their ways as she is, and they are not immune to facts as she is. I write to dissuade them from believing her nonsense.
Regards,
w.
I fear all you are doing is giving her the limelight which is all she did it for in the first place. You get this a bit with uni students when they work out just how broken classical physics is and they can play games with those less educated.
She is a young student because she fell into 2 traps
1.) With the cold end she made the mistake of wanting to have matter that sets up the problem that the cold end had to know matter existed. She has not dared touch that anymore because it ends the game.
2.) As per above if you don’t have a gradient then the energy can clump anyway it likes so there is nothing to stop it melting pieces off. Her only answer she could give was she wanted to see data.
Personally my advice ask Anthony to delete the article and deny her the two minutes of fame.
Willis:
Like many over at PSI, Zoe confuses the case of static equilibrium (no power transfers) with dynamic steady state conditions (constant power transfers). This means that she, like the others, goes off the rails before she even gets going.
To anyone familiar with basic thermodynamics and heat transfer, it should be obvious that this problem is the second case of dynamic steady state.
I’ve said it many times: reading crackpot sites like PSI (or Alex Jones, Sott, Slanderin’ Sou from Bundangawoolarangeera, David “Avocado” Wolfe, etc.) actually seems to kill brain cells.
(However, I might be confusing cause and effect.)
It’s a resistor
Sorry but I haven’t read all comments, so I don’t know if anyone has already said this. And Willis, you have triggered a lot of comments with this one.
The outward radiation on the sunny side. plus the transmission rate through the block, must equal incoming radiation. We’re assuming no reflection, presumably.
The rate of transmission through the block is proportional to the temperature gradient.
The sunny-side temperature and the temperature gradient give you the temperature at the dark side, and the outward radiation there must equal the transmission rate.
Three linear equations for three variables. All you have to do is put in the numbers.
One interesting thing is that – depending on the parameters – the dark side doesn’t have to be very cold. The higher the conductivity of the material, the cooler the sunny side is and the warmer the dark side is.
My solution:
– Cold face: – 53.05°C (219.95 K)
– Hot face: + 108.16°C (381.16 K)
Checking:
– Heat radiated cold face: 0.95*5.8E-8*(219.95^4) = 128.96 W
– Heat radiated hot face: 0.95*5.8E-8*(381.16^4) = 1163.05 W
– Heat tranmitted by conduction: (108.16 + 53.05)*0.8 = 128.97 W (= heat cold face)
– Total heat radiated: 128.96 + 1163.05 = 1292 W = 0.95 * 1360 W (= heat received from the sun)
Is it cheating to assume that any initial difference in T at either end will cause the cube to spin and this the whole thing can be simplified to a single face of 2 square metres?
Well, of course it is. But it makes my sums easier.
Before doing the numbers, I’d like to take this as a learning opportunity for myself, to talk about it in words first. Again, I’m attempting to learn, so tell me where I go wrong:
* The concrete block has one side (hot side) with a constant maximum energy input, and one side with a constant maximum rate of cooling.
* The power of energy going in cannot ever equal the power of energy radiating out, because there is ALWAYS energy coming in on one end (no cooling at the surface area), and ALWAYS energy going away from the other end (constant cooling at THAT surface area).
* The physical structure of concrete allows it to transfer energy from hot side to cold side ONLY at a given rate — this is the conductive heat transfer, and it has a particular value for concrete.
* Conductive heat transfer is the rate at which the hot end can supply energy THROUGH the block, .. TO the surface area at the cold end.
* If the cold end is ALWAYS cooling, then the cold end can never achieve the temperature of the hot end. Rather, it will achieve the highest temperature possible with the given amount of conductive heat transfer offset by the constant cooling rate. This will always be a lower temperature at equilibrium.
* Now the cold end, at the lower temperature will now be radiating energy at a lower rate than the hot end. This cold-end radiation, though, is NOT the conductive heat transfer. Rather, the conductive heat transfer is the CONSTANT quantity of energy delivered to the cold end to MAINTAIN its radiating temperature to the cold ambient space around it.
* Input flux is greater than conductive heat transfer … is less than output flux … is also greater than conductive heat transfer. The hot-end higher-energy area transfers some of its value constantly to the cold-end lower-energy area to MAINTAIN that lower energy, which enables the value of the cold end’s temperature at equilibrium.
* A watt is a joule per second. Joules-per-second constantly occurring for a number of seconds delivers a number of joules to the cold end, while the cold end constantly looses joules per second cooling. The conductive heat transfer overcomes the cold end’s cooling enough to maintain the cold end’s temperature, which temperature itself is higher than the conductive heat transfer maintaining it.
* The cold end, now at its lower-than-hot-end temperature COULD now act as a source of heating for, say another concrete block, with its own conductive heat transfer, which would be a smaller value than the first block’s conductive heat transfer.
* The conductive heat transfer of the first block is what maintains the cold-end temperature. THAT cold end temperature has its own flux, but this cold-end flux is NOT a conductive heat transfer or a source of conductive heat transfer, until we know what is next to it.
Okay, where did I go wrong? Thanks.
Yours truly,
Either-making-a-fool-of myself-or-providing-useful-insight
(you can call me “foolsight” for short)
Thank you Robert.
Since you ask, I was wondering about this:
“The power of energy going in cannot ever equal the power of energy radiating out, because there is ALWAYS energy coming in on one end (no cooling at the surface area)…”
The hot side absorbs 95% of the incident radiation, and heats up.
Thermal energy begins to conduct through the block, and conducts faster as the heated side gets hotter.
But as soon as the front side absorbs any energy, it begins to emit radiation at 95% emissivity.
And this also increases as it gets hotter.
So the front surface is cooling by conduction through the block, and by radiation from the surface.
No?
Nicholoas M, regarding your February 28, 2020 at 3:26 pm comment,
that sounds logical to me.
The temperature will vary at every point on and in the block. On the surface of the block facing the sun, it will be warmer in the centre of the square and cooler at the edges. So the formulation of the question, with its assumption of constant surface temperatures over a single surface, is incorrect.
You could assume that each surface of the block was composed of perfect thermally conductive material, and totally thermally insulated at the edges from the other surfaces. Then you would have to know the thermal resistance between each of the surfaces. Hard to do without a computer.
Why do you say it is warmer in the center?
The side pointing to the sun will be hotter. The sides pointing away will be coldest. Therefore the edges of the sun-pointing side will be nearest to the coldest sides, so the temperature there will be colder, due to the shorter thermal pathway. If the temperature at the edges of the sun-pointing side is /are colder, the temperature in the middle of the sunpointing side must be warmer.
If the material of which the cube was constructed had infinite thermal conductivity, then all would be the same temperature.
” Therefore the edges of the sun-pointing side will be nearest to the coldest sides, so the temperature there will be colder, due to the shorter thermal pathway.”
No…the surface is flat and the sides are insulated. The center of the front is the same distance from the center of the back as the edges of the front are from the edges of the back side.
The entire front surface is one meter from the entire back surface.
If the sides were not perfectly insulated then yes.
Part of exercises like this is to pay attention to every part of the problem as described.
https://www.youtube.com/watch?v=AiZWFrb76pM
Here we see a video of a radiator being heated.
The source seems to be about 60C,
so we’ll say it is 333K
The room temperature is, say, 25C or 298K
Now remember according to ideological mathematics:
CHF = CSR
CSR = 5.67e-8*(333^4-298^4) = 250 W/m2
Well, what do we see in the video?
The output temperature = input temperature, despite
the huge size difference of the output.
Tcold = Thot !
CHF == ZERO !!!
But not according to ideological mathematicians
Because CSR MUST EQUAL CHF
Set k to 1 ( we don’t know or care what it is )
CHF = k (333 – Tcold) = 333 – Tcold = 250
Tcold = 63K
No wait, we can’t do that (I did it on purpose). Let’s drop
the room remperature:
333 – Tcold = 5.67e-8 * Tcold^4
Tcold = 214K or -59C
Is that what we see? NO!
Why do people think Boltzmann and Planck were idiots who couldn’t notice CSR=CHF?
Just kidding, k is important to know.
Steel radiator has k between 12 and 30
The result is 296K to 314K
But we see the temp go to virtually 60C, or 333K.
“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?”
Well, according to the Apollo 13 fraud, freezing cold throughout.