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.
I’d like to propose a different experiment.
You have a one meter cube box which is completely covered with unobtainium. (is that name trademarked by the Pandorans?)
On the inside wall of the is covered with gold leaf, which is very reflective. 1 centimeter in is another cube with a layer of vantablack on the outside, and a layer of gold leaf on the inside. Inside that box is another box of the same specification with vantablack on the outside and gold on the inside continuing until you have a small 2 centimeter box in the middle.
The space between each of the cubes is filled with air at a temperature of 100 C at sea level pressure. The walls with vantablack and gold also start at 100 C.
What will happen as vantablack absorbs any emissions and re-emits those emissions equally inward and outward. Given that gold is highly reflective, and vantablack is highly absorbing, would the energy of this system tend to rise as you get to the center and lower at the outsides until some steady state occurs?
The question my be ill formed by my ignorance of blackbody radiation and laws of thermodynamics, but the question has bugged me for decades. Thanks.
Jeff, IF I understand your experiment, eventually the entire apparatus would be at ambient temperature. Neither gold leaf nor vantablack is a perfect relflector or absorber, so the end result would be that they would all trend to the same temperature.
At least that’s my answer, but I’ve been wrong a few times. Well, more than a few times if you insist on brutal honesty … but that’s my answer.
I say that temperatures will equalize in part because if it were not true, if there were a persistent temperature differential, then we could use it to run a heat engine forever … perpetual motion. No can do that.
w.
I think the air rather negates the though experiment since conduction and convection would equalise temps. If you evacuate the the whole thing, I think that is maybe more like what Jeff wanted to imagine.
If we also replace imperfect materials with perfect ones we remove the get-out clause and get back to basics.
In that case I would say that the black surface would emit given total power, the ungoldium would prefectly reflect it and it would be perfectly absorbed by the ideal black surface. Each cavity would remain in just the same state as if started.
Even if there was a temperature difference between the outside and the innermost cavity this would remain unaltered since there is no mechanism : radiative , conductive nor convective to transmit heat between them.
Any other set up short of this ideal would eventually lead to equal temperatures throughout: 2nd law.
Zoe, I woke up in the middle of the night and I thought “Dang! How did I miss that?”
What I realized was that in the midst of all of your dodging, and redirecting the conversation, and plain and fancy tap dancing, and talking about “ideological mathematics” and such, I’d totally overlooked a simple fact:
You never answered the question in the head post.
So, since you hold yourself out as an expert in these matters, much more knowledgeable than us mere mortals, could you please use your “non-ideological mathematics” and tell us how hot you think the two sides of the block will end up?
As requested in the head post … please show your work.
Many thanks,
w.
My thoughts went like this. Starting at ambient temperature 100 C of everything, the gold would emit at some frequency of IR. The vantablack would absorb 99.99 percent o that. Half the energy would go to the gold on its other side, the other half would be re-emitted back at the gold that emitted it first. Gold would reflect most of it at around 95% iirc. The decrease in energy of the outermost cube would be replenished by contact with the air with the air decreasing ever so slightly. The radiation doesn’t affect the air temperature and gets emitted to the black surface. In mean time the energy transmitted to the next layer of gold gets emitted to the next black layer. The air slowly gets cooler. And so forth.
In this particular scenario with the perfect insulator on the outside. The total energy of the system stays the same. The energy moves about depending on the properties of the stuff it comes in contact with. Were the box just full of air, the individual molecules would be transferring energy with each collision with some locations occasionally warmer than others. The energy moves about, but the whole system stays the same. I was applying stuff we know such as if you paint your house black, you’ll be hotter than if it is covered in mirrors because the radiation behaves differently depending on what it comes in contact with. If you take the universe as a whole, the total energy of the system isn’t changing, but in our little part of it, we make heavy uses of the energy potential differences to extract work from it. So I came up with this closed system idea to see if one could create an energy difference in one part of an area to be converted to work, and then back to heat and then start over again.
So when you say “if there were a persistent temperature differential, then we could use it to run a heat engine forever … perpetual motion. No can do that.” I was simply adding the word “yet.”
This is a thought experiment because I don’t have the resources to perform the actual experiment, but there are people who do. I’d like to see the actual experiment performed. Like you, I’m highly sceptical of the idea, but can’t seem to shake the idea. Either way, we’d learn something and have something to throw at all the perpetual motion proponents who like to scam people with their ideas. Like you, I’ve been wrong enough times to not take thought experiments too seriously. The real world has a nasty habit of popping bubbles.
“This is a thought experiment because I don’t have the resources to perform the actual experiment, but there are people who do.”
There are people that can make a set of nested boxes that do not touch each other?
Who are those people?
Besides, vantablack only has the properties you are thinking of for some set of wavelengths and perpendicular rays.
And it does not absorb 99.99%, it is 99.96, and that is for visible light with perpendicular rays.
The numbers are less for other wavelengths and angles.
It would not matter anyway.
There is no magic box with nested cubes that never touch.
And it would have to have a perfect vacuum, also impossible.
And gold does not reflect perfectly, just very well, and again it does so best at certain wavelengths.
No one has made a perpetual motion machine YET because never is a long time.
It is impossible, because nothing is perfect.
No process is perfect, and if you can invent a better version of Maxwell’s Demon, it will not work any better than his version, so why go to all the trouble of nested boxes and exotic materials?
Just have a little guy that only let’s slow molecules (or the fast ones…either will do the trick) past a door he opens and closes real fast.
As you have laid this out, there is no temperature difference, thus there is no (net) heat transfer, the system is already steady state and will remain exactly as you have described it.
Zoe, I woke up in the middle of the night and I thought “Dang! How did I miss that?”
What I realized was that in the midst of all of your dodging, and redirecting the conversation, and plain and fancy tap dancing, and talking about “ideological mathematics” and such, I’d totally overlooked a simple fact:
You never answered the question in the head post.
So, since you hold yourself out as an expert in these matters, much more knowledgeable than us mere mortals, could you please use your “non-ideological mathematics” and tell us how hot you think the two sides of the block will end up?
As requested in the head post … please show your work.
Many thanks,
w.
Hi, Willis.
I answered the question with the very first post I made:
“Zoe Phin February 28, 2020 at 9:55 am
Assuming there is no matter past the cold end, the final temperature of both sides will be equal.”
Both sides will be ~120C.
Now answer my questions:
https://youtu.be/Var3o_eko9U
Does the top and bottom of the pan not get to the same temperature?
Doesn’t that make CHF = ZERO?
And CSR = whatever bright yellow is?
And don’t tell me there’s cond/conv/ective losses all over the place. There’s also sideways radiative losses. All those things are not advantages, but disadvantages to reaching equilbrium.
Sorry, Zoe, but you didn’t answer it because you didn’t show your work.
As to your video, you ask “Does the top and bottom of the pan not get to the same temperature?”
No, they don’t.
w.
toolbox :
k=70 , A=0.1 , T1=502, T2=500 , s=0.01 Calculate ! Heat transfer = 1400W ( right ballpark for the cooker ).
Do you really think your thermal imager will detect that kind of temperature difference on the scale you are using?!
No, the top and bottom of the pan does NOT get to the same temperature but your pathetic “experiment” is not able to resolve that fact because you did not design an experiment capable of doing so.
The only empirical evidence I see is that you have no idea what you are doing.
Greg,
Yeah, I saw this coming. Deny that the temperature is the same and fabricate your own numbers.
Hey, poophead, why don’t you model this with Willis’ method. Go ahead. No excuses!
Hey poophead, why are insults such an essential part of science for you? Go over the CAWG camp, that is one of their qualifications.
Oh dear, you used the D word , see what a I mean?
I did not “fabricate” the numbers any more than you fabricated those in your example. I used what are reasonable figures to illustrate what will be happening with your frying pan. If you had half a brain you would have done that before designing your experiment and would have ensured that it was capable of proving something instead of shouting from the rooftops that you had “empirical evidence ” when you have nothing.
The calculation shows that, if there was a finite heat flow consistent with the accepted view of thermal conduciton, your pathetically incompetently designed “experiment” would be incapable of detecting it.
As such it is not “empirical evidence ” of your claim but clear empirical evidence that you do not understand squat about physics or how to design an experiment to test a hypothesis.
Zoe Phin March 1, 2020 at 9:56 am
Wonderful. More tap dancing. ANSWER THE DAMN QUESTION!!! How does the cold end know that there is no matter past the cold end?
w.
Well, it is light, so it must be able to see, eh?
How else could it know where it was going?
And a distant star has to not only be able to see, before it dares to shine any light on the Earth, but it must be able to anticipate what will be where, in tens, to hundreds, to billions of years. Obviously.
Hubbell Deep Field has objects that shone their light at the telescope over ten billion years ago, before the solar system formed, and before the large stars that went supernova to seed the cloud that became out solar system and us with heavy elements, had themselves formed and exploded.
And it was way before we decided to build that telescope and put it there…of course, before the Earth was even a glimmer in the solar proto-disc’s eye. Yeah, before the elements that went into the make up of the disc were even created in those supernovas.
That kind of makes a mockery of the speed of causality…but what the hey…light cannot shine at nothing…
So says Zoe!
Pharaoh had nothing on this gal.
Zoe, one other comment. You asked if I’d seen the online calculator for this. When I looked at it, I understood why you make the claim that you do.
If we set up the online calculator with the setup you use in your problem, with one wall at 70°C and the other wall at 50°C and all the rest, we indeed get a flux of 2 watts just as you say.
However, you then make the claim that there are only two watts flowing through the wall, but the cool side of the wall is radiating at a Stefan Boltzmann level around 590 W/m2. You use this as evidence for your theory that the conductive heat flus is different from the cold side radiation.
What you are not taking into account is, why is the cold side at 50°C and why is the hot side at 70°C?
The answer, of course, is that they are held at those temperatures by whatever is heating the rooms on either side of the wall. That energy, the additional energy needed to maintain those temperatures, is NOT taken into account in your formulation, and it leads to your incorrect conclusion.
I’d hoped to make this obvious by restating your problem in outer space. There, we can see that there is nothing heating the cold end of the block. And as a result, all of the energy being radiated by the cold end has perforce must, come from within the block itself.
And this is the reason why the conductive heat flux through the block has to equal the amount radiated out the far end. This requirement forms one of the two simultaneous equations that everyone solved to get the correct answer.
But heck … give us your answer, using your “non-ideological mathematics”, and we can all see how you would calculate the temperatures of the two sides of the block in the head post.
Regards, I’m off to burn some fallen tree limbs in the back yard, I’ll check back in a bit.
w
Willis,
Answer my question:
https://youtu.be/Var3o_eko9U
Time: 01:57
Does the top and bottom of the pan not get to the same temperature?
Doesn’t that make CHF = ZERO?
And CSR = whatever bright yellow is?
Your ideological math:
The first thing you would’ve done is compare the surface area of the pan to that of the planar fire head touching the pan.
Bam! You’re already on the path to the WRONG answer.
Bam, you ignore everything that was said go back to your totally uncontrolled frying pan again.
Willis has just restated what I pointed out to you about 3 days ago when I initially commented on your blog. You are spuriously thinking the cold end is cooled by SB radiation and there is NO REASON to suppose that. Indeed it is incompatible with the rest of the conditions you laid down.
You confused yourself an made a simple error based on a spurious assumption. It’s that simple.
The arbitrary values you chose do indeed give Tc as 50 deg C , as I said you have that much correct. Sadly, the good bit it ends there.
The engineers toolbox is the closest you have come to supplying an example of this kind of “text book” calculation and the conspicuous difference is that they say nothing about SB radiation at all. So your claim about your original exercise remains a false claim.
Your refusal to address this by this stage makes it pretty clear you realise this, that you have no answer or counter argument and so your last line of defense is to divert and hope it works.
500+ comments indicates that is a failed strategy.
One definition of insanity is repeating the same behaviour and expecting a different outcome. You are doing a lot of repetition here.
“You are spuriously thinking the cold end is cooled by SB radiation and there is NO REASON to suppose that.”
LOL. Projection. Actually you and Willis believe that, and I don’t. If there is no matter past the cold end, there no heat transfer. The 557 W/m^2 is a radiating potential.
Why is that sophists ignore empirical evidence in the video? Oh that’s right, they’re low life sophists – it’s all they can do.
Zoe Phin March 1, 2020 at 9:12 am
Um … I hesitate to ask, knowing that I’m just gonna end up going down a Zoelian rabbit hole but … how does the cold end know that there is no matter past the cold end? Radar? ESP?
And is it an “all or nothing” kind of deal? I mean, if there is one molecule of matter out there, does the cold end suddenly begin to radiate full bore? Or is the amount of radiation proportional to the amount of matter?
Next, does the matter have to be in a perfectly straight line perpendicular to the flat end, or anywhere within the field of view?
Next, if you pick an arbitrary line and follow it far enough out into space, sooner or later you’ll hit matter … does that count? (This is particularly true if the answer to the previous question was “anywhere within the field of view …).
Finally, could you point us to one, just one, text of any kind that says that solid objects above 0K only radiate when there is other matter around?
Regards,
w.
“radiating potential” another made up term of your new physics world.
No radiation to space ? Another invention of your new physics.
More insults.
I’ll repost what I just posted about you “empirical evidence”.
toolbox :
k=70 , A=0.1 , T1=502, T2=500 , s=0.01 Calculate ! Heat transfer = 1400W ( right ballpark for the cooker ).
Do you really think your thermal imager will detect that kind of temperature difference on the scale you are using?!
No, the top and bottom of the pan does NOT get to the same temperature but your pathetic “experiment” is not able to resolve that fact because you did not design an experiment capable of doing so.
The only empirical evidence I see is that you have no idea what you are doing.
” If there is no matter past the cold end, there no heat transfer. ”
So what happens to the 2W arriving at the cold end, why does it not get hotter with all that thermal energy arriving with nowhere to go ?
That’s the problem with rewriting one of the laws of physics you end up needing the rewrite the rest and still come up with a self-consistent , coherent theory.
Good luck with that.
So now your problem is where is happening to the 2W ?
Willis,
“how does the cold end know that there is no matter past the cold end? Radar? ESP?”
Yes Willis, “ESP”. You never read how Planck derived his law?
Waves only form between matter.
If the separation distance between objects is 1 meter, no 2m wavelength will form.
Why do you think not all photons are possible? What do you think is the basis for Quantum Mechanics.
According to you, photons are emitted willy nilly, so ALL photons should be possible.
Supposing there is some matter about 13 billion light years away, how long will it take you “radiation potential ” to resonate with that and start emitting?
Just wondering because we don’t know the rest of your new physics yet.
Willis,
‘”As to your video, you ask “Does the top and bottom of the pan not get to the same temperature?”
No, they don’t.’
My pixel analysis shows you to be a liar. You didn’t analyze anything before you spouted that convenient self-serving lie.
There’s dozens and dozens of real-time thermal videos on youtube showing you that you’re wrong. You ignore it all and call yourself honest. Hilarious!
But please don’t kill yourself. That would be a very awful thing.
The bottom side of the pan is in contact with an open flame with a temperature of ~3560°F, and the top of the pan is in contact with a cold slab of some cold food-like substance and room air that is probably in the range of 68-78°F.
It takes time for heat to flow through any material, whether you believe in that or not.
And you somehow have the idiotic notion that the two surfaces are the same temperature?
It is obvious at a glance that you make zero sense, but it takes some thought to understand the true extent of your ignorance and belief in an unphysical reality.
“Um … I hesitate to ask, knowing that I’m just gonna end up going down a Zoelian rabbit hole but … how does the cold end know that there is no matter past the cold end? Radar? ESP?”
How about field radiation pressure? You know, that which you denigrate, while not explicating a satisfactory replacement mechanism for absorption / emission regulation in accord with 2LoT, therefore you tacitly must claim all matter emits as though its in a 0 K ambient, and absorbs as though its in an ∞ K ambient?
You only deny radiation pressure, 2LoT, the Work-Energy Theorem, Stefan’s Law and due to that, pretty much everything else. LOL
You wrote: “Heat can’t flow from cold to hot, but radiated energy absolutely can.” That violates 2LoT, and is self-contradictory. Heat is definitionally an energy flux, so you’ve essentially just written “energy can’t flow from cold to hot, but radiated energy absolutely can”… as though radiated energy is a ‘speshul’ kind of energy that can violate 2LoT. Of course, in this context, “heat” is “radiated energy”… so you’ve self-contradicted in a particularly ironic way. LOL
Energy is a measure of the ability to do work… just how, exactly, can a photon of lower potential do work upon an object of higher potential? Or are you going to follow the climate catastrophist guidebook and claim radiative energy transfer “has nothing to with the photon’s ability to do work”… you’ve already tacitly done so in your denial of the Work-Energy Theorem in your claim that radiated energy can violate 2LoT.
Bet you didn’t know that Rice University measured the temperature of a single molecule by having a deep understanding of 2LoT as relates to radiative energy transfer, huh. That, if you care to look it up, proves you wrong on your blather about “Heat can’t flow from cold to hot, but radiated energy absolutely can.”
You’re yet another of those ‘colleeg edumacated idjits’ with equations you use like they’re black-boxes (plug in the numbers and out pops an answer… is it correct? Who knows. It’s an answer, though!), sans any deep understanding of the concepts behind those equations, and you’ve bought into a wide swath of the climate catastrophist weltanschauung.
I’d politely encourage you to educate yourself, but “you are far too arrogant and self-important to follow my polite request” (your words), so instead, I’ll raucously laugh at your perpetual and self-imposed confusion. LOL
You do no one any good deed in leading them astray while purporting to educate them… educate yourself first… you don’t even have a grasp on the basics.
It will equalize between the two sides!
Smart man
Robert,
Walk outside on a sunny day.
Find a stone wall, or just a stone, that has the sun shining on one side only.
Now, feel the sunny side.
Then feel the other side.
Then wait several hours for the stone to warm up some more.
Feel both sides again.
The think about what you just said.
Come back and offer your opinion again based on the actual world and how things actually work.
Zoe Phin March 1, 2020 at 9:56 am
Wonderful. More tap dancing. ANSWER THE DAMN QUESTION!!! How does the cold end know that there is no matter past the cold end?
w.
Willis,
The answer is: mu.
That’s the proper answer to a question that already presupposes something.
Have you stopped beating your Wife?
Your question presupposes the Corpuscular Theory of Light as valid. It’s not.
Come on, Willis, why are not all photons possible (the premise of Quantum Mechanics)?
Think.
Zoe, it is YOUR theory that the cold end will not radiate (despite being above zero K) if there is no matter beyond the cold end. Not mine. Yours.
I’ve simply asked, how does the cold end know that there is or isn’t matter there? Simple question about YOUR theory.
I “presuppose” NOTHING. It’s your theory, so if there are presuppositions, they are yours.
So once again I ask that you stop the ridiculous tap-dancing and ANSWER THE SIMPLE QUESTION ABOUT YOUR OWN THEORY.
w.
Willis,
Why are you such a liar?
Of course you presuppose that matter MUST emit to nothing.
What’s your scientific basis for that?
Zoe Phin March 1, 2020 at 2:06 pm
Listen, you vile little pissant. I don’t lie. You must be thinking of your friends, or perhaps you are looking in a mirror, but you’re not looking at me.
I am an honest man, and I tell the truth as best I know how. Obviously, you have little experience with my type of human.
I was brought up under what in my family was called the “Captain’s Code”. “The Captain” was my great-grandfather, a Mississippi riverboat captain. One part of his Code goes as follows:
Now, this isn’t the 1870’s when he was alive, so I don’t kill anyone for that. However, I’ve done my best to live by that maxim, and I don’t put up with your kind of ugly crap from anyone, man, woman, or child.
I’m done with you. Your endless insults have finally brought me to do something I rarely do, which is to flat give up on someone because they’ve proven that they have no honor. You’ll have to peddle your garbage to someone else. You’ve cancelled your vote with me. All I’m going to do with your madness from here out is to point and laugh.
w.
Zoe,
You have a Nobel Prize waiting for you, so why not claim it.
And you need to go and rewrite a whole bunch of Wikipedia article as well.
I am sure everyone will be grateful for your having corrected these erroneous statements, such as this one:
“Thermal radiation, also known as heat, is the emission of electromagnetic waves from all matter that has a temperature greater than absolute zero.[3] It represents the conversion of thermal energy into electromagnetic energy. Thermal energy consists of the kinetic energy of random movements of atoms and molecules in matter. All matter with a temperature by definition is composed of particles which have kinetic energy, and which interact with each other. These atoms and molecules are composed of charged particles, i.e., protons and electrons, and kinetic interactions among matter particles result in charge-acceleration and dipole-oscillation. This results in the electrodynamic generation of coupled electric and magnetic fields, resulting in the emission of photons, radiating energy away from the body through its surface boundary. Electromagnetic radiation, including visible light, does not require the presence of matter to propagate and travels in the vacuum of space infinitely far if unobstructed.”
https://en.wikipedia.org/wiki/Thermal_radiation
The fact is you have no idea what the hell you are talking about, and everyone here knows you are one of the stupidest people ever to show up.
People like you are why humanity was stuck believing nonsense for all of time up until logic replaced doubletalk.
Nicholas,
No, I think you deserve a Nobel Prize. Quoting Wikipedia uncritically is the height of genius.
Look up the sources.
^ K. Huang, Statistical Mechanics (2003), p.278
^ K. Huang, Statistical Mechanics (2003), p.280
^a b S. Blundell, K. Blundell (2006). Concepts in Thermal Physics. Oxford University Press. p. 247. ISBN 978-0-19-856769-1.
Zoe, you said:
“Quoting Wikipedia uncritically is the height of genius.”
But what I said was the opposite of what you claim.
I said:
“I am sure everyone will be grateful for your having corrected these erroneous statements, such as this one…”
Does that sentence sound uncritical to you?
As anyone can see, I said the opposite of what you claim.
So maybe that is what is the matter with your brain…you have the reading comprehension of my cat. And not the smart one…the dumb one.
Zoe Phin March 1, 2020 at 8:07 pm
“… Nicholas, No, I think you deserve a Nobel Prize. Quoting Wikipedia uncritically is the height of genius … .”
—
What Nicholas quoted was consistent with Richard Feynman’s Quantum Electrodynamics 1961 lectures–which you should read.
Nicholas,
You’re an immature child. Your argument is still: Wikipedia said so, so it’s true. You didn’t demonstrate that waves grow from nothing to something over long distances.
If you look up Wikipedia “photon”, then you get several alternative views. And since there are alternative views that which you quoted is not the final word.
This aint about me Zoe…it is about you.
You know, the one who has no idea what the hell she is talking about.
I take back the part about you having the same reading comprehension as my stupidest cat.
Obviously you have no comprehension at all.
And you hallucinate…wildly.
I know why you have to pretend that I said what you hallucinated.
Because if you did not, you would actually have to offer your rewrite, which is what I actually said.
And obviously you have as much ability to rewrite physics as the lizard my cat is playing with.
Face it Zoe…you are a complete failure.
You cannot even spin tall tales very well, let alone bullshit your way through a coherent explanation for any of the malarkey you have invented.
Greg,
You blab so much I need to start a new thread.
“k=70 , A=0.1 , T1=502, T2=500 , s=0.01”
How nice you fabricated your own numbers in order to justify your ideology.
I screen-captured the video @ur momisugly 1:53, just as the pan was being removed. I compared the pixel color just below the burger to just above the fire.
They are exactly the same!
As you move up above the burger, it does decrease very slightly, and perfectly reasonably, as that part of the pan is much farther from the fire, and had time yo cool.
You claim the resolution is not good enough and that’s a complete self-serving lie.
I counted 53,915 unique colors in the image.
The resolution is just fine.
LOL, as if there was just one video on youtube showing top T matching bottom T in all sorts of mediums (metal, water, wood, hard plastic), instead of DOZENS and DOZENS.
“I compared the pixel color just below the burger to just above the fire.”
How nice you fabricated the zone you look at to justify your ideology.
“just above the fire” ? What the hell kind of sampling method is that?
What is the specified resolution of the device? Sure, you will have noise and can pick as many convenient pixels as you want. Find two that are the same and say “bingo!” I’ve proved my case.
How many different colours are there is those white blobs? The whole pan and the burner looks the same temp : white, you’ve just saturated the device. You have no resolution at all at that end of the scale, that’s why they look the same “temperature”.
You are pretending that both sides of the pan are the same temp but you have no shot of the underside of the pan, so you are not even measuring what you claim to want to know.
Go away and design an experiment which would be capable of detecting a difference of less than a degree at that temperature and start again.
Anyway you are so ridiculously disingenuous since your big mouth has got you into a position where you can’t back down. You have to just keep going whatever, so there is absolutely no point in continuing to converse with you.
It’s always interesting looking at new ideas but your attitude is beyond a joke. As Willis said you have made it pretty plain for everyone to see that can not be taken seriously and you fell for the bait.
Bye-bye “poophead”, it was fun playing with you.
The camera is designed to work -20C to 650C.
It is not saturated. The pixel color is #FDFCE7, not #FFFFFF
“How nice you fabricated the zone you look at to justify your ideology.”
Center of the fire and as close to the center of the pan as possible. By Willis’ theory there should actually be a radial gradient as well.
“Go away and design an experiment which would be capable of detecting a difference of less than a degree at that temperature and start again.”
The camera is capable of 0.05C degree discernment. The software that comes with the camera rounds to the nearest 0.1.
The camera man chose not to display temperatures.
Question: What experiment validated Willis?
Oh it’s in a textbook? Well, I guess we’re just supposed to presume it’s been validated, right? No textbook has ever had falsehoods in it?
Sure Greg, all those videos simply have noise and saturated the device. Yeah right.
Why do people like you exist?
No evidence will uproot your ideology that has no evidence, other than theoretics, to support it.
Are you a Maoist like Willis was? They’re just as convinced of their own B.S.
Zoe’s solution with both sides at a temperature of 120 °C is more or less possible in a universe in which the hemisphere visible from the backside of the concrete bar is empty. For the same reason the Sun would not radiate if it were alone in space and no other bodies could absorb its radiation (Tetrode’s star) the backside would not radiate. That does not agree very well with the fact that the frontside is heated by the Sun.
Teerhuis, you blithely repeat Zoe’s curious claim that if something is alone in space it doesn’t radiate. Neither of you have provided the slightest scrap of evidence that this is true, and it seems totally bizarre.
As I asked Zoe, how would some inanimate object know whether it is “alone in space” or there was some kind of matter around it? What if the “matter” is one other star 50,000 light years away? Would the sun radiate in that situation?
w.
Willis,
You are an ideological twit. You were asked to explain several times why some photons can form and not others. You don’t give a hoot about science, just pushing your ideology. Why don’t you just read how Planck derived his formula? He couldn’t have done it believing matter just sends photons to anything.
What we learn from Planck is if the separation distance between MATTER is 1 meter, no 2 meter wavelength photon will form, got that? There is NO 2 meter photon. That means the objects “know” the distance between each other and “know” not to send a 2m photon.
But in your theory, all photons are possible because objects don’t care what’s around them. They are dumb in your view, and just send whatever. Well this view does not yield EMPRICIAL results, so this view is wrong.
You got that?
Willis,
May be Zoe’s curious claim is from a comment of mine on her website. I read the argument about Tetrode’s star on Math Pages:
https://www.mathpages.com/home/kmath610/kmath610.htm
No, I did not take it from you.
I read science.
A smart person can understand that standing waves can only form with a “rope” that’s attached on the other end.
But Willis is not a smart person, and believes the rope simply grows in length and can wave with fixed amplitude anyway, and then attach itself to something. Yup, he’s a growing rope believer. How silly.
Zoe, you are expressing a belief in something called an luminiferous aether, which Michelson and Morley demonstrated was unlikely to exist, and by the 1920s it was considered prove that there was no aether.
Explain redshift and blue shift with your rope analogy.
You seem to fixate on analogies and transmute them into exact parallel situations inside your mind.
Read and learn.
Most people here probably knew all of this when they were teenagers.
You are not smarter then everyone else, you are dumber.
A veritable cement head.
https://en.wikipedia.org/wiki/Luminiferous_aether
Michaelson and Morley’s famous 1887 experiment did not get a null result as theologians of Corpuscular Light Theory will have you believe.
In fact the experiments were perfected and carried out again, also showing serious results.
But of course you don’t know that.
Zoe,
Planck assumed a cavity with perfectly reflecting walls, then there are only standing waves in the cavity. But later it was proposed that the ‘oscillators’ in the walls absorb and emit photons, so the photon gas is not restricted to standing waves.
How about a supernova, Teerhuis?
Does it only explode when every photon has a destination printed into it’s itinerary?
If a gamma ray burst shoots out to the whole universe when two black holes collide, can they only collide once every gamma ray photon has it all worked out where it will end up?
LOL@Klimate Katastrophe Kooks March 10, 2020 at 10:10 am
I keep asking you the same question and you keep ignoring it … so I’ll ask it again.
If your theory about “field radiation pressure” is correct, how can two flashlights shine directly at each other? Per your theory, the “field radiation pressure” from the stronger flashlight should prevent the weaker flashlight from shining at all.
But obviously, that’s not happening … why not?
First, heat is NOT radiated energy. Heat is the spontaneous flow of net energy from a warmer to a cooler object. Or to use the Encyclopedia Britannica definition:
Here’s another definition:
Note that NOT ALL ENERGY IS HEAT. Heat is a particular form of energy—the energy that flows spontaneously from warm to cold.
If we have two bodies within each others’ “view field”, each one is transmitting radiant energy to the other object. However, the flow of heat is the NET flow of energy, which always goes from the warmer object to the cooler object. So no, there is no violation of the Second Law of Thermodynamics. Radiant energy goes both ways, but in conformity with the 2LoT, the heat flow only goes one way, from a warm object to a cool object.
Regards,
w.
LOL@Klimate Katastrophe Kooks March 11, 2020 at 10:34 am Edit
Heat is indeed an energy flux. But all energy fluxes are NOT heat. This is Thermo 101 stuff. Get a textbook and stop making a public spectacle of your ignorance.
w.
Comment on Zoe and Willis:
“how does the cold end know that there is no matter past the cold end? Radar? ESP?”
The emission and absorption of a photon is one single event, despite that in our reference frame there can be billions of years between both events. Try to travel with a photon. There is no such thing as a ‘free photon’. The Sun radiates so its photons will be absorbed no matter how far the absorbing bodies.
If it is assumed that the Sun can radiate and can heat the front side of the bar then the backside can also radiate.
You are ignoring the case where there is nothing in the path of the photon, ever.
Look at the Hubbell Deep Field.
See all the places with nothing there?
You are taking the case of a photon that hits something and inventing the idea that there is no other case possible, with no evidence.
Nicolas,
In the reference frame of the photon there is no time and distance. You cannot know very accurately what is beyond the observable universe. But I admit that this is not a very practical way to describe spectroscopy. It is much more useful to describe the emission of a photon as the result of a coupling of a particle with a vacuum-state (see Feynman’s work).
Chicken and egg.
Whether you call it a speed of a distance, the photon cannot know where it will end up when it leaves.
Does it have a little homunculus riding along with it, peering out a porthole prior to embarking, and knowing that in billions of lightyears of distance, objects that do not even exist yet will bend it path through multiple lensing events, and eventually impact some atom of matter than has not even created yet?
The entire notion of knowing where it will end up before it leaves can be disproven by reduction ad absurdum.
”’speed or a distance…
Willis:
“Listen, you vile little pissant”
How Lovely.
This is what happens when you try to educate someone that doesn’t understand their ideological posession.
No experiment and no reasoning will suffice. And expect to be accused of the same thing.
Whatever. Like I said Willis, you can have your ideological nitwits that believe they understand science, and I’ll stick with Planck and empiricism.
He is frustrated with you, Zoe, because what you are writing is nonsense, and because you steadfastly refuse to learn.
You are only hurting yourself, but it is a common problem. I often encounter people who erect impenetrable barriers to their own learning, by refusing to ever admit, even to themselves, that they were confused or mistaken, about anything of consequence.
They apparently think that admitting an error would be admitting weakness, and that is something that they will never do. The truth is just the opposite. Everyone makes mistakes; it is how you handle them which reveals your strength, or weakness. A readiness to admit errors, correct them, and learn from them, is a sign of strength. It is evidence of intellectual integrity. It is the unwillingness to admit errors that proves weakness.
– Jack B. Sowards (writer), as the character Lt Cdr Data (Brent Spinner), TNG S2E2
https://archive.is/MlOiz#selection-3467.9-3469.1
https://youtu.be/8eDYVtPwWiM?t=43
Hilarious!
Even though there’a dozens of thermal imaging videos on youtube refuting your claims and supporting mine, you ask me to acknowledge my “error”.
Wow. What a pathetic sore loser … projecting his gross incompetence.
Zoe says: “If the separation distance between objects is 1 meter, no 2m wavelength will form.”
This is small point in larger discussions, but it goes to show once again Zoe is misinformed. A standing wave between two objects/surfaces/ends a distance D apart can have wavelengths of 2D/n.
So when the distance is 1 m, the allowed wavelengths are
2m, 1m, 2/3 m, 1/2 m, 2/5 m, ….
Yes, 2 m is indeed allowed.
Yes, Tim, I dumbed it down so that I don’t have to retype the same thing on my smartphone the first 8 times I tried to explain reality to Willis.
Willis would mock what you just said and call it ESP.
Nice try liebaby.
That one was truly laughable.
You ‘dumbed it down’ by giving wrong information? How is that supposed to strengthen your case?
Tim,
When Willis ignored my claim the first 2 times I told him this on my blog, he ignored it. Then I started writing it wrong, so he could have a chance to refute it. This would at least have shown he read something. He did not.
Thanks for agreeing that not all photons can form and therefore objects don’t emit photons willy nilly.
You have refuted Willis. That supports my case.
I have refuted one of your claims about cavity radiation (standing waves within a closed, reflective chamber) — correcting your incorrect claim about the allowed wavelengths. Since Willis is making no claims about cavity radiation, then you have refuted nothing he said.
Most of the world is not a cavity resonator, but there are still rules about the allowed wavelengths of light emitted by materials. These are related to the quantum mechanics of atoms & molecules & crystals (not the quantum mechanics of cavity resonators).
Here’s another video:
https://youtu.be/LckKhwJJgtE
According to Willis et al:
Since the incoming surface area is much smaller than the outgoing radiation surface area, and because of much convective surface area, we should expect to see a very steep “steady-state” gradient form where the top is at the very least twice (more like 4x) as cool as the bottom.
And yet reality shows different.
And remember, Willis is an “honest” guy.
Moderators:
I posted a longish comment here last night at around 8:00 EST that has not appeared.
Any sign of it in the moderation bin?
I did get a notice it went to moderation, although reading through it I could not see why.
Thanks,
Greg and Willis,
The Saints are jealous of your seemingly infinite patience.
But I am wondering if there is a such thing as being too patient?
At some point it becomes obvious one is dealing with a pathological mind, and it becomes unethical to humor them.
Such people live for the confrontation.
Consider the antivaxxer guy who insists there is no evidence whatsoever than any vaccine has ever done anyone any good, ever, at all.
This is the same thing.
Imagine the nerve of this brat who calls Willis a liar?
Who claims Greg does not know actual physics?
I personally know people like this, have my whole life, and I know that what really motivates them is awareness of their own ignorance, and crushing feelings of inferiority coupled with a huge ego.
The only thing they understand is being spoken to in the same language they use on others.
Trying to be logical with them is futile, because what they need psychologically is to engage with those that make them feel dumb on what feels like to them to be an equal footing.
So treating them as if they are smart and logical but misguided only feeds their pathology, in my experience.
Just sayin’.
And JMO.
Thanks for your consideration Nicholas. As I said in my last post I will not be conversing with this offensive idiot any longer.
Part of the problem is that she has mouthed off and insulted everyone, from Einstien downwards, for so long it is now impossible for her to backtrack and say, oops I was wrong. She is obliged to keep this going until everyone else gets bored and goes away.
Pathological , quite possibly. I remarked on a couple of occasions that she is displaying behaviour typical of a flat-earther. Such people are not amenable to reason. To think they are is to misunderstand their motivation for holding whacky ideas.
When someone start calling people things like “poophead” because they are not getting their own way in what is supposed to be a scientific discussion, you have to realise that you are dealing with someone with the mental maturity of a pre-school infant.
At that point I draw the line.
Quite clearly there is absolutely zero chance that she will accept anything anyone else presents to her, so to maintain the same desperate hope that logical discourse may arrive at an agreed understanding with this person would be foolish.
It was fun solving Willis’ problem and it is always useful having things you take for granted challenged from time to time, so the effort was not totally wasted. I’m unlikely to waste a similar amount of effort in the future as a result of this “learning experience” , so I’ll regard it as investment for the future.
Thanks for your thoughts.
LOL, quite a lot of projection.
“Quite clearly there is absolutely zero chance that she will accept anything anyone else presents to her.”
Not true. I will accept empirical evidence.
I presented empirical evidence, and the ideological mathematicians refuse to accept it. Hence your ranting is just projection.
I can’t believe the depths of denial of some people. It’s honestly quite disgusting to watch, and I hoped that a little bit of shaming would help. I guess I was wrong. It just cemented their beliefs even further.
As new comments rolled in, I saw the same pattern emerge:
Everyone avoids observational experiments I presented. They want to stay in fantasyland, where they can be right.
The only person who did not, Greg, had to fabricate his own numbers.
Imagine that: people who avoid reality and then want you to admit you’re wrong.
They dare not show any empirical evidence, attack you 50 to 1, and then call you rude as they do the same thing.
Whatever. Hopefully my fight with these fantasy dwellers helped somebody.
Let’s take a look at this video:
https://youtu.be/LckKhwJJgtE
Electrically Heated Water. Conduction based.
How would Willis fans solve this problem?
Energy In = Energy Out
Conduction from Bottom => Radiation(All 3D direction) + Conduction + Convection
It’s not easy to solve, and maybe I got it wrong (on purpose; I wanna see some nitpicking on purpose) but it’s easy to realize that just like his block, a steep gradient will form.
How would I solve it? Easy:
Conduction from Bottom => Radiation at the Top
Conductive Heat Flux => 0
Obviously our two approaches differ greatly. His predicts a gradient and mine: thermal equilibrium.
Now what does the video show?
Uhuh. So when are my attackers finally going to acknowledge reality? Never. Because you see … I’m just wrong. That’s the premise.
“Obviously our two approaches differ greatly. His predicts a gradient and mine: thermal equilibrium.”
The coffee pot shows a small but clear temperature gradient at the end of the video — warmer at the bottom and cooler at the top.
Well yeah, the video cuts off. I assumed my audience can think where it’s going.
CHF approches ZERO and clearly CSR != CHF.
The Willis method predicts a steep gradient, and CSR = CHF. We don’t see anything anywhere near that. Quite the opposite.
So your “empirical evidence’ is only wishful thinking of what you *hope* might happen in the future (but never does).
“easy to realize that just like his block, a steep gradient will form.”
No it is not easy to see. If it is so easy, I am sure you could show a few lines of calculations supporting your assertion .
PS. Remember that Willis was considering a system that ONLY had external thermal radiation as input, ONLY had conduction within the system, and ONLY had thermal radiation as output. This system has an internal electric heater as input, conduction, convection, and radiation within the system, and conduction, convection and radiation as outputs. Only a fool would think these are in any way analogous.
For this system, the “Willis method” (also known as the “Every Engineer and Physicist Everywhere method”) would include all the additional pathways for heat transfer available and would indeed predict a rather gradual gradient here — exactly as observed.
Tim
“and would indeed predict a rather gradual gradient here”
Don’t lie, Tim.
The Willis method would dilute the incoming surface area by the outgoing surface area, and create a very steep gradient.
In my video, the cooler top is the AIR above the water.
The top has a hole that allows convection. But even so, let’s say the small T difference you observed was 2C. (80C vs 82C)
k-value of air is ~0.025.
L=0.1, A=0.1
CHF = 0.05, CSR = sig(353.16)^4
Hmm, not even close to what the Willis method would predict.
“The Willis method would dilute the incoming surface area by the outgoing surface area, and create a very steep gradient.”
That doesn’t even make sense!
You haven’t even defined exactly what “problem” you are considering.
A) the gradient from the heated bottom to the unheated top?
B) the gradient from the heated interior to the unheated exterior?
C) other???
For (A), there is not just conduction through the ceramic walls of the pot transferring heat upward, there is also convection of the air within the pot, steam boiling away from the bottom and condensing at the top, and IR within the pot. These additional methods of heat transfer would make the interior nearly uniform in temperature (not a ‘steep gradient’ as you claim).
For (B), their apparently is a gradient across the walls. The interior is 100C (boiling water) but the exterior is ~ 80C based on the IR images. That gradient would be most equivalent to Willis’ scenario — heat applied on one side of a solid material (electric heating element in the interior), heat escaping form the other side of the solid material (IR and conduction to the air) and heat conducting thru a solid material. And here there is a definite ~ 20 C gradient in less than 1 cm.
Tim,
You won’t even do the math or show your setup because you know you’re lying.
“In my video, the cooler top is the AIR above the water.”
Are you comparing the solid concrete cube to the gaseous air above the water? Your calculation for CHF would seem to be
CHF = k A (Th-Tc) / L = 0.025 * 0.1 * 2 / 0.1 = 0.05
Where “L” would be the length from the hot side (the water) to the cold side (the lid); 10 cm is the right magnitude.
“A” would be the cross sectional area of the pot. Your estimate here is WAY off — the cross section is closer to 10 cm x 10 cm = 0.01 m^2. Consequently, the conductive heat flow would an order of magnitude smaller than you estimate.
That is all well and good — as far as it goes. But …
1) there is convection of the air within the pot (nothing to do with the ‘hole at the top’ that you mention). This will transfer orders of magnitude more heat than conduction through the air.
2) there is water boiling from the bottom, and then condensing on the sides. This also will transfer orders of magnitude more energy than conduction through the air.
3) there is radiation from the room back to the top of the pot. The net IR loss from the top of the pot is about 1/2 of what you calculate.
The fact that you don’t seem to even recognize any of these factors shows just how much you don’t know about heat flow in the very example you chose to present.
“The fact that you don’t seem to even recognize any of these factors shows just how much you don’t know about heat flow in the very example you chose to present.”
I do recognize them, and that you pointed out all the reasons there should be a steeper gradient than what we see according to your philosophy. And yet …
“Consequently, the conductive heat flow would an order of magnitude smaller than you estimate.”
You emphasized my point even further!
Thanks.
“you pointed out all the reasons there should be a steeper gradient than what we see ”
No, I pointed out the reasons the gradient should be SMALL, not LARGE. Only your continued misunderstandings make you expect that we would predict a large gradient for your coffee pot.
Tim,
“No, I pointed out the reasons the gradient should be SMALL, not LARGE.”
The pot is being delivered enough so that the water can boil, yes?
If you’re arguing that CHF is SMALL, then you’re arguing that CSR is LARGE.
CSR != CHF
QED
Your hypothesis is that CSR = CHF. You admitted CHF must get really small, so why purchase an electrically heated water boiler, if it can only produce an equally small CSR? The kind that can’t boil water at the top and have it escape as steam.
The video is for water, and water only. Air is unfair, as due to convection, the mass is not constant. As for the ceramic container, it’s also heating an unfixed mass.
Still need your response to this one Zoe:
“… OUT OF THE SURFACE, NOT IN THE GROUND.”
https://wattsupwiththat.com/2020/02/28/the-hot-and-cold-of-space/#comment-2928112
Is the wall of a mineshaft not a “surface” which emits IR into the same gaseous mixture as the air above the Earth’s surface? Please explain why (allegedly) only some of the IR photons emit from the “surface” of a mine’s walls but not the whole 335 W/m^2, which you claimed emits from the earth’s surface?
Please explain in plain English so I can easily understand your reasoning, thank you.
I don’t understand what the point of your question is.
It’s hot in a mine shaft. 335 is for the surface. It is much greater at depth. Certainly mineshafts wouldn’t be hot, if geothermal delivered 92 mW/m^2.
I think you understand it perfectly well and wish to avoid further embarrassment.
Do a day tour to the ore body of a deep mine given you’re so observations oriented.
The reason I have trouble understanding you, is because you’re arguing my point, while pretending it somehow refutes me. It looks to me to be some type of reverse psychology trick.
Space is a (near) vaccuum, so it doesn’t need to be insulated. It already is, just like a thermos bottle.
Willis, Thot = 388.5 K, Tcold = 222.8 K. The heat flows are 1227 watts radiating away from the hot side, and 133 watts radiating away from the cold side.
I think my answer is a bit different than the others because I used T = 4 K for the temperature of space.
Works for me, Loren. You’ve gotten the answer that just about everyone here got. Well, everyone but Zoe, who claims that both sides (the heated and the unheated side) of the concrete block will have the same temperature, absolute zero, viz:
Shaking my head …
w.
Willis, you call your self an honest man, yet you did not give the answer I gave – ~120C both sides.
Zoe, I quoted your answer. It appears you’ve given two answers. And yes, I am indeed an honest man, which is why I quoted your actual answer without altering it in any way.
In fact, your first answer was
Now, that’s no answer at all … but of course, you claimed it was an answer:
Zoe Phin March 1, 2020 at 8:08 am Edit
Finally, you gave this answer, which I quoted the conclusion of above:
So you aren’t even telling the truth about your own answers.
BZZZZT! Nice try. Vanna White, what prizes do we have for our failed contestants?
w.
Willis,
I gave one answer.
The thing you quoted was not an answer, you LIAR.
Loren,
What experiments convinced you?
Getting back to the question of whether it is hot or cold in space…
I think it is clear enough, but no one has to take my word for it.
Let hear what the man who wrecked the Enterprise, almost killed Kirk, and BTW also was the inventor of Corinthian Leather, let’s hear what he has to say:
https://youtu.be/5vwHLMs04XA
Getting back to the question of whether it is hot or cold in space…
I think it is clear enough, but no one has to take my word for it.
Let hear what the man who wrecked the Enterprise, almost k!lled Kirk, and BTW also was the inventor of Corinthian Leather, let’s hear what he has to say:
https://youtu.be/5vwHLMs04XA
I think that ultimately, this is where Zoe goes wrong. She incorrectly attributes this idea to Willis (and by extension, to all engineers and physicists):
“He believes that
Hot Side Radiation = Conductive Heat Flux = Cold Side Radiation.”
where (with a few simplifications in coefficients) Zoe defines
HSR = σ(Th)⁴
CSR = σ(Tc)⁴
CHF = Th-Tc
What we actually believe in this scenario is that
Hot Side Heat = Conductive Heat Flux = Cold Side Heat
The ‘HSH’ is the NET heat into to the hot side. This is not the same as ‘HSR’. In fact they have opposite signs. ‘HSR’ = σ(Th)⁴ is the radiation FROM the hot side, not radiation TO the hot side. The hot side input (call it HSI) is fixed at 1360 W/m^2 and is not a function of Th.
(HSI-HSR) = HSH = CHF = CSR
[ PS CSR is the same as CSH, since the only heat leaving the cold side is radiation = σ(Tc)⁴ ]
No, the issue that your cult believes matter would split its energy between matter and space. I don’t believe that, and so I correctly identify that by Willis’ way HSR = CHF.
If you believe in backradiation, you must believe in backconduction as well.
Both are retarded views, actually.
Light has a pressure, remember?
Willis’ block can’t backradiate to its source, because such a pressure would cancel out what it receives.
So you guys must really believe the block gets 130W on the hot side. The hot side can’t be 383K by this. And so you have 221K on both sides, and CHF != CSR.
What you’re trying to debunk is very well thought out. But it means nothing to your because you don’t understand EM pressure and reality in general. Your premise of backradiation is false.
https://www.britannica.com/science/light/Radiation-pressure
Zoe Phin March 3, 2020 at 9:36 pm
Zoe, I neither said nor do I believe that the hot side radiation is equal to the conductive heat flux. That’s not true.
Note that, unlike you, I don’t call you a “liar”. That’s a bridge too far.
I simply know that you are mistaken. I said that AT STEADY STATE conductive heat flux (CHF) must be equal to the cold side radiation (CSR). Not the hot side. The cold side.
Now, I asked you several times—if at steady state the conductive heat flux is greater than the cold side radiation … where does the excess heat go? It can’t go into heating the block, it’s at steady state.
And if at steady state the heat flux is less than the cold side radiation … where does the extra heat come from? The “ether”?
Each time, you’ve tap-danced around the question … and you may well do so again. However, be very clear that people see that you have been extremely unwilling to answer the question, and have drawn the obvious conclusion.
w.
Gosh, Willis, you’re attempts at ridicule are hilarious.
You continue to compare a differential with an absolute.
Do you know the difference between profit/loss and assets/liabilities?
Just as I predicted, you have dodged the question, and come up with some nonsensical side-track.
Zoe, do you truly not realize that people clearly see that you don’t have the courage to actually engage with the issues? Are you completely unaware that people are pointing at you and laughing at your plain and fancy tap-dancing to avoid answering a simple question?
w.
Oh Willis, you are so stupid.
Why do you compare kinetic energy to a difference of kinetic energies of two locations. Why?
Your definition of “tap dancing” is me trying to reason with a fool that can’t see or refuses to see his foolishness.
Aaaand, once again Zoe doesn’t answer the question …
This is my surprised face.
w.
PS—You ask “Do you know the difference between profit/loss and assets/liabilities?”
In my last job, I was the Chief Financial Officer for a business with sales of $40 million per year … so yes, dear Zoe, I do understand the difference, and once again your arrogance comes back to bite you in the fundamental orifice.
Willis don’t you understand that you’re were arguing that profits must equal total liabilities?
Why are you comparing a difference of kinetic energies at two different locations to kinetic energy at one of them?
You keep accusing others of what you do: tap dancing around questions asked of you.
I have answered enough of your questions. I have presented empirical evidence, and you, and you, didn’t even bother looking for any.
All this shows that your integrity is not that high.
More tap-dancing and hand-waving, more personal attacks, and in all of it, Zoe STILL hasn’t answered the simple question about conservation of energy …
w.
Suppose you have a hot rock — say 100 C. This rock can be placed in a room where everything is either 80 C, 20 C, or -40 C
In which room will the rock cool most quickly to 90 C?
a) the -40 C room.
b) all will take the same time.
How does your answer change if the walls are still those temperatures, but all the air can be pumped out to eliminate conduction and convection through the air, and so that radiation is the primary means of cooling?
a) the -40 C room.
b) all will take the same time.
c) the rock cannot cool.
Intuition and simple calculations confirm that the rock will cool quickest in the coolest surroundings — with or without air in the room.
But if you believe “Willis’ block can’t backradiate to its source, because such a pressure would cancel out what it receives”, ie because the ‘hot side of the concrete’ is cooler than the sun, then you also believe that the walls can’t ‘backradiate’ (also simply known as ‘radiate’) to the warmer rock. If there is no radiation from cooler walls to warmer rock, then the rock would radiate identically in all rooms and cool identically.
PS. The sun only takes up about 0.001% of the hemisphere facing the ‘hot side’ of the cube. This means that in about any direction, the whole question of “can’t backradiate to its source” is moot. The hot side radiation almost never is radiating back to its source, so it has no problem, at all radiating to the 99.999% of its view that is 2.7 K.
This is not a backradiation problem.
Willis’ block starts at zero kelvin, right?
The rock here is the hottest thing, and it’s not a constant raw energy source.
You inverted everything and demonstrated nothing.
It only looks like a backradiation problem to you because you believe in two way photon flow.
There is no such thing. The potential for heat flow is one way. And it’s greatest for greatest T diff.
You can’t believe in radiation pressure and two way photon flow.
That’s the point.
I choose the ideas that lead to the least contradiction, best if not at all.
Don’t forget that the walls are blocking the rock from cooling to 3K space somewhere behind the walls.
The rock must heat up. LMAO
“The sun only takes up about 0.001% of the hemisphere facing the ‘hot side’ of the cube.”
And that’s why the sun’s radiation from source was reduced to 1361 from 63000000. And that 1361 is now 100% of its view. And you want to send 1162 (was it? I won’t check) right back in 1361’s face.
1 metre by 1 metre by 1 metre concrete block floating in outer space.
The block is insulated on four sides, a perfect insulator, no heat at all is lost from the four insulated sides.
the emissivity “epsilon” and absorptivity across the spectrum are both 0.95.
the thermal conductivity “k” of the concrete is equal to 0.8 watts per metre per kelvin (0.8 W/m K^-1)
it gets full-time sunshine on the front side at a rate of 1360 watts per square metre (W/m2).
what will be the temperature T_hot of the hot side and the temperature T_cold of the opposite cold side?
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At a 1 molecule thinness the temperature on both sides would be equal hence half as hot as expected if the back surface was also insulated. 65.67C
At a million metres the back surface would be at a very low temperature just above 62 K.
This would be enough to drain the minute amount of energy that makes it across the block.
Th surface of the block receiving radiation has to heat up to a higher temperature to force heat across the concrete gradient. The maximum it can heat to is double the energy it absorbs.
The soldering iron I think someone referred to it as.
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129.77 C or 512.92K is the Temp of the hot side.
Similar range to the surface of the moon different albedo.
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The cold side is more difficult. The bulk of the thermal mass built up by absorption of energy is at the heated end which radiates most of the radiation back out.
The small amount that “conducts” 0.8 watts per metre per kelvin finally gives that level to the other side which immediately radiates it into space giving it a temp of -210.4 C or 62.75 K
–
1 metre by 1 metre by 1 metre concrete block floating in outer space.
The block is insulated on four sides, a perfect insulator, no heat at all is lost from the four insulated sides.
the emissivity “epsilon” and absorptivity across the spectrum are both 0.95.
the thermal conductivity “k” of the concrete is equal to 0.8 watts per metre per kelvin (0.8 W/m K^-1)
it gets full-time sunshine on the front side at a rate of 1360 watts per square metre (W/m2).
what will be the temperature T_hot of the hot side and the temperature T_cold of the opposite cold side?
–
At a 1 molecule thinness the temperature on both sides would be equal hence half as hot as expected if the back surface was also insulated. 65.67C
At a million metres the back surface would be at a very low temperature just above 62 K.
This would be enough to drain the minute amount of energy that makes it across the block.
Th surface of the block receiving radiation has to heat up to a higher temperature to force heat across the concrete gradient. The maximum it can heat to is double the energy it absorbs.
The soldering iron I think someone referred to it as.
–
129.77 C or 412.92K is the Temp of the hot side.
Similar range to the surface of the moon different albedo.
–
The cold side is more difficult. The bulk of the thermal mass built up by absorption of energy is at the heated end which radiates most of the radiation back out.
The small amount that “conducts” 0.8 watts per metre per kelvin finally gives that level to the other side which immediately radiates it into space giving it a temp of -210.4 C or 62.75 K
–
Musing. Reflectance, or albedo, seems to break the rule of equal and opposite reaction. Why is this?
If 10% say of incoming energy is reflected into space where is the 10% of energy that should be going in the other direction!
Puzzled.
Willis, this is an excellent example of the fact that the energy leaving the physical domain cannot be specified as a boundary condition in mathematical models: https://judithcurry.com/2018/05/22/energy-budgets-climate-system-domains-and-internal-variability/
Application of results from the divergence theorem, outlined in that post, leads to the equation for overall conservation of energy for the system in this problem: sum of energy out = energy in.
Dan Hughes, thank you for that reference. I remember seeing your piece on Judith Curry’s website, but now I have gone back and read it all more carefully, along with Willis’s comments and replies to others. About the “boundary condition” notion applied to climate models in respect to greenhouse gases, I have long thought, “That can’t be right.” The variable emitter of longwave energy out to space is supplied by a variable-performance heat engine. The concept of the atmosphere as a radiative insulating blanket is only a part of the picture. Willis has been doing great work over the years, observing and describing emergent climate phenomena such as thunderstorms, and the governor concept.