The Hot And Cold Of Space

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. 

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Richard
February 28, 2020 6:14 am

Psst! Hey there, nerd at the desk to my right- could you move your left arm so I can see what you’re writing? I have no idea how I got into this exam, and I don’t want to look like a total moron.

Reply to  Richard
February 28, 2020 4:43 pm

The Stephan-Boltzmann Constant = 5.670 374 419 x 10^-8 W m^-2 K^-4 Exact.
See Numerical value 5.670 374 419… x 10-8 W m-2 K-4>NIST Fundamental Physical Constants CODATA 2018
https://physics.nist.gov/cgi-bin/cuu/Value?sigma|category=physchem_in

Bob boder
February 28, 2020 6:31 am

Willis

can you give me the answer on the back page, i got another class i have to get to.

Douglas Pollock
Reply to  Bob boder
February 28, 2020 7:14 am

Stefan-Boltzman, Watson, Stefan-Boltzman.
Assuming an emissivity coefficient = 1 (at steady state the block behaves as a perfect black body, according to the statement), then:
T = 393.54 °K
T = 120.4 °C
Impossible to write the equation here.

Kelvin Vaughan
Reply to  Douglas Pollock
February 28, 2020 8:28 am

That’s what I made it on the hot side and -270.4K on the cold side.

gbaikie
Reply to  Kelvin Vaughan
February 28, 2020 9:11 am

-270.4 K would not be radiating anything. So it would be very similar to having the backside completely insulated, and remove this insulation after 10,000 hour of the sunlight shining on front.
And before removing insulation the back side should be about 120 C. After insulation on back side is removed, the surface rapidly cools. And with insulation removed for 10,000 hours, most of block will be hot {120 C} and it seems backside will be warmer than -270 K, seems likely it’s about 100 K.
Oh, -270.4 K is wrong it should -270.4 C to make any sense.

Reply to  gbaikie
February 28, 2020 12:44 pm

-270.4 K could be the outside temperature of the unobtanium sides. 😉

Mike Maxwell
Reply to  Douglas Pollock
February 28, 2020 5:02 pm

Since 0 C = 273.15 K, you have one side at 393.54 K, and the other at 120.4 + 273.15 = 393.55 K. That’s the same to almost 5 significant digits. I assume that’s what you intended?

mcswell
Reply to  Douglas Pollock
February 28, 2020 5:05 pm

Front and back the same to almost 5 significant digits?

Tyler
February 28, 2020 6:50 am

42

Reply to  Tyler
February 28, 2020 7:44 am

Lol
+1

Jeff Mitchell
Reply to  Gary D.
February 29, 2020 9:23 pm

So are you saying 1 or 43?

John A Shutt
Reply to  Tyler
February 28, 2020 1:05 pm

..and thanks for the fish?

ironargonaut
Reply to  Tyler
February 28, 2020 8:31 pm

I think Tyler is correct. Since we know that is the answer to the ultimate question. that no one was able to answer. Judging from the myriad responses below it would appear to meet that criteria. So, I say we name him President of the Galaxy as his reward for the right answer.

Ian E
Reply to  Tyler
February 29, 2020 1:41 am

Correct – but, what I need to know is, What happened to the CO2 released whilst the concrete was setting – was it left on Earth???!!!

Tom in Florida
February 28, 2020 6:54 am

T-hot 260F
T-cold -280F

David S
Reply to  Tom in Florida
February 28, 2020 10:17 am

You have the hot side being colder than the cold side?

Nicholas McGinley
Reply to  David S
February 28, 2020 2:22 pm

Is 260 lower than -280?

George Tomaich
February 28, 2020 6:56 am

Hot side 126dC; cold side 107dC.

ATheoK
Reply to  George Tomaich
February 28, 2020 9:23 pm

126dC?

dC = dangCold?

Nick Schroeder
February 28, 2020 6:58 am

Space – the Hotter Frontier

One of the heated issues underlying greenhouse theory is whether space is hot or cold.

Greenhouse theory says that without an atmosphere the earth would be exposed to a near zero outer space and become a frozen ice ball at -430 F, 17 K.

Geoengineering techniques that increase the albedo, the ISS’s ammonia refrigerant air conditioners, an air conditioner in the manned maneuvering unit, space suits including thermal underwear with chilled water tubing, UCLA Diviner lunar data and Kramm’s models (Univ of AK) all provide substantial evidence that outer space is relatively hot.

But outer space is neither hot nor cold.

By definition and application temperature is a relative measurement of the molecular kinetic energy in a substance, i.e. solid, liquid, gas. No molecules (vacuum), no temperature. No kinetic energy (absolute zero), no temperature. In the void & vacuum of outer space the terms temperature, hot, cold are meaningless, like dividing by zero, undefined. Same reason there is no sound in space – no molecules.

However, any substance capable of molecular kinetic energy (ISS, space walker, satellite, moon, earth) placed in the path of the spherical expanding solar photon gas at the earth’s average orbital distance will be heated per the S-B equation to an equilibrium temperature of: 1,368 W/m^2 = 394 K, 121 C, 250 F.

Like a blanket held up between a camper and campfire the atmosphere reduces the amount of solar energy heating the terrestrial system and cools the earth compared to no atmosphere.

This intuitively obvious as well as calculated and measured scientific reality refutes the greenhouse theory which postulates the exact opposite even incorrectly claiming the naked earth would be a -430 F ice ball.

Zero greenhouse effect, Zero CO2 global warming and Zero man caused climate change.

gbaikie
Reply to  Nick Schroeder
February 28, 2020 10:05 am

–Nick Schroeder February 28, 2020 at 6:58 am
Space – the Hotter Frontier

One of the heated issues underlying greenhouse theory is whether space is hot or cold.–

Space is neither hot or cold. The lunar surface, which very close to vacuum of space, is about 120 C in sunlight and after 2 weeks the surface is about 100 K.
But the top surface of Moon surface is about 0 C just before before the sun goes down- or sun at low angle above horizon is not lunar surface which is level {a rock vertical to surface can be he heated by the sun by a lot {120 C}.

–Greenhouse theory says that without an atmosphere the earth would be exposed to a near zero outer space and become a frozen ice ball at -430 F, 17 K.–
No greenhouse theory doesn’t say this. But people might imagine that Earth without an atmosphere could be similar to the Moon. But the Moon is quite different than Earth and the Moon has slow rotation. And if Moon had enough atmosphere to make any difference {a Mars atmosphere would not make much difference} then the daytime lunar surface temperature would be cooler and night time temperature would be much warmer.
But what is as important as Earth’s dense atmosphere is that 70% of Earth is covered by oceans. Or if Moon had thick atmosphere and an ocean covering 70% of the surface, it would quite similar to Earth.

And like Earth is matters where the Land areas are, within the global ocean.
Or we currently in an Ice Age, this is related to where the land is related to the ocean, in non Ice Age conditions on Earth, Earth has much higher average temperature than 15 C.

Bryan A
Reply to  Nick Schroeder
February 28, 2020 10:17 am

Yes, without an atmosphere, the Earth would be hot. Similar to but not identical to the Moon on the Day Side. The Moon’s daylight side remains in Solar Radiation for about 356 hours before It’s Tidally Locked orbit begin to remove the effects of Solar Radiation for a period of 356 hours. The Earth will only heat from Solar Radiation for a period of around 12 hours at the equator before cooling for the next 12 hour period.
So the Earth will not get as hot as the Moon during daylight hours or cool as significantly during the nighttime hours

Nicholas McGinley
Reply to  Nick Schroeder
February 28, 2020 2:47 pm

Nick,
One hundred kilometers up, at the edge of space, there are about one million million million molecules per cubic meter.
Further up at the space station, there are a hundred times less…only ten trillion per cubic meter.
Halfway to the Moon, away from the influence of the atmosphere, there are still a lot of molecules.
Some 7,000,000 per cubic meter.
I do not know where this empty place with no temperature is.
And what does it mean to say at a temperature of absolute zero, there is no temperature?
You said what the temperature is in the same sentence you said there was not one.
If we are doing science, let’s do science.
If you want to demonstrate that 7 million molecules per cubic meter is not a lot of molecules…do that.
But if the point is that with no molecules there is no temperature, and yet there are molecules, what exactly are you saying?
The upper atmosphere of the Earth is called the thermosphere cause it is hot, and it gets hotter when solar activity is high.
This is because the average velocity of the molecules is high, even though there are not a lot of them.

BTW…here is enough air there (at the height that low Earth orbit objects are placed) to have caused the crash of the Spacelab satellite when the atmosphere heated up and expanded more than anticipated back in the 1970s, IIRC.

I am wondering if anyone who knew little to nothing about this subject would be able to learn anything by reading these comments.
I feel like I just got stupider.

Space has no temp because it has no molecules. But it does have some.
Space is hot because stuff in the unfiltered sunshine above the Earth gets real hot on one side?
Huh?
That has nothing to do with the temperature of space.
Even here in Earth we do not count the temperature of an object in direct sun to be an actual air temp.
All measurements are specifically understood to be, and measured in, the shade.
Because otherwise you are measuring a radiation effect, not anything to do with the motion of the molecules.

I am not complaining or anything…you guys comment however you want.
I just do not understand what the point is.
Personally, I like to say things that inform people.

Bryan A
Reply to  Nicholas McGinley
February 28, 2020 10:24 pm

There are 3.3455×10^22 molecules of water in a gram of water. 334,550,000,000,000,000,000,000 (334 sextrillion molecules…
A gram of water is approximately equal to +/- 1/4 teaspoon.
7,000,000 molecules of water takes up 1/4.9357143e+15 (1/4,935,714,300,000,000) 1-5 quadrillionth of a teaspoon. 7,000,000 is a very large number unless you are talking about a very small object
7,000,000 water molecules will fit on the Head Point of a Needle

Nicholas McGinley
Reply to  Bryan A
February 29, 2020 3:10 pm

Your examples would be more apt if you used gas molecules in air.

Crispin in Waterloo
Reply to  Nick Schroeder
February 28, 2020 4:36 pm

Nick S

You are making a conceptual error of great importance:

“Greenhouse theory says that without an atmosphere the earth would be exposed to a near zero outer space and become a frozen ice ball at -430 F, 17 K.”

Greenhouse gas theory makes no such assertion. The Greenhouse gas theory (as expounded by the IPCC and Gavin and lots of others who are conceptually misdirected) reads more like this:

“Greenhouse theory says that without any GHG’s in the atmosphere the near-earth average air temperature would the same as the average temperature of the surface of the moon, which has no atmosphere.”

“Greenhouse gas theory” has nothing to say about planets without any atmosphere at all. Greenhouse gas theory is limited to the effects of greenhouse gases. Greenhouse gas theory is subject to the laws of physics, not a progenitor of them.

It the Earth had an atmosphere containing no GHG’s at all, Gavin says (contravening physical laws governing convective heat transfer) the air would be as cold as the surface of the moon. That is how screwed up is the conceptual framework within which NASA/GISS operates.

Reply to  Crispin in Waterloo
February 28, 2020 5:43 pm

Thank You.
I have had the same thoughts since I first heard of this CC Propaganda.
As a Nuclear Engineer I learned about flow in a pipe, and Nucleate Boiling. Even with flows of hundreds of feet per second a “film” will persist on the surface of the pipes in coolers or heat exchangers. This film is an insulator, that is, it slows down the transfer of heat from the pipe to the liquid or vice versa. This same effect is present in the atmosphere surrounding the earth. The reason for dimples on golf balls is to minimize this persistent film on the golf ball. Relatively, the Earth is a thousand times smother than the dimpled gulf ball thus the atmosphere is relatively stagnate close to the surface creating an insulator several feet thick of poorly conducting AIR. three to six feet of air is a very good insulator. Period. This is clearly shown in the temperature gradient of atmosphere. There are many charts on the internet. This will be present with or without CO2. It has been much to long since I took advanced thermodynamics courses so someone else can calculate the mu or thermal resistance of this effect.

Crispin in Waterloo
Reply to  Uzurbrain
February 29, 2020 6:08 am

Uzurbrain

I caution against describing any “stationary” layer or film or clinging when it comes to describing a surface with a flowing fluid (air is also a fluid). The reason is that what you describe is a representation of reality, is a way to think about a problem, not that there is an actual stationary layer, even one molecule thick.

I had the company of a newly graduated young French engineer in Dakar during the design of a concrete cooking stove and the modeling of the equilibrium outside surface temperature was that day’s task. The engineer had been taught the most recent approach to making this calculation which was to assume that there was a stationary “layer” that was conductive only, with no convective heat transfer at all, and that the default thickness was 0.1mm. This “layer” does not exist, it is only a mathematical convenience that approximates what really happens and is, as they say, good enough for government work.

Willis’ proposed problem is interesting for a student: how literally should we take the problem description? The material is given as “concrete”. Heat flow through concrete is like what happens with a porous matrix filled with fluid. If we take it that the 1 metre cube was cut from a long 1×1 m beam, the faces would have a number of sliced stone pieces bound in a limestone matrix. The emissivity of the face is given as 0.95 (average) and that is fine, however a good student will ask what the aggregate size is. The radiation from both faces will be affected by a factor being the aggregate size divided by the length (because the heat conductive properties of the aggregate and the matrix are different). The effective length of the block will be less than one metre if the heat conduction is the average of 0.8 W /sq m/K. In reality (not in effect) the aggregate will mine heat from within the body and radiate it to space more effectively that would a homogenous material (which concrete is not).

For an engineer to ignore the difference, they should first calculate the effect of the granularity and then show that the difference is or is not significant. If it was like a hydro dam containing 76mm aggregate, the difference is probably significant given that the input power was provided to 4 significant digits and the constants are presumed to be perfect.

LOL@Klimate Katastrophe Kooks
Reply to  Crispin in Waterloo
March 1, 2020 5:58 pm

All gaseous molecules absorb radiation under some circumstances. For instance, N2 and O2 are actually statistically significant absorbers of radiation due to collisional perturbation at the same time as they are impinged upon by a photon. The collision perturbs their magnetic dipole, allowing them to absorb radiation, which they ordinarily wouldn’t do.

There’s a recent study somewhere online, I’ll try to find it. I think I saw it on NoTricksZone.

{ Ten Minutes Later… }

Here it is:
Scientists: Oxygen & Nitrogen ‘Radiatively Important’ Greenhouse Gases With IR Absorption Temps Similar To CO2
https://notrickszone.com/2020/02/10/scientists-oxygen-nitrogen-radiatively-important-greenhouse-gases-with-ir-absorption-temps-similar-to-co2/

Of course, any molecule which can absorb radiation must also be able to emit that radiation.. and given the predominant heat transport means of convection, a N2 or O2 molecule would likely absorb a photon, become excited in one of its available quantum states with some of that energy equipartitioned into translational energy, thus be convectively transported higher in the atmosphere, then release that energy in the form of a photon… so gases actively “pump” energy upward via convection.

And given that the mean free path length of radiation increases with altitude, this means there is more upward terrestrial LW flux than downward. There is no “greenhouse effect” in the atmosphere as the leftists claim it to be. Once again, the leftists and climate loons have turned reality on its head and gone off squawking about the end of the world, when in reality it’s exactly opposite to what they claim. I’ve noticed they tend to do that a lot.

If Earth had no “greenhouse gases”, then no gases could radiatively emit energy to space. Radiative emission to space is the *only* means by which the planet can shed energy. So Gavin Schmidt claiming the atmosphere sans “greenhouse gases” would be cold is, yet again, diametrically opposite to reality. In reality, the gases would heat up via conduction upon contacting the surface and convect upward… but they couldn’t emit that energy to space. The whole of the atmosphere would heat up, with a very small lapse rate.

Reply to  LOL@Klimate Katastrophe Kooks
March 3, 2020 10:34 am

The problem I have is that the AGW Crowd keeps talking about the absorption od energy in CO2 and ignoring all of the rest of the energy hitting the earth. The percentage of energy from sun in the IR spectrum is 1/10^23 of the energy spectrum emitted by the sun. Even one tenth of 10^23 times the IR energy absorbed is definitely significant.
They also ignore all of the energy from cosmic ray interactions, One source said that the typical cosmic ray impact releases more energy into the atmosphere than all of the power generated on Earth! ! The article also claims that there are hundreds of thousands cosmic ray strikes strikes per hour. What is all of that energy doing? The article was describing Cosmic Rays and had no discussion at all on “Climate Change.” Surely the location of the Earth in our Galaxy affects the number of Cosmic rays hitting us. Also the activity on the Sun also affects the magnetic field around the earth and also affects the number of cosmic rays hitting the Earth. I can hear this activity as background noise while using my Amateur Radio. There are times when this noise is more than an S9 signal. Even the noise from Jupiter rises and falls in a predictable manor. And that “Noise” is strong enough at times to operate a one transistor radio. [Radio JOVE – NASA]

Reply to  Uzurbrain
March 3, 2020 12:59 pm

Henrik Svensmark a professor at Danish Technical Univeristy has been investigating the cosmic ray formation of cloud nuclei, after he found out, that the weather changes seen in phase with the solar cycle of approx 11 years could not be explained by TSI wich only vary approx 0.1 pct. He found that the solar wind blows away the cosmic rays and in low solar activity periods it do not. So cosmic rays forms cloud nuclei, which form the clouds, which controls the temperature of the earth.

Robert of Ottawa
Reply to  Nick Schroeder
February 29, 2020 1:56 pm

Oh come on, assume a spherical Earth.

Reply to  Nick Schroeder
March 1, 2020 7:56 pm

Nick Schroeder wrote, “any substance capable of molecular kinetic energy… placed in the path of the spherical expanding solar photon gas at the earth’s average orbital distance will be heated per the S-B equation to an equilibrium temperature of: 1,368 W/m^2 = 394 K, 121 C, 250 F.”

From the fact that there are actually many such objects which have average temperatures much lower than that you should realize that your statement cannot possibly be correct.

For example, the average temperature of the moon (which has no atmosphere) is much lower than the average temperature of the Earth (which does have an atmosphere), even though they receive the same solar irradiance.

In fact, there is not single, predictable temperature for an object one AU from the Sun, because there are many factors which affect such an object’s temperature.

Differences between the radiation emitted by an object orbiting one AU from the Sun, and the radiation absorbed by that object, can cause the object’s temperature to vary greatly.

One obvious difference is “color.” Solar radiation is mostly much shorter wavelengths than radiation emitted by objects orbiting the Sun. An object which is has high emissivity for long wavelength radiation, but high reflectivity to the shorter wavelengths which dominate solar radiation, will end up being cooler than a similar object which has low emissivity for long wavelengths, but low reflectivity to the shorter wavelengths that dominate solar radiation.

Another difference is that radiation is absorbed by our orbiting object only on the side facing the Sun, but it is emitted in every direction. Manmade satellites are often engineered to make use of that fact. An object which is reflective on the side facing toward the Sun, but black on the side facing away from the Sun, will stay much cooler than an object of uniform color.

Another difference is rotation. Consider the Earth and its Moon. Although the Earth and Moon receive the same level of solar insolation, even if the Earth had no atmosphere, and even if its albedo was identical to that of the moon, the Earth’s average temperature would still be higher than the average temperature of the moon.

That’s because the Moon rotates only 1/27-th as fast as the Earth. That means each side of the Moon is heated by the Sun for 27 times as long as each side of the Earth is heated; and then cools for 27 times as long, as well.

That means that, even if the Earth had no atmosphere, the temperature extremes on the Moon would be much greater than on the Earth. The highs would be higher, and the lows lower.

But since radiative emissions are proportional to the 4th power of surface temperature, there would be more rapid energy loss during the Moon’s higher highs, so the difference between high temperatures on the Moon and Earth would be less than the difference between low temperatures on the Moon and Earth. In other words, the Moon’s average temperature would be lower than the Earth’s average temperature.
 

Nick Schroeder continued, “Like a blanket held up between a camper and campfire the atmosphere reduces the amount of solar energy heating the terrestrial system and cools the earth compared to no atmosphere.”

That’s not working very well for Venus, is it?

Dodgy Geezer
February 28, 2020 6:58 am

Psst! …. Do you do it by S = UT + 1/2 AT squared….?

Red94ViperRT10
Reply to  Dodgy Geezer
February 28, 2020 7:20 am

Nope. Cuz U is a kluged term to combine the heat transfer resistance against radiation, conduction and convection. This problem deals only with radiant heat transfer, so all you need is emissivity.

Paul
Reply to  Dodgy Geezer
February 28, 2020 10:55 pm

If T1 = temp (K) at sunny side and T2 is the temp at opposite side then heat will flow by conduction to the cold side until T1=T2.
Then T2 and T2 will be 255.15K

Leo Smith
February 28, 2020 7:04 am

I would say it would end up at half the sum of the temperature of space and the object that is illuminating it

I am not sure that the thermal conductivity is particularly relevant. That will just act to raise the upstream a bit and lower the downstream a bit

What is the albedo of ‘grey’?

Nicholas McGinley
Reply to  Leo Smith
February 28, 2020 3:10 pm

Distance to the illuminating object would not matter?
The Sun is thousands of degrees.
Nothing gets 1/2 that hot in Sunshine, 90 something million miles away.

Clyde Spencer
Reply to  Nicholas McGinley
February 28, 2020 5:52 pm

Nicholas
The distance is unimportant because the flux of the source is not given, just what is impinging on the concrete block, which coincidentally is the same as what arrives at the top of Earth’s atmosphere. By deduction, the distance is probably 93 million miles; however, it could be different if the source were different.

Nicholas McGinley
Reply to  Clyde Spencer
February 28, 2020 7:35 pm

Yes, I understand the problem as stated perfectly well.
I was responding to this sentence:
“I would say it would end up at half the sum of the temperature of space and the object that is illuminating it…”
Which seems to be saying that only the temperature of the source of the incoming energy need be considered in order the known the temp.

Nicholas McGinley
Reply to  Clyde Spencer
February 28, 2020 7:36 pm

And thank you for the reply, Clyde.

Clyde Spencer
Reply to  Leo Smith
February 28, 2020 5:48 pm

Leo
You asked, “What is the albedo of ‘grey’? Probably about the same as the moon.

Robert W Turner
February 28, 2020 7:12 am

Answer: the block never existed because it started at absolute zero. But let’s say it did exist at 0.01 K, then it never warmed up because it is surrounded by a perfect insulator and receives no insolation.

MarkW
Reply to  Robert W Turner
February 28, 2020 7:30 am

It is only surrounded by the perfect insulator on 4 sides.

Robert W Turner
Reply to  MarkW
February 28, 2020 7:40 am

Oh I see.

Bob boder
Reply to  Robert W Turner
February 28, 2020 8:17 am

What Chair, i mean block?

Joe Campbell
February 28, 2020 7:13 am

Willis: What’s the absorption coefficient, alpha, of the surface of the concrete,?…

Frenchie77
Reply to  Joe Campbell
February 28, 2020 7:21 am

Yeah, was just gonna ask that. You can’t solve this problem without knowing how much solar irradiance is absorbed. We can assume that the absorption is constant across the incident wavelengths, but the problem should at least provide an absorption a.

Otherwise, I assume a = 0 and then it remains at the initial state.

Frenchie77
Reply to  Willis Eschenbach
February 28, 2020 9:39 am

Sorry Willis, but if that was the case spacecraft would roast their electronics.

I’d put some links here but am not sure if that is allowed. Nevertheless, just google spacecraft thermal emissions and absorption materials and I am sure that you’d find examples.

I run a lot of research on this and can confirm they are different. We are getting some very nice materials (with lwoer cost ) soon to be qualified that have emissions >0.9x while their absorption is <.1, sorry can't be more specific. This is key to improving thermal radiator designs.

Clyde Spencer
Reply to  Willis Eschenbach
February 28, 2020 6:05 pm

Willis
You assumed that there was no dispersion of emissivity with wavelength, i.e. the behavior is the same with visible-light as with IR. So, we have a contradiction. If the absorptivity in the visible region is 0.7, then the albedo has to be 0.3; however, if the absorptivity is 0.95. then the albedo has to be 0.05! You can’t have it both ways unless you change the parameters of the problem by allowing the visible-light behavior to be different from the IR behavior.

Clyde Spencer
Reply to  Willis Eschenbach
February 28, 2020 5:58 pm

Joe, Frenchie, and Willis,
If the absorptivity is 95%, then the reflectivity is 100%-95% or 5%. This is then not a grey body, but a black body, literally, with a reflectance on the order of magnitude of coal, or that mythical Arctic “Dark Water.”

LOL@Klimate Katastrophe Kooks
Reply to  Willis Eschenbach
March 1, 2020 6:30 pm

Kirchhoff’s Law is a ratio, not an equality.

In Kirchhoff’s original parlance:
E/A = e
Not E=A.

In modern parlance:
Eν/αν= f(T, ν)

In other words, the ratio between emissive power and absorptivity is equal to specific intensity.

In describing his original formula, Kirchhoff wrote “emissivity”, when he meant “emissive power”. They are not the same.

Only at thermodynamic equilibrium is emissive power and absorptivity equal. If they were always equal, an object could never change temperature.

You’ll note the IPCC assumes in its equations that emissive power = 1 and thus absorptivity = 1, which is clearly unphysical. CO2 is not a hypothetical perfect blackbody.

The correct formula: h = (e (A) (σ)(Ts ^4 – Ta ^4)) / (A * ΔT)
The incorrect IPCC formula, applied to gray bodies: h = ((σ)(T^4)) / (A * ΔT)
Note the lack of e in the IPCC formula… they assume emissive power = 1 and thus absorptivity = 1 at thermal equilibrium, a hypothetical perfect blackbody.

Red94ViperRT10
February 28, 2020 7:15 am

I once owned the textbook to be able to solve this. You may have just copied one of the example problems. But I’ll be darned if I know where it is at the moment. And at least 30 minutes of good hard thought just to narrow down where to look for it.

Tom Bakewell
Reply to  Red94ViperRT10
February 28, 2020 8:13 am

Maybe the book is John Lienhard’s “A Heat Transfer Textbook”

Red94ViperRT10
Reply to  Tom Bakewell
February 28, 2020 9:32 am

I’d have to find it to tell you the author.

Mike McMillan
Reply to  Red94ViperRT10
February 28, 2020 6:46 pm

The chart of thermal conductivities has the formulas at the bottom of the page.

https://www.engineeringtoolbox.com/thermal-conductivity-d_429.html

Roy W. Spencer
February 28, 2020 7:16 am

Hint: Make heat budget equations for the front and back sides, each with heat gains equal to heat losses. Two equations in 2 unknowns (the temperatures).

Joe Campbell
Reply to  Roy W. Spencer
February 28, 2020 7:32 am

Roy: +100

LdB
Reply to  Roy W. Spencer
February 28, 2020 7:52 am

ROY the problem is the sides .. he has tried to get around them by say they are insulating but unless they don’t conduct as well they will have a gradient and they will re-radiate thru the back face anyhow. You basically end up with a soldering iron tip situation the front face and the sidewalls make up a larger area than the back face so it’s hotter than you would directly calculate.

LdB
Reply to  Willis Eschenbach
February 28, 2020 8:54 am

That doesn’t stop the material re-radiating as a conduction. So let guess you are trying to use this forumla for the conduction
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thercond.html#c3

Note the last part of the statement

Algebraic methods can be used for the calculation of conduction heat transfer across plane walls, but for most geometries the heat transfer must be expressed in terms of the thermal gradient.

What is happening is the material re-radiates each point in the material becomes it’s own heat source. By insulating you stop the external bleed but it doesn’t stop the same thing happening internally. So I am sorry there simply is no way to remove the sidewalls from the calculation.

The only true answer to your question is a rather complicated gradient calculation.

LdB
Reply to  LdB
February 28, 2020 9:19 am

Due to the delay these are going to end up in wrong order. However I suggest you read why you need a gradient calculation even on the conduction start by READING THE LINK.

It would appear you want the naive calculation which will pop up above this one which is make the dam thing a large flat sheet because you want to use the flat sheet approximation.

This is a bit like D..slayers not getting the whole re-radation thing, if yuou haven’t worked out the concept by now I am not sure I can help.

Loren Wilson
Reply to  LdB
February 28, 2020 8:08 pm

These walls are the equivalent of saying that the heat flow is in one direction only, which makes the problem much easier. No finite element analysis required. I’ll pull out my heat transfer book tomorrow and give it a try.

LdB
Reply to  Willis Eschenbach
February 28, 2020 9:13 am

Willis if you do want the naive answer then just make it a sheet 20m x 20m sheet and you want the formula for the middle square meter so the side faces are 10m away and we can ignore the gradient. You don’t need unobtainium then either because you entitled to use the flat sheet approximation.

Reply to  LdB
February 28, 2020 11:18 am

Willis,
Yeah, sure, I’m misleading people. Nice!

If CSR = CHF, you would think Kirchoff, Boltzmann, and Planck would have noticed it. But no, when the walls of the cavities they inserted carbon or lamp soot stabilized (CHF=0), B & P discovered the laws of thermal emission. CSR != CHF

But no, they are dumb and you are a genius who figured it all out with your armchair physicist assertions.

Can you walk and chew gum at the same time?

Molecules can move and emit radiation at the same time, and this emission doesn’t slow them down.

Greg
Reply to  LdB
February 28, 2020 12:57 pm

“Molecules can move and emit radiation at the same time, and this emission doesn’t slow them down.”

There you violating conservation of energy again.

Reply to  LdB
February 28, 2020 3:11 pm

Greg,

“There you violating conservation of energy again.”

EM radiation is a REFLECTION of Kinetic Energy, not its SPENDING.

Molecular motion disturbs EM fields, causes radiation.

If radiation were to eliminate motion, then you couldn’t get motion at all, and thus no radiation.

I can observe you walking without you slowing down, so to speak.

Reply to  Roy W. Spencer
February 28, 2020 10:10 am

Great idea, Roy. But Willis wants to include conductive heat flux, otherwise, the solution is too easy (394K).

Hot Side Radiation = Conductive Heat Flux = Cold Side Radiation

HSR => [ CHF ] => CSR

Conduction Formula: q = KA(Th-Tc)/L
Radiation Formula q = ɛσT⁴

Simplify:

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

Ed Bo
Reply to  Zoe Phin
February 28, 2020 1:47 pm

Zoe:

You are demonstrating your complete inability to analyze what is really a very simple problem. You obviously have never taken even an introductory thermodynamics course, because you can’t get even the most basic analysis correct. You can’t do even trivial energy balance (1st Law) calculations.

To satisfy the 1st Law for the cube as a whole in steady-state conditions, HSR + CSR = AbsorbedSolar.

To satisfy the 1st Law for the far end of the cube in steady-state conditions, CHF = CHR.

Those are the constraints of the problem. Many of us here did this exact analysis, as we learned in the first couple of weeks of our first thermo courses. It’s not that hard.

Crispin in Waterloo
Reply to  Ed Bo
February 28, 2020 4:40 pm

Agreed Ed Bo

It is a first term mechanical engineering thermo problem. It introduces the concept of an albedo for the receiving and re-radiating surface, heat conduction at a given rate and an albedo for the cold side radiating into a space with no return.

Reply to  Crispin in Waterloo
February 28, 2020 6:20 pm

So why hide the evidence?
Show it in the book.

Crispin in Waterloo
Reply to  Crispin in Waterloo
February 29, 2020 10:28 am

Zoe

I am not sure what question you are asking. In order to each the elementary principles of heat transfer there are set problems with many simplifications in order to concentrate the mind on one or two factors. Another task is to calculate the equilibrium temperature of a hollow cube which is insulated and has a window of given dimensions, and a heat source inside with power [Watts].

The question posed by Willis is of this instructional type. The answer has been provided by several contributors. There are on-line calculators for such problems that have all the necessary factors included. These problem usually involve air or water flow and are much more complex than radiative problems.

Remember that the great author of heat transfer textbooks, Adrian Bejan, looked at the GHG-based global warming calculation and said that is as so simple it was not even interesting. His book on Convection Heat Transfer has problems that are extraordinarily difficult that explain things like the development of sets of thunderstorms on hot days and how convective cell structure changes with temperature, and why (meaning the physical and mathematical basis for the changes).

I can add that the temperature response of the atmosphere to an increase in CO2 concentration is very modest, and we are far better off spending our time developing “forever” energy production methods than we are worrying about ending the use of a set of declining resources. We have to have this problem solved within 200 years, 5 or 6 generations.

Reply to  Zoe Phin
February 29, 2020 7:25 am

Zoe says: “Great idea, Roy. But Willis wants to include conductive heat flux, otherwise, the solution is too easy (394K).”

Yes, with no heat conduction through the cube, the front side is 394K. And the back side is 0K. Good work!

With NO conduction, the front is 394K and the back side is 0K.
With a TINY conduction, the front will get slightly cooler and the back will get slightly warmer.
With a bit more conduction, we get the answers provided by multiple people.
With PERFECT conduction, then the front and back are the same.

Stan Robertson
Reply to  Roy W. Spencer
February 28, 2020 4:10 pm

I forgot to knock off the 5% reflection and just took the 1360 W/m^2 as the flux absorbed at the front surface. Your two equations are a bit messy, but can be solved iteratively in about a dozen steps. I came up with 388.5 K at the front and 222.7 K on the cold side. Anyone who can write the correct equations can easily verify the solution without doing much work.

Stan Robertson
Reply to  Stan Robertson
February 29, 2020 10:10 am

If you take off the 68 W/m^2 reflected from the front, it changes the front and back temperatures to 383.8 K and 221.4 K, respectively.

Roger Taguchi
Reply to  Stan Robertson
February 29, 2020 2:15 pm

To Stan Robertson: I agree (383.3 K and 221.4 K).

Douglas Pollock
February 28, 2020 7:19 am

Stefan-Boltzman, Watson, Stefan-Boltzman.
Assuming an emissivity coefficient = 1 (at steady state the block behaves as a perfect black body, according to the statement), then:
T = 393.54 °K
T = 120.4 °C
Impossible to write the equation here.

Anthony Mills
February 28, 2020 7:20 am

The solar absorptance of rough concrete is 0.60. The total hemispherical emittance at 300K is 0.91.Your problem statement requires the use of a gray body model with an emittance of 0.95.Your result will be meaningless.The essential feature of radiation heat transfer problems involving solar radiation is the non gray behavior of most surfaces .Check an engineering heat transfer textbook.

Ben Vorlich
February 28, 2020 7:25 am

Will,
I reckon that the hot side to cold side temperature difference has to be large enough to transmit the 1360W. As you’re using 1square metre for the area and 1 metre for the length of transmission then the temperature difference is 1360÷0.8. Which is 1700’C

So if the cold side doesn’t heat then its
Hot = 1427’C
Cold _ – 273’C

Next I will have to work out how the block heats up which is more tricky and will take a bit of tbought

Tom Bakewell
February 28, 2020 7:40 am

@Red94ViperRT10

Maybe John Leinhard’s excellent “A Heat Transfer Textbook” ?

LdB
February 28, 2020 7:41 am

Most have got the front side because it is relatively easy with SB

I suspect whatever answer you derived is actually wrong Willis because this requires a complex Integral.
The issue is the graphic and question text Willis misleads there are 5 cold sides not 1 🙂

The back side if you ignored the actual 4 sides would be straight forward it is just thermal conductivity thru 1 square meter . The sides create the problem they get progressively colder the further they are away from the hot face creating a gradient along them. The front face to back face conduction is what 98 watts/sec so the side face re-radiation are also going to be significant. Each of the hotter gradient along the sides will also radiate out the back face because it is the furthest away and hence coldest.

So my question is do you really want the hard answer or were you trying to do the naive answer ignoring the sides?

LdB
Reply to  LdB
February 28, 2020 8:09 am

I should say adding a perfect insulator around then sides does not simplify the problem they represent because they conduct. As I stated to Roy you have made a soldering iron tip and it’s a little more difficult than the naive answer.

Greg
Reply to  LdB
February 28, 2020 10:19 am

“The front face to back face conduction is what 98 watts/sec ”

I have a general rule that when people get the units wrong , they probably don’t understand the physics.

LdB
Reply to  Greg
February 28, 2020 10:53 am

So do I. So either you don’t get why it has to have a time component or you don’t get the difference between say KW and KWh.

Greg
Reply to  LdB
February 28, 2020 12:51 pm

Firstly it’s kWh not KWh. Second you have the equivalent of kW/h not kWh, I rest my case.

LdB
Reply to  Greg
February 28, 2020 4:43 pm

ROFL

ironargonaut
Reply to  Greg
February 28, 2020 8:12 pm

In case you are wondering why he is ROFLing…”The kilowatt-hour is a composite unit of energy equal to one kilowatt (kW) of power sustained for one hour. ” in other words kW per hour or kW/h. Even I knew that one.

LdB
Reply to  Greg
February 28, 2020 11:11 pm

He was trying to play like some do with English and spelling, that they are more superior. We have a non flattering word for him in the industry, its the old case of those who can’t teach.

Crispin in Waterloo
Reply to  Greg
February 29, 2020 11:43 am

Units:

“… in other words kW per hour or kW/h”

The Watt has a time component so there are never W/hr (Watts per hour) in any form.
kWH, often written KWH is one thousand Joules per second for an hour. 1 kWH is 3.6m Joules of energy moved or dissipated or released in the course of one hour.

Any number of Joules per second divided by 3.6 million = kWH which converts the energy component (Joules) and time component (seconds) into kJ and Hr.

Always some to WUWT – learn or teach something every day without finger pointing.

WXcycles
Reply to  LdB
February 28, 2020 11:40 pm

” … The sides create the problem they get progressively colder the further they are away from the hot face creating a gradient along them. …”

LDB, the entire block is going to have the same gradient, illuminated side to shadowed side. I don’t know why you’re getting snarled-up on a side-issue. 😉

Ed Zuiderwijk
February 28, 2020 7:43 am

Hot 125C
Cold -145C

Philip
February 28, 2020 7:46 am

One comment, which may or may not be helpful:

The only part of the block that matters is the face.
The other three sides receive nothing and radiate nothing.
So rather than being 1m deep, it could be infinitely thin.
The only difference that would make is the length of time taken to reach steady state.

Radiation in = radiation out.
What temperature does an object with a known emissivity have to be to emit X watts/m^2

Bob boder
Reply to  Philip
February 29, 2020 9:11 am

A block has 6 sides

Citizen Smith
February 28, 2020 7:47 am

Group problem solving in history.

Mr.
Reply to  Citizen Smith
February 28, 2020 9:50 am

So correlation IS causation, exactly as AGW hypothesises, and King Arthur proves via witch = duck deduction.

Take that, den1ers!

yet another Mike
February 28, 2020 7:50 am

emissivity of .95, with constant influx of heat on one side (1360 should eventually heat the core to critical temp of planet explosion right ? Blowing out the sunny side. leaving a much cooler half meter thick block which will then start reheating until in half the time cause the reduced thickness remaining block to once again blow off the heated side. ad infinitum eventually exfoliating the block to a thin post card of concrete. cause the ocean stole the heat. and stored it at 4 degree above freezing in the deep.
Alternative answer: The stray CO2 molecule that arrived on the heated face super heated the concrete block driving it to critical temp near infinite Kelvin and it totally vaporized.
Wait the heated side caused the block to start rotating eventually reaching overspin condition and threw off a large mass of material from the prior heated side forming a smaller less dense satellite which began orbiting its reduced spinning source and they lived happily ever after, until the smaller satellite got its faced locked toward the block remanent.
Or something like that.

Ben Vorlich
February 28, 2020 7:58 am

Willis
Sorry spell checker got your name last time

I went off on a tangent mentally there. The 0.95 emissivity gives the answer to the cold side. So the temperature difference cold side to Absolute Zero has to be
1431’C using the same calculation as before
So my bizarre answer is
Hot = 3131’K
Cold=1431’K

But most likely wildly incorrect

I look forward to seeing the solution

Johanus
February 28, 2020 8:01 am

Simple. Use Stefan-Boltzmann equation to compute temperature for given energy flux and albedo.
Then apply Fourier’s Law (not to be confused with Fourier Series, but Fourier had to discover the series to solve the equation) to compute heat conducted through the cube with the given dimesions and thermal conductivity.

I’ll leave the details and final result as an exercise for you students. :-]

Matthew Sykes
February 28, 2020 8:03 am

120C
44C

Matthew Sykes
Reply to  Matthew Sykes
February 28, 2020 8:28 am

Typo, that is 4.4 C

Matthew Sykes
February 28, 2020 8:10 am

Oh, conductivity is w/m.k not w/m/k

GHreg
Reply to  Matthew Sykes
February 28, 2020 11:22 am

firstly watt = W , not w ; kelvin is K not k .

secondly W/(m.K) is identical to W/m/K , whoever gave you w/m.k is wrong on three counts.

Matthew Sykes
Reply to  GHreg
February 29, 2020 6:31 am

The formula for conductivity is QL/A deltaT. This tells you the units. Wm/m^2K.

W/mK.

If you divide m by K you end up flipping K to become WK/m.

Crispin in Waterloo
Reply to  GHreg
February 29, 2020 11:33 am

“w/m.k…”

W/m•K is correct.

The • character is produced by holding the Alt key down and typing 0149 on the number pad.

Matthew Sykes
Reply to  Crispin in Waterloo
March 1, 2020 4:12 am

Or full stop when written. It replaces the ‘x’ symbol for multiplication of course, to avoid confusion with variables called ‘x’.

I spent many years studying and working in engineering (Mechanical, also Thermodynamics and fluid dynamics). it is watts / (meter x temperature).

And yes, James Watt was the inventor of the steam engine, I am fully aware they symbol for power and temperature are in upper case.

I also speak French, and m well aware of the use of the alt+numeric keypad number to generate characters form the ASCII table, alt+132 for example.

In fact I am am now a software engineer, I am fully aware of the ASCII table in it’s entirety.

In software it would be W/m*K. The asterisk is used for multiplication in software calculations.

‘^’ is used for the power of, so it can also be written: W*m^-1*K^-1 if you understand software notation.

In fact I know rather a lot about all of this sort of stuff.

Ed Bo
Reply to  Crispin in Waterloo
March 1, 2020 6:21 pm

Using W/m*K for conductivity is ambiguous at best, but actually wrong.

The rules of algebraic precedence have multiplication and division at the same level of precedence, with operations from left to right. So in that format, it is equivalent to (W/m)*K, and the K ends up in the numerator, which is wrong.

W/m/K is correct, as is W/(m*K).

Matthew Sykes
Reply to  Ed Bo
March 2, 2020 6:07 am

Operator precedence in the C language is hardly relevant.

10/10/10 is not the same as 10/ 10 x 10

Ed Bo
Reply to  Ed Bo
March 2, 2020 8:10 am

I was referring to the standard algebraic rules of precedence, which have been followed for centuries. There is a reason all serious programming languages follow them.

You say: “10/10/10 is not the same as 10/ 10 x 10”

My point exactly. “W/m*K” is not the same as “W/m/K”. The second is correct. The “K” term must “end up” in the denominator.

Rick C PE
February 28, 2020 8:16 am

Don’t have time to work this out, but as the block warms up the energy absorbed will decrease. It is proportional to the difference of the 4th powers of the emitting and receiving surfaces (i.e. sun, block) per S-B. I think we can assume a constant sun surface temperature. I suspect Willis intended his hypothetical block to have a 0.95 absorbtivity as well as a 0.95 emissivity.

Also, Dr, Spencer is correct, I think, that it requires solving two simultaneous equations.

Ulric Lyons
February 28, 2020 8:19 am

389K and 327K, as long as there no gas inside conducting heat from the front to the back.

Ulric Lyons
Reply to  Ulric Lyons
February 28, 2020 6:13 pm

Oh it’s a solid block! silly me.

February 28, 2020 8:21 am

Once the block is in equilibrium, the hot and cold side would be the same temperature.

The perfect insulator guarantees this, so long as the conductivity is greater than zero.

February 28, 2020 8:24 am

Correction. I thought the cube was insulated on 5 sides.

Reply to  ferdberple
February 28, 2020 9:23 am

correction to correction. Got the same answer using 4 sided insulation, by treating block as a part of much larger sphere. In the absence of a greenhouse effect, the material is predicted to be isothermal.

Ed Bo
Reply to  ferdberple
February 28, 2020 1:24 pm

Fred:

Like many, you are confusing static equilibrium with dynamic steady-state conditions. This problem is clearly the second case.

The sun-facing side is (net) absorbing power, and the back side is (net) outputting power. So there will be a temperature difference between the two ends that leads to ongoing conduction.

Gino
February 28, 2020 8:24 am

First, kirchoffs law states that absorbtivity equals emissivity. There fore, use boltzman to calculate the absorbed surface energy.

The use boltzman to calculate the required surface temperature to emit that energy flux, assuming radiation to a 0 degree sink. That will give you Tc of the concrete.

Next apply fourier’s law of conductivity to determine the temperature gradient across the block of concrete (conduction coefficient provided by Willis). That will give you the temperature on the hot side.

This neglects surface back radiation to other cold sinks that may be available on the hot side of the block because that side will emit as well unless there is no “visible” low temperature sink.

Frenchie77
Reply to  Gino
February 28, 2020 10:06 am

Sorry gents, you misunderstand kirchoff’s law, maybe better is you are misapplying it. Google the sort of bible of material properaties, nasa ref pub 1121, dated april1984, “Solar absorption and thermal emittance of some common spacecraft thermal coatings.”
It is not the most up-to-date complete list, but it is a good reference to begin with.

Needless to say, we design many materials to have quite differnt a and e.

Johanus
Reply to  Frenchie77
February 28, 2020 11:10 am

@Frenchie77

No, it appears you do not understand that Kirchoff’s Law states that absorption and emittance are equal at the same wavelength.
Pub 1121 defines ‘a’ and ‘e’ at different wavelengths (i.e. black-body temperatures): absorption => 5800K and emittance => 300K. In other words, sunlight is primarilly absorbed at wavelengths less than 1 micron, but emitted from 5 to 35 microns.

… room-temperature emittance measurements (300 K) were made using an infrared spectrophotometer with an attached heated cavity (Holhraum), (reference 5). This infrared source was used both as a reference and for illuminating the sample. Spectral reflectance measurements were made over the wavelength region from 5 to 35 micrometers. This region contains approximately 90 percent of the energy of a 300 K blackbody radiator …

Frenchie77
Reply to  Johanus
February 28, 2020 1:31 pm

Well, do you think that the concrete is going to get as hot as the sun? Hence, why I think you are misapplying it here.

Matthew Sykes
February 28, 2020 8:26 am

125.46 C Hot
3.27 C Cold

AleaJactaEst
February 28, 2020 8:33 am

Is it anywhere near a black hole?

Steve Fitzpatrick
February 28, 2020 8:35 am

Hi Willis,
To solve this problem you need to know both the optical absorbance and the infrared emissivity.

Kurt Linton
February 28, 2020 8:36 am

You can’t all be right but you CAN all be wrong.

Moderately Cross of East Anglia
February 28, 2020 8:39 am

Is this problem something to do with why no one makes concrete frying pans? Otherwise while I am sure it is fascinating as an academic exercise I will stick to my iron, non-Teflon coated, steak pan.

Nicholas McGinley
Reply to  Moderately Cross of East Anglia
February 28, 2020 3:33 pm

In other news, the Chinese lunar explorer fried an egg on the surface of the moon.
The wanted sunny side up, but it wound up scrambled.

February 28, 2020 8:46 am

Your cube is completely possible as a model of the Eartha atmosphere without GHG, because the ‘sides’ join each other in a sphere. You don’t need unobtainium.

And in an atmosphere without GHG (or circulation) the temperature is isothermal.

So the hot and cold side must be the same temperature. So my original answer was right, but for the wrong reason.

Ulric Lyons
Reply to  ferdberple
February 28, 2020 9:15 am

He could have a front and a back fixed apart with poles at the corners and with no sides at all, that saves obtaining the unobtainium.

Bob boder
Reply to  ferdberple
February 28, 2020 9:24 am

is that true on mercury?

Bob boder
Reply to  ferdberple
February 29, 2020 4:30 am

No it’s the same as the earths crust, is the surface the same temperature as the core?

Alex
February 28, 2020 8:47 am

Simple students problem from 2nd course.
Why?

Kurt Linton
Reply to  Willis Eschenbach
February 28, 2020 9:22 am

“steady-state” is what’s not possible.

LdB
Reply to  Willis Eschenbach
February 28, 2020 11:32 am

The ISS and most satellites seem to managed it but apparently it is impossible because Kurt says so.

Greg
Reply to  Willis Eschenbach
February 28, 2020 10:33 am

Hey Willis, there is no helping Zoe. She’s off into flat-earther mentality. The rest of the world is mad and only here little clique of fans” seem to be able “understand” this amazing discovery. This thread has its own merits but don’t imagine it will help Zoe, she’s needs to be special and ain’t going to give it up.

I explained where she mistakenly assumes that the “cold end” of her bar is cooled by emission simply because it is not specified. That is her fundamental mistake. She avoids addressing that. I explained that her idea the you can have “two different fluxes” at a surface violates conservation of energy. She can’t deal with that either.

She is convinced that a flow of 2.5W/m^2 in the bar can lead to 557W/m^2 at the other end. I suggested she patent the technique. If I can have over 1kW of space heating from 5W of electrical input, I’ll pay good money for one those concrete bars.

Greg
Reply to  Willis Eschenbach
February 28, 2020 11:30 am

Willis, I also like to start from your generous view of someone. My current view comes from trying patiently to explain to Zoe where she went wrong and her stubborn refusal to engage in a logical argument to arrive at a commonly agreed position.

She seems very young and I expect it will take her about the same length of time to work this one out as it took you with the Mao thing.

It’s a fair point that this may be enlightening to others, that is also why I put some time into her blog trying to straighten it out. I final realised I was peeing into the wind.

Steen Rasmussen
Reply to  Willis Eschenbach
March 1, 2020 1:52 am

Hi Willis!
It would be nice to see a new quest:
1. You have a ball shape size like earth, with no condutivity recieving an energy radiation of 1361w/m2, albedo = 0,3 – but now rotating with a speed of 360degrees/24hours. T-hot and T-cold?

2. Same as 1. but now divided into 3 zones. Zone a= +-30degrees longtude, zone b=+-60degrees logitude – zone a and zone c = +-90 degrees logitude – zone a – zone b

kind regards
SteenR

Alex
February 28, 2020 8:59 am

Your “perfect insulator” is just a perfect mirror.

The solution is straightforward.
The transverse size of the block does not matter.
We have incident Power flux on the front surface.
There is a temperature Tf established at this surface.
The front surface radiates a power flux according to the Stefan-boltzmann.
The back surface equilibrates at temperature Tb.
So, there is power flux inside the block due to the temperature gradient.
The back surface also radiates due to the Stefan-boltzmann.
We have two temperatures to find and two equations: total power fluxes at the both surfaces must be zero.
The system is well defined and solvable.
Will not work out numbers.
Too trivial.

Nicholas McGinley
Reply to  Willis Eschenbach
February 28, 2020 3:41 pm

Willis,
I agree,.
Not actually solving the problem does not win the cookie.
Anyone can say “I know how to do it but it is too easy to bother with.”

Personally, I do not know the equations for solving this, and there is no point in me looking them up because other people here know much more about this than me.
But I am struck by how many people here claim to know exactly how to do this exercise and yet disagree with each other.
I do have a question…maybe I overlooked something.
Do you not need to know the specific heat of concrete to be able to say how hot it gets and thus how much it radiates from the sunny side of the block?
Or maybe that cancels out by the time thermal equilibrium is reached?
But what if it is rotating?

Clyde Spencer
Reply to  Nicholas McGinley
February 28, 2020 6:37 pm

Nicholas
You said, “But I am struck by how many people here claim to know exactly how to do this exercise and yet disagree with each other.” Yes, I’m also struck by the fact that there is often disagreement over what seems to be basic science or engineering, between people who claim or act as though they experts in the field. I’d be happier if there was more “consensus” on basic physics calculations.

Steve Case
Reply to  Alex
February 28, 2020 9:46 am

Alex February 28, 2020 at 8:59 am
The system is well defined and solvable.

There’s a very funny story about an Engineer, a Physicist, a Mathematician and a midnight fire in the bathroom wastebasket.

Clyde Spencer
Reply to  Alex
February 28, 2020 6:31 pm

Alex
Yes, if the Unobtainium has an absorptivity of zero, and hence an emissivity of zero, [at all wavelengths] then you have your “perfect mirror” with a reflectivity of 100%. Your can get that with a material with a complex refractive index with a very high real index of refraction (n>>10) and an imaginary (as in sq rt of -1) extinction coefficient approaching infinity. With nothing in empty space to conduct to, all the internal energy has to be confined to the channel between the front and back, and can therefore only exit by radiation.

Pablo
February 28, 2020 9:06 am

60.2ºC for both

Divide incoming by 2

rbabcock
February 28, 2020 9:09 am

I would post the correct answer and the associated math behind the numbers.. but I don’t want to appear superior, so I won’t. Good Luck!

February 28, 2020 9:12 am

1what would happen if you replaced the earths atmosphere with a perfectly transparent material 1 meter thick?

Sunlight would warm the surface of the earth, and the earth would warm the transparent material. The question now becomes, is there a difference in temperature between the inside and outside of the transparent material.

If there iis a difference how does one explain the prediction of an isothermal atmosphere. Inn the absence of GHG?

leitmotif
February 28, 2020 9:20 am

Using Stefan-Boltzmann T-hot = 398.8K

If the thermal conductivity “k” of the concrete is equal to 0.8 watts per metre per kelvin (0.8 W/m K^-1) then the T-cold side must be finally emitting 0.8×398.6 W/m2 which is 318.9W/m2.

Put that 318.9W/m2 back into the S-B equation with the same emissivity of concrete and T-cold = 277.4K

Clyde Spencer
Reply to  leitmotif
February 28, 2020 6:49 pm

leitmotif
You appear to be touching on an essential part of the problem that is not well defined. If the rate of conduction is less than the rate of arrival, then the surface facing the source of energy will continue to heat, irregardless of the emissivity of the face radiating. This strikes me as being similar to the classic differential calculus mixing problems where one needs to know most of the variables. Conductivity seems to be a key parameter here. If the thermal conductivity is too small, it becomes a ‘bottleneck’ and the front surface will continue to heat, while the back surface reaches an equilibrium determined by the conductivity and emitting flux. I’m not sure that the “Thought Problem” is adequately posed to calculate the result.

DMacKenzie
February 28, 2020 9:34 am

Assume a Tcold for the unlighted side.
Calculate how much heat is radiated to space from Tcold. 0.95*5.67e-8*Tc^4
Call this Qrcold
This amount of heat had to be conducted through the concrete from Thot
So by heat conduction from hot to cold (Thot-Tcold)*0.8 is equal to Qrcold….
So your calculated Thot is Tcold+(Qrcold/.8)
Calculate how much heat is radiated to space from Thot …..0.95*5.67e-8*Th^4
Call this Qrhot
Qtotal =Qrhot+Qrcold
Select new Tcold until Qtotal is 1360
Answer…I got Tcold=222.7 Kelvin, Thot=388.3 Kelvin
This assumes the “albedo” of the concrete is 0 in keeping with Willis’s definition of the problem. Actual albedo of concrete is about 0.5, which makes about 15 degrees difference.

MichiCanuck
February 28, 2020 9:35 am

This problem is equivalent to the infinite sheet problem, where the sheet is 1 m thick (see Carslaw and Jaeger, or Crank). Since it’s steady state, the solution for the temperature profile has a zero time derivative. That solution will have linear dependence on T with respect to depth within the sheet. So the solution will be one that satisfies the boundary conditions. On the hot side, you have a flux in (from the sun), a flux out (S-B thermal radiation) and a flux through to the cold side. These have to sum to zero (taking direction into account). The flux through will be determined by the temperature difference between hot and cold and the thermal conductivity. On the cold side, there is flux from the hot side and flux out to space (I’m neglecting leftover radiation from the big bang, but if you like, you could factor that in). The 2 fluxes on the cold have to sum to zero. I haven’t done the calculation, but it should be straightforward.

Of course, the above assumes that the thermal conductivity is temperature independent (often not true) and it assumes that the block is totally opaque (also sometimes not true at some wavelengths), so there is the possibility of reality biting one in a “fundamental” way.

MichiCanuck
Reply to  MichiCanuck
February 28, 2020 10:11 am

Plugging in the numbers and assuming constancy of heat conduction, no big bang, epsilon emit = epsilon absorb) etc., I get T(hot) = 383.3 K (110.1C) and T(cold) = 221.4 K (-51.7C).

commieBob
Reply to  Willis Eschenbach
February 29, 2020 11:15 am

That makes a bunch of us including Roy if you take Glen’s comment as being correct.

As a sanity check, I set the emission/absorption at 1. The hot side was 383.7 K and the cold side was 219.4.

commieBob
Reply to  commieBob
February 29, 2020 2:03 pm

Yes Willis. As I write this there are 453 comments on your story. The number of comments that show a more-than-rudimentary understanding of thermodynamics is truly impressive. Is there an alarmist blog that could muster such a performance?

Clyde Spencer
Reply to  MichiCanuck
February 28, 2020 6:53 pm

MichiCanuck
You assumed “constancy of heat conduction.” Is that reasonable? I think that the rate of conduction would be driven by the temperature differential between the two faces. It further assumes that the material is capable of any rate of conduction.

MichiCanuck
Reply to  Clyde Spencer
February 29, 2020 6:42 pm

It’s unlikely to be realistic, but it’s a necessary assumption given the problem as stated. For many materials, some of the heat conduction is actually via radiation across internal voids or because the material isn’t perfectly opaque at important wavelengths. I remember hearing an interesting talk a long time ago about how important blocking IR radiation was to making effective fabrics for cold weather survival suits. Once blackbody radiation enters the picture, non-linearity rears its ugly head. The problem’s still soluble, but the temperature profile across the block/sheet would no longer be linear. You’d also have to use numerical methods to obtain a solution to even the steady state equation.

One dramatic effect that always impresses me about a temperature dependent parameter is electrical conductivity in metals. If you’ve ever seen an RF coil used to heat up a metal bar, you’ll have seen that it heats very slowly at first. That’s because at low temperature, its conductivity is high and Ohmic heating isn’t very effective. However, as temperature climbs, resistance in the metal increases and heating speeds up. Eventually, it only reaches steady state when heat losses equal the absorbed RF energy. In air, conduction and convection can be important. In a vacuum, it’s good old black body radiation that has to do the cooling, which isn’t all that effective until things get quite toasty.

Adrian
February 28, 2020 9:40 am

h=temp hot surface
c=temp cold
s=stefan’s constant

(1)Incident radiation=.95×1360 (some reflected)
(2)Emitted rad hot=.95sh^4
(3)Emitted rad cold=.95sc^4
(4)Conducted heat=k(h-c)/1 {temp gradient 1m}

Balance at 2 surfaces
(1)=(2)+(4) at hot
(3)=(4) at cold

Solve numerically: 2 equations in 2 variables h and c – ignore complex and negative temps 🙂

Michael Hammer
Reply to  Adrian
February 28, 2020 1:22 pm

I was going to post the same thing but you got in first. Exactly right and easier to solve numerically than exactly. Guess C, use equn (3)=(4) to calculate H and then use H to to calculate (1)-(2)-(4). Adjust C until (1)-(2)-(4) equals zero. A couple of minutes in excell

David Dibbell
February 28, 2020 9:41 am

Nice exercise Willis. I get Thot = 383.3 K, Tcold = 221.4 K
Work in Excel shown here in a screenshot with formulas written out. As Roy Spencer notes simulaneous equations are used.
Qh and Qc are watts out for the hot face and cold face.

comment image?dl=0

Procedure: Enter trial Th, calculate Qh out via S-B, calculate initial Qc out by conservation of energy, calculate initial Tc via S-B, calculate resulting Qc out by conduction and iterate trial values of Th until the two Qc values converge.

Roy W. Spencer
February 28, 2020 9:43 am

Instead of solving it algebraically, I programmed the heat budget equations for the front and back surfaces in Excel. I whipped it up quickly, but I get 388.52K for the front surface and 222.74K for the back surface. I had to specify a heat capacity (I used granite), but the final answer does not depend on that, the heat capacity only affects the time it takes for the front and back surfaces to come into thermal equilibrium (with granite it took over 30 days to approach equilibrium).

Reply to  Willis Eschenbach
March 1, 2020 12:28 pm

Here’s an R solution if Word Press doesn’t chew up the code:

epsilon = 0.95
sigma = 5.670374419e-8
R = 1360
k = 0.8
err = function(T.hot){
T.cold.1 = (R / epsilon / sigma – T.hot ^ 4) ^ (1/4)
T.cold.2 = T.hot + (epsilon * sigma * T.hot ^ 4 – R) / k
(T.cold.1 – T.cold.2) ^ 2
}
T.hot = nlm(err, 389)$estimate
T.cold = T.hot + (epsilon * sigma * T.hot ^ 4 – R) / k

Reply to  Roy W. Spencer
February 28, 2020 11:54 am

Roy,
Do you think Boltzmann and Planck were dummies?

Why didn’t they get this solution?

LdB
Reply to  Roy W. Spencer
February 28, 2020 12:00 pm

I would accept either if you use flat sheet laminar flow and just to be clear because Zoe seems messed up on this there is a gradient thru the block.

Glen
Reply to  Roy W. Spencer
February 28, 2020 6:45 pm

Dr. Spencer,

The temperatures you posted indicate no correction of incoming heat duty from 1360 watts to 1292 watts with 0.95 emittance. If you correct to 1292 watts, you should get the 383.3 K and 221.4 K temperatures that David posted above.

Roy W. Spencer
Reply to  Glen
February 29, 2020 3:08 pm

I was going by Willis’ original post, which suggested the heated side was absorbing 1360 W/m2.

son of mulder
February 28, 2020 9:45 am

I made hot side 385.5K and cold side 222K

Anders Rasmusson
February 28, 2020 9:50 am

At the hot side, at 389 K, there is 1227 W/m2 of radiation to space and 133 W/m2 of conduction to the cold side of the cube, at 223 K, from which there is 133 W/m2 of radiation to space.

Kind Regatds

February 28, 2020 9:55 am

Assuming there is no matter past the cold end, the final temperature of both sides will be equal.

Greg
Reply to  Zoe Phin
February 28, 2020 10:36 am

Ah, the old “radiation needs to know where it will land before it leaves” hypothesis. You obviously are ready with an explanation of how radiation emitted 13 billion year ago “knew” it was going to find Hubble space telescope before it left home.

Please lets us know how that works.

Reply to  Greg
February 28, 2020 12:32 pm

‘the old “radiation needs to know where it will land before it leaves” hypothesis’

Yes, very one Boltzmann and Planck used to derive their laws.

Tell me the new junk science that can’t derive them.

http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html

Please tell me which scientists originated what you claim, and why did you believe him?

Greg
Reply to  Zoe Phin
February 28, 2020 1:43 pm

So how does light from 13bn years ago “know” it is going to “resonate” with HST before leaving home? You never answer a question. You always divert with a different question.

Reply to  Greg
February 28, 2020 3:04 pm

Greg,
The space between the EM tethered objects got stretched, if you believe in the big bang.

There’s no EM rope trying to find an end. All was tied up in the beginning, then got stretched.

Reply to  Greg
February 29, 2020 11:47 am

Zoe is right.
It is a popular misconception that light leaves A at time zero and arrives at B at times 1. Even calling c the “speed” of light is an inaccurate colloquialism. Light travels at the speed of – well – light, and thus experiences no time.

Light is outside of time. An EM photon does not experience the passing of time.

Take Betelgeuse, 700 light years away. Did light leave Betelgeuse 700 years ago? No. Light took no time at all to get from there to here. But time over at Betelgeuse is 700 years different from here.

It’s hard to get your head around, but this is correct cosmology and a full understanding of what Einstein illucidated. Time and space are one. The “light year” is actually a very good term since the distance between earth and Betelgeuse can just as well be described as 700 years as a number of meters.

Sorry but it’s just WUWT old boy science to talk about light taking time to travel a distance. Light experiences no time so it takes no time for get anywhere. As Matt O’Dowd of PBS SpaceTime YouTube channel explains very nicely, the speed of light is not even about light. It’s the speed of causality.

https://youtu.be/msVuCEs8Ydo

Bob boder
Reply to  Greg
February 29, 2020 3:35 pm

Phil Salmon

So in Willis’ example if the universe the sun and block are in is curved in such a way that light from the sun is on one side 93,000,000 miles away and on the other side 13 billion light years away in Zoe’s system when we place the block in its position when does the far side of the cube start to cool or radiate, instantaneously or after 13 Billion years?

LOL@Klimate Katastrophe Kooks
Reply to  Greg
March 1, 2020 8:06 pm

Phil Salmon wrote:
“Light experiences no time so it takes no time for get anywhere.”

That’s so wrong that “wrong” is the wrong word to describe how wrong it is. LOL

First, we’ll explore what a photon “is”.

A photon is the interaction of the electronic and magnetic fields, oscillating in quadrature, geometrically transformed into a spiral (because a sinusoid is a circular function and a circular function spread axially over space-time is a spiral). When we see electromagnetic energy as a sinusoid on our oscilloscopes, we’re literally looking at a shadow of reality, because we don’t have 3-D oscilloscopes, and the oscilloscope only looks at the electronic field, not the magnetic field.

The photon, being massless and circularly polarized when considered singularly, doesn’t carry its energy in its linear momentum, it carries it in its angular momentum.

comment image
http://staff.washington.edu/bradleyb/spiralsynth/fig3.1.gif

This is why, when traveling through transparent mediums of differing refractive indexes, the photon energy does not change, while the apparent photon speed does. If its energy were carried in its linear momentum, photon energy would necessarily change as it transited different mediums.

E^2 = p^2 c^2 + m^2 c^4
pc is the magnitude of the momentum vector. Since c is fixed in vacuum, p must change for the photon’s energy to change.

p=ħk
where k is the wave vector (where the wave number k = |k| = 2π/λ), ω = 2πν is the angular frequency, and ħ = h/2π is the reduced Planck constant.

The energy and momentum of a photon depend only on its frequency (ν) or inversely, its wavelength (λ):
E= ħν = hc/λ

This is why Planck’s constant has units of angular momentum (J-s), and the reduced Planck constant represents the quantum of angular momentum.

———-

The photon itself (if it is traveling at c, which it may not necessarily do… it is dependent upon the medium it is transiting. c is for vacuum.) _experiences_ no time, which is why it is persistent and doesn’t entropy into the background zero-point quantum vacuum field (at least, until after it impinges upon invariant-mass matter or experiences a sufficient gravitational field (because light doesn’t travel in a straight line, it follows the path of least time, and gravity is a manifestation of the warpage of space-time, which is why light ‘bends around’ large celestial objects, because invariant-mass matter expands the surrounding space-time and slows down time.).

But just because the photon _experiences_ no time doesn’t mean no time has _elapsed_.

Only for entities at c would the photon seem to not experience any time… for objects at anything less than c, that photon does indeed take time to traverse space. It’s called the theory of *Relativity* for a reason.

If “Light… takes no time for get anywhere.”, then causality is irreparably broken.

Light travels at 299792458 m/s in vacuum… it transits 299792458 meters for every second in our frame of reference.

Now we have to describe what we mean by ‘vacuum’…
hard vacuum (no invariant-mass matter, quantum vacuum zero point field exists)
perfect vacuum (no invariant-mass matter, no QVZP field)

In fact, it is the quantum vacuum zero point field which is the medium which limits the speed of light. In a hypothetical perfect vacuum (which cannot exist, because the metastability of invariant-mass matter is dependent upon the existence of the QVZP field, thus we could never construct a machine capable of a perfect vacuum without the atoms / molecules of that machine’s vacuum-facing components undergoing beta capture and thus transmutation [1][2][3][4]), there would be no speed limit except for the energy field the photon itself lent to the perfect vacuum in transiting that vacuum.

[1] https://journals.aps.org/prd/abstract/10.1103/PhysRevD.11.790

[2] https://web.archive.org/web/20190713220130/https://arxiv.org/ftp/quant-ph/papers/0106/0106097.pdf

[3] https://web.archive.org/web/20190713225420/https://www.researchgate.net/publication/13330878_Ground_state_of_hydrogen_as_a_zero-point-fluctuation-determined_state
“We show here that, within the stochastic electrodynamic formulation and at the level of Bohr theory, the ground state of the hydrogen atom can be precisely defined as resulting from a dynamic equilibrium between radiation emitted due to acceleration of the electron in its ground-state orbit and radiation absorbed from zero-point fluctuations of the background vacuum electromagnetic field, thereby resolving the issue of radiative collapse of the Bohr atom.”

[4] https://web.archive.org/web/20180719194558/https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20150006842.pdf
“The energy level of the electron is a function of its potential energy and kinetic energy. Does this mean that the energy of the quantum vacuum integral needs to be added to the treatment of the captured electron as another potential function, or is the energy of the quantum vacuum somehow responsible for establishing the energy level of the ‘orbiting’ electron? The only view to take that adheres to the observations would be the latter perspective, as the former perspective would make predictions that do not agree with observation.”

LdB
Reply to  Zoe Phin
February 28, 2020 11:23 am

LOL this I have to hear.

So how does the emission leaving the cold end know ahead of time there is matter out there somewhere?
Does it send out some special faster than light communication asking for all matter to respond?
Or perhaps it just omipotent know?

The ISS is up there right now and I am seriously wondering why it isn’t all the same temperature given your statement.

Rick C PE
Reply to  Zoe Phin
February 28, 2020 2:05 pm

Zoe: It might help to use the form of S-B:
Qrad = e A sigma (T1^4-T2^4)

(In Willis’ problem A =1 , e = 0.95 , Sigma = 5.67 E-8)

Where T1 is the warmer T2 is the colder surface temperature.
You will find that T1= 398.59 K and T2 = 0, you get 1360 watts/m2
With T1 = 398.59 and T2 = 388.5 you get 132.5 watts/m^2, the heat Flux to the hot side of the cube.
With T1 = 222.7 and T2 = 4 you get 132.5 watts/m^2, the heat Flux from the cold side of the cube to space.
And 0.8 * (388.5 – 222.7) = 132.5 watts/m^2, the heat Flux from the warm side to the cold side.

So the temperatures are all reasonable and the 3 heat flows are equal as required for a system in thermal equilibrium. I think this exercise has been a good demonstration that being an excellent computer programmer is not enough to get correct answers. You must also understand the problem and apply the correct solution method.

Reply to  Rick C PE
February 28, 2020 4:28 pm

Rick,
You missed the part where 90% of energy from the source was rejected, while pretending absorptivity is 0.95.

You can’t solve HSR = CHF = CSR

So you solve the cold end, CSR = CHF, and ignore HSR = CHF or you reject 90% of incoming HSR to get your equality, but then your Thot is fake.

Rick C PE
Reply to  Zoe Phin
February 28, 2020 6:04 pm

Zoe: No, you missed the basics of thermodynamics. Everything with a temperature above absolute zero (which is literally ‘everything’) will radiate energy. If you have two objects at different temperatures, there will be a NET heat flux from the warmer to the colder. The 90% of the heat you’re talking about is not rejected, it is absorbed raising the surface temperature of the block from 0 to 388 K which results in 90% 0f the heat absorbed being radiated back out to space. The net rate of heat flow will be proportional to the difference in the 4th powers of each object’s absolute temperature. Thus, a 1000 C radiant source surface can heat a 975 C surface, but not very much. But both surfaces would radiate a lot of heat to a cold object. They don’t stop radiating energy when you position them to face each other. However, the same 1000 C source will heat a 20 C surface by 100’s of degrees. This is really basic thermodynamics stuff. Take a course – there are lots of them.

DMacKenzie
Reply to  Rick C PE
February 28, 2020 10:48 pm

Rick, you should just ignore Zoe. She/he is just prankster blogger.

LdB
Reply to  Rick C PE
February 28, 2020 11:37 pm

Correct DMac, I thought she was serious and just being dense but yep I worked that out as well.

Reply to  Willis Eschenbach
February 28, 2020 10:57 am

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.

LdB
Reply to  Zoe Phin
February 28, 2020 11:44 am

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?

Reply to  LdB
February 28, 2020 1:05 pm

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.

Greg
Reply to  Zoe Phin
February 28, 2020 1:54 pm

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.

This is not “back radiation”, it’s radiation. It happens from all bodies in all directions, no matter who is looking.

Bob boder
Reply to  Zoe Phin
February 28, 2020 2:25 pm

Because each step is radiating in every direction.

leitmotif
Reply to  Zoe Phin
February 28, 2020 3:37 pm

“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
Reply to  Zoe Phin
February 28, 2020 7:46 pm

“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. 🙂

Smart Rock
Reply to  Zoe Phin
February 28, 2020 11:15 pm

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”.

Reply to  Zoe Phin
February 29, 2020 11:31 am

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.

leitmotif
Reply to  Zoe Phin
February 29, 2020 3:34 pm

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.

Nicholas McGinley
Reply to  Zoe Phin
February 29, 2020 5:27 pm

“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?

LOL@Klimate Katastrophe Kooks
Reply to  Zoe Phin
March 1, 2020 8:31 pm

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

Reply to  Zoe Phin
March 4, 2020 8:14 pm

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.

Reply to  Zoe Phin
March 5, 2020 1:49 pm

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?

LOL@Klimate Katastrophe Kooks
Reply to  Zoe Phin
March 5, 2020 5:53 pm

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

LOL@Klimate Katastrophe Kooks
Reply to  Zoe Phin
March 5, 2020 7:48 pm

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
Reply to  Zoe Phin
March 5, 2020 9:01 pm

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

Reply to  Zoe Phin
March 5, 2020 9:22 pm

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:
comment image

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
Reply to  Zoe Phin
March 5, 2020 10:04 pm

To Willis Eschenbach:
You left out a bit on your image… I’ve helpfully included the requisite information. You’re welcome. LOL

comment image

LOL@Klimate Katastrophe Kooks
Reply to  Zoe Phin
March 5, 2020 10:11 pm

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

comment image

LOL@Klimate Katastrophe Kooks
Reply to  Zoe Phin
March 5, 2020 10:29 pm

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.

LdB
Reply to  LdB
February 28, 2020 5:06 pm

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.

Reply to  LdB
February 28, 2020 6:26 pm

“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.

Reply to  LdB
February 28, 2020 5:30 pm

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.

WXcycles
Reply to  Zoe Phin
February 29, 2020 12:54 am

” … 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.

Bob boder
Reply to  Zoe Phin
February 29, 2020 4:25 am

Zhoe

If what you say is true the earths crust would be the same temperature as the core.

Reply to  Zoe Phin
February 29, 2020 9:29 am

“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.

leitmotif
Reply to  Zoe Phin
February 29, 2020 2:52 pm

Zoe Zhoe Zoey

They can’t even get your name right and it’s only 3 letters! :-DDD

WXcycles
Reply to  Zoe Phin
February 29, 2020 5:36 pm

“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.

leitmotif
Reply to  Zoe Phin
February 29, 2020 4:04 pm

“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.

Rick C PE
Reply to  Willis Eschenbach
February 28, 2020 11:04 am

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.

Greg
Reply to  Rick C PE
February 28, 2020 1:55 pm

Part of the challenge was to explain how you got the answer.

DMacKenzie
Reply to  Rick C PE
February 28, 2020 7:19 pm

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.

Alex
Reply to  Willis Eschenbach
February 28, 2020 11:49 am

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.

son of mulder
Reply to  Willis Eschenbach
February 28, 2020 2:12 pm

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.

Another Joe
Reply to  Willis Eschenbach
March 6, 2020 4:33 am

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.

February 28, 2020 10:01 am

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.

ghl
Reply to  Willis Eschenbach
February 28, 2020 9:06 pm

Willis
I wonder, based on style and muddy thinking, wether Zoe is actually Sou of Hotwhopper losing traffic.

MichiCanuck
Reply to  Zoe Phin
February 28, 2020 10:37 am

You seem to have ignored the input radiation on the hot side.

Greg
Reply to  Zoe Phin
February 28, 2020 10:45 am

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.

Reply to  Greg
February 28, 2020 12:00 pm

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.

Writing Observer
Reply to  Zoe Phin
February 28, 2020 1:53 pm

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.

4 Eyes
Reply to  Zoe Phin
February 28, 2020 2:45 pm

Zoe, a potential is not a flux. The object radiating does not care where the energy is going.

Reply to  4 Eyes
February 28, 2020 3:26 pm

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

Greg
Reply to  4 Eyes
February 28, 2020 3:52 pm

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.

Reply to  4 Eyes
February 28, 2020 5:17 pm

“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!

Editor
Reply to  Zoe Phin
February 28, 2020 6:03 pm

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.

Rainer Bensch
Reply to  Mike Jonas
February 29, 2020 7:02 am

Good one.

leitmotif
Reply to  Zoe Phin
February 29, 2020 4:19 pm

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.

Hugs
Reply to  Greg
February 28, 2020 1:40 pm

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.

Reply to  Hugs
February 28, 2020 5:24 pm

And where does CHF fit in?

Gordon Dressler
Reply to  Zoe Phin
February 28, 2020 11:14 am

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.

Reply to  Gordon Dressler
February 28, 2020 11:52 am

Uhuh,
So only CSR = CHF, HSR can remain completely disconnected?

Reply to  Gordon Dressler
February 28, 2020 12:20 pm

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!

Gordon Dressler
Reply to  Zoe Phin
February 28, 2020 3:08 pm

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.

Bob boder
Reply to  Zoe Phin
February 28, 2020 11:39 am

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.

Matthew Schilling
February 28, 2020 10:03 am

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.

Greg
Reply to  Willis Eschenbach
February 28, 2020 10:54 am

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.

Marcus Allen
Reply to  Willis Eschenbach
February 28, 2020 11:19 am

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.

3x2
Reply to  Marcus Allen
February 28, 2020 2:53 pm

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.

Greg
Reply to  3x2
February 28, 2020 3:58 pm

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.

3x2
Reply to  Willis Eschenbach
February 28, 2020 12:56 pm

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.

3x2
Reply to  3x2
February 28, 2020 3:07 pm

(many words sound the same … to, too, two but have very different meanings and only context separates them in speech)

Clyde Spencer
Reply to  3x2
February 28, 2020 7:13 pm

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.

3x2
Reply to  3x2
February 28, 2020 9:42 pm

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.

It doesn't add up...
Reply to  3x2
February 28, 2020 4:54 pm

But compare “nanometre” and “manometer”.

Gordon Dressler
February 28, 2020 10:20 am

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.

Steve Fitzpatrick
Reply to  Gordon Dressler
February 28, 2020 10:32 am

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.

Gordon Dressler
Reply to  Steve Fitzpatrick
February 28, 2020 4:00 pm

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.

Greg