# Geothermal Denial

Climate “scientists” look at Earth’s geothermal heat flux, see that it’s small (~0.1 W/m²), and then conclude geothermal can’t possibly predominantly explain Earth’s surface temperature. This is plain wrong. I came up with an illustration to demonstrate my point a while ago:

This is a fictional planet. I did this on purpose to accentuate my point. The question here is: Why is the surface 1000°C?

Mainstream climate “scientists” would see the small geothermal heat flux (0.1 W/m²), ignore it, and conclude it must be because Solar + Atmosphere delivers 148,971 W/m² to the surface. It couldn’t have anything to do with the 1010°C a hundred meters below the surface. Oh no, it can’t be that!

But did I say this fantasy planet even has an atmosphere? What if not? The sun only delivers 165 W/m² … Where’s the rest of the energy coming from to make the surface 1000°C?

And if there is an atmosphere … where did the atmosphere get 148,806 W/m² to give to the surface?

Climate cranks come to the rescue and claim that due to infrared absorbing gases in the atmosphere, the Sun’s 165 W/m² gets auto-magically boosted to 148,971 W/m², because that 165 W/m² can’t escape to space. You know what I have to say to that?

Some people have serious problems accepting the truth: The surface here is 1000°C because it’s 1010°C a hundred meters below the surface. Simple.

Now why is that so hard to accept? Ideological presupposition. That’s why!

Let’s take a look at a recent discussion here:

Geothermal heat is about 0.1 W/m². Solar absorption is around 161 W/m². All solar is lost on a regular basis and heat loss by the surface is very (!) dynamic. Which means that a little bit more energy from below (for example 0.1 W/m2) is easily lost, together with the dynamic 1610 times higher ‘standard heat loss’.

Wim Röst

I said that her claim, that the heat flux leaving the ground was thousands of times larger than the heat flux passing through the ground, was physically impossible.

Willis Eschenbach

So what they are both saying in our context is that a 0.1 W/m² geothermal heat flux can’t “support” a 148,971 W/m² emission from the surface, so I must be wrong!

In Willis’ view, there must be equality between geothermal heat flux and surface emission.

He believes this is needed to preserve conservation of energy. But what is he really doing?

He’s equating a heat flux between two locations to an absolute energy flux equivalent at one location. Is that conservation of energy? No!

Energy is energy, and heat flux is the energy transfer from hot to cold, i.e. a DIFFERENCE of energies at two locations. How can you compare a differential to an absolute? It makes no sense. But don’t believe me, check your own eyes:

The top of the water represents the planet’s surface.

If what Willis et al were saying was true, we should expect a steep thermal gradient from the bottom to the top of the water column – so that conductive heat flux equals emergent radiation. But what actually happens?

As you can see, the top and bottom of the water column becomes the same temperature. In other words, the conductive (“geothermal”) heat flux becomes 0 W/m².

The top of this water column is capable of emitting εσT⁴, or (5.67e-8)*(273.15+83)^4 =~ 912 W/m²

In this case, 0 W/m² has no problem “supporting” 912 W/m² ! That’s infinite times!

So why can’t a 0.1 W/m² geothermal heat flux “support” a 148,971 W/m² emergent flux?

Of course it can. It’s not a problem at all. These people are simply confused on the physics.

Now imagine they had to explain this infrared electric kettle hot water video without mentioning the heat source below. How would they do that? Well…

The top of the kettle (“sun”) emits 20°C worth of radiation to the top of the water (419 W/m²) . The water vapor and carbon dioxide in the air above the water prevents radiation from leaving to colder space, and so auto-magically the top of the water becomes 912 W/m², or 83°C. Simple!

Now do you see it?

Greenhouse Effect == Geothermal Denial.

Note: Real planetary subsurface has a density gradient, and so you will see a small geothermal heat flux. There is no density gradient in this water example, so the conductive heat flux goes to zero.

Now we take my argument down to Earth, literally. Why is Earth’s surface ~15°C?

Because geothermal “delivers” 0°C to the surface. Add insolation and subtract latent and sensible heat, and you get your 15°C. Simple. No Greenhouse scam necessary.

More details in other articles, such as here, here, and here.

Enjoy 🙂 -Zoe

Happy New Year, Everybody!

https://phzoe.com

## 123 thoughts on “Geothermal Denial”

1. Max Polo says:

Great explanation. It seems so simple and obvious, once you get it 🙂

Liked by 1 person

2. gbaikie says:

“Now we take my argument down to Earth, literally. Why is Earth’s surface ~15°C?”
It’s not.
Earth’s average land surface air temperature is about 10 C.
Earth average ocean surface air temperature is about 17 C.
Average tropical ocean surface air temperature is about 26 C
The rest of ocean surface air temperature is about 11 C
Surface air temperature can warm to 50 C and can cool below -50 C

I would Earth surface temperature is average temperature of the entire Ocean which is about 3.5 C.
Why does surface air temperature become warmer than about 3.5 C?
It seems it’s due to sunlight, sunlight can make surface air be around 50 C, the lack of sunlight can make surface air
become colder than -50 C.

But I say the surface temperature is 3.5 C, because amount heat absorbed ocean, determines Earth’s global climate average air temperature, which is said to be about 15 C.’
Or Earth’s global climate is called an Icehouse climate. We are in an Ice Age. Icehouse climate is another word for Ice Age. I like the term, icebox climate, as refrigerator is another word for icebox.
Another global climate is called a hothouse climate or another word for hothouse is greenhouse.
A greenhouse climate as a warm ocean and average ocean temperature, instead of 3.5 C could be warmer than 15 C.
Most of Earth’s history is not being in an Icehouse or Hothouse climate- these are extremes of Earth’s climates.
Though some believe Earth has had snowball climate.
But I don’t, and you don’t.
And I believe “Greenhouse Effect theory” is not a theory, and it’s best described as a stupid cargo cult. Or if being polite, it’s pseudo science.

In our ice age, our current state, is being in interglacial period {a warmer period between glaciation periods- in which has most of duration spent in Our Ice Age.
The CO2 thing, was to “explain” why we have glacial and interglacial periods- and it failed to explain it.
End of story.

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3. Rob says:

Zoe,

1) You ask why the surface of your “fictional planet” is 1000 C. The answer is that you arbitrarily chose the number. You could have just as easily chosen the surface to be 500 C, or -10 C. Similarly, you could have chosen the geothermal heat flux to be any number you want. That is to say, your example doesn’t demonstrate anything about physics, because all of the numbers are chosen arbitrarily rather than relating to each other through some physical laws.

2) You claim that in a boiling kettle, conductive heat flux through the water column is much less than emergent radiation from the column. But this is not a problem, because convective heat flux is far greater than conductive heat flux in this situation. I also don’t think you can say the temperature is constant throughout the column from this picture — maybe you can say it differs by less than 5 C by eye.

3) You are overcomplicating conservation of energy. The distinction you try to draw between “a heat flux between two locations” and “an absolute energy flux equivalent at one location” is meaningless. Heat is a form of energy, and it is perfectly valid to look at how much heat flux is crossing into some volume. Energy conservation can be written as a continuity equation ( https://en.wikipedia.org/wiki/Continuity_equation#Energy_and_heat ). Integrating over a volume, the rate of change of energy within the volume equals the surface integral of energy fluxes, including heat flux. If 390 W/m^2 is leaving the earth’s surface, and only 160 + 0.1 W/m^2 is entering, then it would be cooling rapidly. But this is not the case.

– Rob

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1. 1) I chose 1010C and 100mW/m^2. The physical principle here is conduction.

2) The end state is ZERO heat flux through the water.

3) Heat only exists because of the hotter side. Heat is not the “form of energy” that needs to conserved. Your link shows heat goes to zero to achieve equiibrium. 0.1 W/m^2 is not “energy in”, energy in is the temperature, so to speak. 390 W/m^2 is easy to achieve because as you go deeper you get an emissive potential of 391,392,393,… W/m^2.

If what you say was true, then SB and Planck law would be stated as “emission from surface depends on material’s internal heat flux”. Why do you think they are never stated this way?

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1. Rob says:

2) I don’t see how you can say there is zero heat flux. There are clear variations in temperature throughout the kettle. Convection can generate large heat transport with minor temperature gradients. What’s more, your video specifically has a warning that says “Temperature readings are approximate”. There is no way to confidently say the temperature gradient is exactly zero everywhere.

3) Heat is a flux of energy, temperature is not. The continuity equation for energy is clear: d/dt(U) = -integral q . dS, or the rate of change of energy in a volume is minus the surface integral of energy flux. In steady state, this means the surface integral of energy flux is zero. Energy flux has units of W/m^2, for example incoming solar radiation is 160 W/m^2, geothermal is ~0.1 W/m^2, outgoing long wave is 390 W/m^2.

The fluxes must sum to zero to satisfy the continuity equation, and the only way for that to occur is backradiation. It’s as simple as that. The internal temperature of the earth only matters in that it produces a (small) heat flux. It is not accurate to say that “energy in is the temperature”, because temperature is not an energy flux. It doesn’t even have the right units, so it can’t fit into this equation!

Granted, one source of energy flux (thermal radiation) can be determined from the temperature by the Stefan-Boltzmann law: j ~ T^4 . But this is not in conflict with the continuity equation for energy. It just one source of flux, which when added to all other energy fluxes must conserve energy.

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1. 2) We’re only interested in the water. Different material has different density and produces a gradient.

3) There’s no integral in the continuity equation. Geothermal heat flux is not in the same plane as surface emission. See:
https://phzoe.com/2020/05/22/equating-perpendicular-planes-is-plain-nonsense/

The continuity equation literally says energy flux goes to zero. Geothermal surface emission – “backradiation” = 0. “backradiation” needs FRONTradiation, and its source is geothermal.

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2. Rob says:

I should have specified this, but I converted the continuity equation to integral form by integrating both sides and applying the divergence theorem. See for example earlier in the article from before https://en.wikipedia.org/wiki/Continuity_equation#Integral_form .

If you’re talking about the direction of heat flux, what matters for the surface integral is just how much heat flux crosses the boundary of some volume of interest. If we’re looking at the earth’s surface, the geothermal heat flux of 0.1 W/m^2 enters directly, and the Stefan-Boltzmann law already accounts for the direction of outgoing radiation — the total outgoing flux, integrated over direction, is 390 W/m^2.

I’m not sure what you mean by front radiation, but I agree that net energy flux must go to 0. Given outgoing 390 W/m^2 longwave radiation and incoming 160 W/m^2 solar, 0.1 W/m^2 geothermal, net zero is only possible with another source of incoming radiation. This is backradiation from the atmosphere.

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So you believe that the atmosphere will emit toward the surface in line with its temperature (not its heat flux), but you don’t believe the surface will emit according to its temperature? Interesting bit of hypocrisy.
What is the heat flux through the atmosphere? Surely there must be one …

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4. Rob says:

Zoe,

In your last link, you say in reference to this equation ( https://phzoe.files.wordpress.com/2020/05/radeqcond-1.png ): “What they do is equate the radiation emerging out of a plane with the internal conductive heat flux.” I think it’s clearer to see that the equation arises from the following logic:

1) In a steady state, conservation of energy requires energy flux in = energy flux out. As I showed in the above posts.
2) energy flux in is the geothermal flux (in the absence of sun and atmosphere)
3) energy flux out is thermal radiation given by the SB law
4) Thus, epsilon sigma T_surface^4 = k (T_surface-T_core)/L

This seems to derive the equation rigorously. Let me know if you see an issue with any of these steps.

The link also argues that conduction does not scale with the area perpendicular to the heat flux. But your formulas for conductive power ( https://phzoe.files.wordpress.com/2020/05/condflux.png ) and radiative power ( https://phzoe.files.wordpress.com/2020/05/sblaw-1.png ) have the same scaling with area.

“What is the heat flux through the atmosphere? Surely there must be one …”
If the atmosphere is in a steady state, net heat flux through any plane is zero. To be in a steady state the downward fluxes (insolation and backradiation) must be precisely canceled by the upward fluxes (outgoing radiation and convection). Of course the atmosphere is not in a perfect steady state. In the morning there will generally be a net inward heat flux as the surface and atmosphere warm, and opposite in the evening.

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5. 1) No, conservation of energy is not the same as steady state. Conservation of energy is Eout=Ein. There are requirements for steady state conditions … such as a density gradient or a surface area difference, etc.

4) Not in the same plane. k and L are ORTHOGONAL to A.

“have the same scaling with area.”
No! The interface area is the same, and so cancel out. You’re left with #4, which I just explained.

“To be in a steady state the downward fluxes (insolation and backradiation) must be precisely canceled by the upward fluxes (outgoing radiation and convection).”

But you forgot what gave the atmosphere this “backradiation” energy in the first place! It was GEOTHERMAL! (and solar).

Geothermal delivers 0C (or 4-5C) to the surface. This is what geophysicists’ diagrams, and other data shows.

Now imagine a cubic meter of dirt just below the surface, with one square meter facing the surface.

The geothermal heat flux is the flow of energy from the cubic meter below it, to it. But it is not the available energy in that m^3. The available energy is just a little smaller than the dirt below it. No big deal.

Our m^3 of dirt is at 0C from geothermal.

Q = m*Cp*T

Using typical values,

Q = 1500*1.5*273.15 = 614,587.5 Joules.

THAT is how much energy you have. That is your maximum ENERGY in. It doesn’t matter that the dirt below has 614.587.6 Joules, thus creating a 0.1 J geothermal heat flow every second.

Being that surface is 0C in our example, every second this m^3 of dirt will transfer 315 Joules to the cubic meter of atmosphere above. And THAT is where the atmosphere gets its energy. Once the atmo has it, it can send 314.9 J back to the dirt. Thus preserving the 0.1J heat flux, every second.

We can extend this further …

The total energy of a 20,000 meter tall column of air at an average temperature of 0C (-18C in reality) above our m^3 of dirt is:

Q = 1.25 * 0.025 * 273 * 20000 = 176 KJ

No problem for geothermal’s 615KJ !

You don’t need to create a magical green sky dragon to blow hot air onto the surface.

The geothermal hot plate plus solar perfectly explain atmo temperatures.

Sheesh! Why don’t people get it?

Conservation of energy is not conservation of heat flow. Heat flows can’t explain temperatures. Temperatures are based on energy, not heat. Obviously!

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6. Rob says:

“Conservation of energy is not conservation of heat flow. Heat flows can’t explain temperatures. Temperatures are based on energy, not heat. Obviously!”

I can only point out that the previous equations provide a unique solution for temperature. Heat flows alone do not determine temperature. However, if we know the system is approximately in a steady state, then we have an equation (energy continuity) stating that the sum of heat fluxes into a given volume must equal zero. If each flux can be expressed as a function of temperature, which is true for outgoing thermal radiation and geothermal conduction (and insolation is just a constant), then this equation uniquely determines temperature.

Regarding the direction of geothermal heat conduction: the most direct way to see this is to look at the law of conduction ( https://en.wikipedia.org/wiki/Thermal_conduction#Fourier's_law ). It states heat flow q = -k grad(T) . The negative gradient points in the direction of fastest decrease ( https://en.wikipedia.org/wiki/Gradient ). For geothermal heat, this direction is outward from core to surface. Your argument assigning a direction using units is interesting, but since it conflicts with the fundamental law of heat conduction, it cannot be correct here.

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7. Conduction: q = kA/L*(Thot-Tcold)

Equating the two cancels out A. Both are now fluxes in units W/m^2. But the conductive /m^2 comes from k/L, where L is at a 90 degree angle to the surface area of emission. Do you not understand that?

Do you still think backradiation explains why Earth’s surface is ~15C or do you now understand the real power of geothermal?

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8. Rob says:

The law of heat conduction provides a vector expression for the heat flux. This vector points from hot to cold, or outward from the core to the surface. This is very established theory, which your analysis disagrees with.

Just to be clear, it seems to me you are saying that heat does not flow from hot to cold (outward from the core of the earth to the surface), but rather sideways. Is this correct?

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9. No, Rob, you are suggesting comparing things at right angles to each other.

I have looked through textbook pdfs and google and can’t find a single example of conduction = radiation. I have seen many people interpret the 1st LoT and “steady-state” to come up with that idea, and yet not a single actual example. You didn’t answer my question, why isn’t blackbody radiation stated as conduction=>emission. Why is it only T that matters?

Why is it that I can explain what heated the atmosphere, but you need a sky dragon?

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10. Rob says:

Zoe,

I am not comparing things arbitrarily. Applying energy continuity shows that the net flux into a region is zero in the steady state. So flux in equals flux out. I’m still not sure what your objection to this is.

You say thermal radiation from the surface is perpendicular to geothermal heat flux. But again, heat conduction is from hot to cold. So surely you must agree that geothermal heat flux is straight outward from the hot earth interior towards the surface? Unless you don’t think heat flows from hot to cold.

“You didn’t answer my question, why isn’t blackbody radiation stated as conduction=>emission. Why is it only T that matters?”

This is a result of Planck’s law, which is derived rigorously. The thermal radiation from a body only depends on its thermodynamic state (in particular, temperature), not how it got there.

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11. Rob says:

Zoe, the sun’s photosphere is approximately in a steady state. The article you link identifies its own problem: “My critics will be quick to point out that I can’t use the conduction formula because the sun only has radiative transfers in the photosphere.” Conduction is a negligible component of heat transfer in the sun, so you must include radiation and convection for an accurate calculation.

Earlier I showed a 4-step derivation of an equation which seems to disprove your geothermal hypothesis ( epsilon sigma T_surface^4 = k (T_surface-T_core)/L ). You responded that you believe the direction of geothermal heat flux is not outward along the earth’s radial direction, but perpendicular to this. I want to ask for clarification one more time, because this seems to be your main objection to the derivation. Are you contending that geothermal heat flux is not along the direction of -grad(T), i.e. not from hot to cold?

Lastly, a mistake in one of my earlier posts: it should say that in steady state the net heat flux into any volume is zero, not heat flux through any plane.

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12. Rob,
I perform dimensional analysis in my article. What do you still not understand?
Geothermal is in the Z direction, and radiation is from an X,Y plane. The /m^2 are NOT the same thing.
Do you also believe the amount of rain falling on a roof depends on the height of the building?

“Conduction is a negligible component of heat transfer in the sun, so you must include radiation and convection for an accurate calculation.”

There is NO formula for heat transfer through a line or radius for either convection or radiation. There is NO method other than conduction for finding heat flux through the sun.

“the sun’s photosphere is approximately in a steady state.”

Rob, you’re making things up in a self-serving matter without any evidence.

So you claim a 63 MW/m^2 heat flux through the sun?

That would mean the core emits 126 MW/m^2. This equals 6007 degrees Celcius. Not the 15+ million degrees that is predicted.

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13. Rob says:

Zoe,

Units do not have a direction. Vectors have a direction, that is the whole point of vectors. Conductive heat flux is a vector. It is determined by Fourier’s law, q=-k grad(T), which points outward in the earth. That is the direction of the heat flux vector, full stop.

To calculate how much rain is incident on a roof, you would take the dot product of the rain vector with the surface normal and integrate over area. The exact same is true for geothermal heat flux incident on the earth’s surface.

The bottom line is that if more energy is leaving a region than is entering, that region must be cooling. I showed this rigorously with energy continuity but it is intuitively clear. Looking at a thin element of the earth’s surface, more power leaves as outgoing IR than enters from solar and conductive geothermal flux. Since it is not cooling, there must be another input.

If geothermal heat flux pointed any direction besides outward, even less would be incident on the surface. So this would only make the energy imbalance worse.

Until this original issue is resolved it doesn’t make much sense to move on to new questions about heat transport in stellar atmospheres.

– Rob

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14. “rain vector with the surface normal”
The height is not NORMAL to the rain. Pay attention.

“Since it is not cooling, there must be another input.”
And what is the input to that input?
Where did the atmosphere get the energy to give the surface?

Rob, you look so silly.
The top m^3 of dirt has 614,587.5 Joules of energy.
The m^3 below it has 615,587.6 Joules. And you’re arguing that only 0.1 J/s can be passed to the m^3 of atmo just above it.

LMAO

How’d that top m^3 get to 614,587.5 Joules in the first place, if according to you, only 0.1 J can be sent up?

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2. Rob says:

“The top m^3 of dirt has 614,587.5 Joules of energy.
The m^3 below it has 615,587.6 Joules. And you’re arguing that only 0.1 J/s can be passed to the m^3 of atmo just above it.”

There’s not much room for argument here, just plug numbers into the law of conduction, q=-k grad T , and you get on the order of 0.1 W/m^2. grad T is ~ 0.025 C / m ( https://upload.wikimedia.org/wikipedia/commons/9/9e/Upper_mantle_temperature_profile.png )

“How’d that top m^3 get to 614,587.5 Joules in the first place, if according to you, only 0.1 J can be sent up?”

Not 0.1 J, but 0.1 J/s. The earth has had a lot more than 6,145,875 seconds (~70 days) to establish a thermal gradient between the surface and the core.

“The height is not NORMAL to the rain. Pay attention.”

As I said, flux of rainfall is integral r . n dA, where r is the rain vector and n is the surface normal. This doesn’t depend on the building’s height.

————

Perhaps an analogy would be helpful. Feel the outside of an ice chest. Regardless of whether the inside is freezing cold or very hot, the outside will be the same temperature as the room you are in. Why? Because the walls are thick pieces of styrofoam, a very good thermal insulator. So the temperature of the outer wall is determined almost entirely by external conditions.

Well, 200 km of solid rock in the lithosphere is a _much_ better insulation than an inch of styrofoam. About 100,000 times better. So the temperature of earth’s surface is determined almost entirely by external factors, i.e. solar input and greenhouse gases. Not the conditions below the insulating layer of lithosphere.

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1. Rob,
70 days is irrelevant. A gradient was formed at the time of creation, including through the top gas layer (atmosphere). It’s not as if the core was the only thing on.

We don’t know at what depth 0.1 W/m^2 refers to.
There is no need for only 0.1 W/m^2 to be transfered to the atmosphere. There can be ~315 W/m^2 transfered no problem. This will cool the top fast. Then each subsequent layer beneath loses some Joules to make up for it. This already happened.

Objects at 273K do not emit 0.1 W/m^2 just because there is a slightly warmer layer beneath.

“This doesn’t depend on the building’s height.”

THAT’S EXACTLY MY POINT!!!

The emission surface is like the area of the roof. Conduction has a 1/L term, which is like the HEIGHT.

OMG, How can you be so thickheaded?

“greenhouse gases”

Ha ha ha what a troll. Geothermal emits exactly what is needed for GHG backradiation. The insulation prevented it from being even hotter, but you still get what you get.

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2. Rob says:

Zoe,

Ultimately it has been shown from Fourier’s law and energy continuity that your geothermal hypothesis cannot be correct. I provided a quantitative proof using fundamental and well known physical equations, so there’s nothing left to argue.

You’ve made a lot of unsupported statements and now some personal attacks, but the proof stands. You really should look at it more closely so you don’t spend any more time on an incorrect hypothesis.

Good luck,
Rob

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3. Rob,
You’re a troll. Your so-called “proof” is the the very thing I debunked. All you did was repeat the same thing over and over again, and then declared yourself the winner. That’s what losers do.

Now here’s your last chance to redeem yourself.
If you believe radiation should equal conduction, why doesn’t the Sun create a conductive flux of 340 W/m^2 in the atmosphere or -165 W/m^2 in the ground?

How come you never answered Max’s question? IIRC, he asked what happens when you bring a cold object to an object in thermal equilibrium? According to you, the cold object can’t be heated at all because the conductive flux is 0.

Also, please tell me what motivates you to be a geothermal denier.

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4. Rob says:

“Your so-called “proof” is the the very thing I debunked.”

I can’t see where you debunked it. Mostly you asked questions about subjects that were related, but had no bearing on the accuracy of the proof. The one directly relevant question you had, about the direction of thermal heat conduction, was already answered.

“If you believe radiation should equal conduction”

I don’t believe this. I believe (and proved) that incoming heat flux = outgoing heat flux for a volume in steady state. Under your hypothesis that the only incoming heat fluxes are incoming solar + geothermal, this implies that solar + geothermal = outgoing heat flux. But this isn’t correct, so your hypothesis is false.

“How come you never answered Max’s question?”

I did!

“Also, please tell me what motivates you to be a geothermal denier.”

I’m a physicist, my motivation is to find out an accurate description of the situation. It’s important to be able to criticize your own theories and recognize when they are proven incorrect.

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5. Rob, who are you kidding? You offered no proof, only assertions. Assertions are not proof.

Eh whatever, I’ll let the readers realize how insincere you are.

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6. Rob says:

Zoe, fair enough, I’ve about said my piece anyway, because I believe the proof I provided is sufficient. I do appreciate the open discussion here.

– Rob

Liked by 1 person

7. Rob, if I thought your argument was any good I would have not written any articles about this topic. It’s because you and the rest offer poor reasoning that I wrote them!

Look at the heat flux in the sun. If you don’t like the conductive method, you can use the radiative method. Just convert tempetatures to emission between two successive meters, and note the difference. It will be tiny! The heat flux is tiny in the sun. There is no steady-state in the sun after 4.5 billion years. All you’re doing is equating two things that are in the same units: W/m^2. But the /m^2 in conduction doesn’t come from the surface area, but k and L, which are at a 90 degree angle to each other.

You are disregarding physics for some strange mathematical ideal.

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4. Max says:

Rob, do you imply that it’s impossible to have a body in thermal equilibrium (zero thermal gradient = zero conductive heat flux) because otherwise it would be impossible to satisfy the continuity equation ? (emergent radiation would always be >> 0). This seems a bold statement.

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1. Rob says:

Max,

Good question. If you have a body in thermal contact with an environment at a different temperature, then yes, it is impossible to have thermal equilibrium. For example, the hot tea kettle from the article is leaking heat to the room temperature atmosphere, which implies there must be heat flux and thermal gradients.

However, rough thermal equilibrium is possible in a well-isolated environment. For example, if the kettle were very well insulated, then its contents could approach thermal equilibrium.

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1. gbaikie says:

Miles of ocean from ocean floor to surface is well-isolated, and more of a case as compared to hundreds meters below the ocean surface from warm surface waters.
One could say, cold polar surface water gets lost in ocean depth, but also ocean geothermal heat gets lost in ocean depths.
A question is which greater {and how much} cold falling water or trapped oceanic geothermal warmed water.
I am going to pick Geothermal heat as being the stronger {and going to say, at least, 2 to 1}.

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1. gbaikie says:

https://www.sciencedirect.com/science/article/abs/pii/0165232X84900089
-Abstract
While some information is available on the internal temperatures of ice sheets and ice shelves, virtually nothing is known of the thermal regime of icebergs after they have left their parent glaciers. According to the theory presented here the original temperature of an iceberg at the time of calving will be retained in the central region of the ice mass for years afterwards due to the low thermal conductivity of the ice. There is considerable evidence suggesting that this core temperature may be in the range of −15° to −20°C for icebergs in the North Atlantic.–
https://agupubs.onlinelibrary.wiley.com/doi/abs/10.1029/93JC00138
–Abstract
Calculations using a two‐dimensional numerical model which simulates the heat balance and temperature distribution of icebergs show that the temperature in the central region of an iceberg is almost unaffected by the thermal conditions imposed on its boundary. Hence the original temperature of the iceberg at the time of calving is retained in its core owing to the insulating quality of the ice–

And wondering about polar sea ice.
So thicker sea ice the more insulation.
So could have milder cold conditions and ice gradually thickens as compared to thin ice formed and getting very cold condition.
Though it seems to me that thicker polar sea ice, causes/allows more colder air above polar sea ice- or thin polar sea ice is unlikely to have colder air above it. But it seems if you have a more ice free summer and it later to form polar sea ice, the later part of year could have colder weather.

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2. Max Polo says:

OK Rob so in your “rough thermal equilibrium” can we consider the heat flux to be, let’s say, < 0.1 W/m2 ? Even if it was ten times this value, it would always be a couple of orders of magnitude away from emergent radiation. No, what you claim simply doesn't work, it's an ideological thing in my view.

Liked by 1 person

1. Rob says:

Max,

Thermal radiation is also suppressed by insulation. Again referencing hot water in a kettle, if the kettle is well insulated, then the outer wall will be room temperature (and won’t burn your fingers). The outgoing radiation will be determined from the outer wall’s temperature, not the water temperature, because radiation emitted by the water has is blocked by the kettle walls.

And of course, the net heat flux is then zero, because the outgoing flux from the room temperature wall is balanced by incoming flux from the room temperature surroundings.

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2. gbaikie says:

–OK Rob so in your “rough thermal equilibrium” can we consider the heat flux to be, let’s say, < 0.1 W/m2 ? Even if it was ten times this value, it would always be a couple of orders of magnitude away from emergent radiation. —
It can be a thousand times more than .01 W/m2, the .01 W/m2 is an average.

An question, if thousand times more, how much more quicker does it become emergent radiation as compared to .01 W/m2?
Roughly speaking one call it an unknown. And make it more dramatic. Say 1 square km has geothermal energy which is 100 watts per square every second for 10 years vs 1 square km having .01 W/m2 per square meter every second for 10 years, is there difference of which will emerge as radiation.
Again, unknown. But talking chances and percent of this heat becoming emergent radiation.
Generally speaking, it's said, heat in the ocean can remain in the ocean for thousands of years.

It is said that around 90% of global warming is being lost to the ocean, or 90% of global warming is not becoming emergent radiation within say 1 year {since we measure global temperature in a year time period- and "would not become lost" if this was delayed by merely one year. One would use word delayed rather than lost.
Reference- which saying it differently:
–Change over time
More than 90 percent of the warming that has happened on Earth over the past 50 years has occurred in the ocean. Recent studies estimate that warming of the upper oceans accounts for about 63 percent of the total increase in the amount of stored heat in the climate system from 1971 to 2010, and warming from 700 meters down to the ocean floor adds about another 30 percent.–
https://www.climate.gov/news-features/understanding-climate/climate-change-ocean-heat-content

So they are saying heat is accumulating in ocean for last 50 years or it's not becoming emergent radiation.
It seems reasonable ocean heat stored in upper level ocean has higher chance of becoming emergent radiation in the next 50 years, and 30 percent has less chance of becoming emergent radiation within 50 years.
But no one predicting when it will become emergent radiation, but in terms chance, the lower it is, seems to indicate a a lower chance.
So let's pull a number from nowhere, say with the .01 watt per square meter geothermal heat, within 10 years, .1 % becomes emergent radiation. Then one could say if geothermal heat was thousand time more {100 watt per square meter} there is no reason that 1000 times more heat becomes emergent radiation within same 10 year period.
And added to it, one could imagine 10 or 50% is due to more heating per square km.
Or if everything is the same, more intense geothermal heat **might** significantly hasten the time needed to become emergent radiation.
You could try to model it, and get some formula that has all kinds variables, but roughly, if have huge volcanic explosion under water or few cubic km of new lava is created within a year, it becomes emergent radiation X times faster than compared to average of .01 watt square meter.
But I would tend guess it's around 20% more within the 10 years or it's seems less likely it's more than 50%.

A similar question might how much heat is retained in ocean if 1 km diameter space impactor hits the ocean.

Though 1 km diameter impactor is extremely rare, so say, maybe 100 meter diameter impactor would be more interesting.

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5. Jarle says:

Strictly speaking, how much energy a volume contains, doesn’t say much about the flux. Degree of insulation is obviously essential. At a constant rate of 0.1J/s, the energy contained in the uppermost m^3 whould take ~70 days to transfer its energy to the atmosphere. Is this so strange? In the spring and summer, heat is passed down through the soil column. In the winter, the pattern reverses course, so maybe the net is expected to be quite small…

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1. Well…
There is no requirement to send just 0.1 J if the atmosphere was 0 Kelvin. More would be sent! The colder it is, the more rapidly it’s initially heated. Since T is 273K in this case,

Q = m * Cp * dT

= 1.29 * 1004 * 273.15 = 353 KJ

This of course will cool the 615KJ. This will propagate down the column. But that is OK.
The atmosphere was warmed LONG ago by geothermal. Now there’s just solar variation.

The main point is that the atmosphere has the “backradiation” capability already built in from geothermal. There is no need to create it from “greenhouse effect”.

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1. Jarle says:

“Back radiation capabilty of the atmosphere”. But surely only where the atmosphere can be warmer than the surface? Like in the Antarctic and maybe a few other location…

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1. I just meant radiation. As for whether it heats the surface, you are correct: only at the the poles.

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6. Jarle says:

What fascinates me about geothermal denial is the inability to realize the obvious. Everyone that has been inside a mine, knows that these places are strickingly hot, and you don’t even have to descend very far. I have been to the St. Kingas Chapel in the salt mine outside Krakow. It is 101m below the surface. The temperature in this spectacular place, in total absence of sunlight, is 15-16C. The idea that this heat should dwindle to utter insignificance upon traveling the last 100m to the surface, when it has passed thousands of kilometers on its journy already, is very strange. As you suggest, Zoe, there has to be a motivation for such a thought. Especialy if it stems from the mind of a trained scientist. From the skeptical people commenting here, I would like to ask for a sound explanation of why they think that it is impossible that geothermal delivers on average ~0C to the surface (as Zoe predicts)

Liked by 1 person

1. Rob says:

Hi Jarle,

Sometimes what seems to be “the obvious” turns out to be incorrect when you examine it carefully. The lithosphere provides an excellent thermal insulation between the asthenosphere (at ~1300 C) and the surface (~14 C). In the middle, there is an approximately linear temperature gradient. This ends up being about 30 C / km.

What’s interesting is that, if you solve the heat equation, this gradient is exactly the same regardless of the thermal conductivity (k) of the intervening rock (as long as it’s fairly low). That means that even with the rather small k of rock, there is still a steep rise in temperature as you go deeper below the surface. Thus the high temperature in mines.

However, the small thermal conductivity means that very little heat power flows upward toward the surface. This is given by Fourier’s law, which shows approximately 0.1 W/m^2 flows upward through the lithosphere toward the surface. But above the surface, much larger heat fluxes are at play — about 240 W/m^2 come in from the sun, and 390 W/m^2 are emitted as thermal radiation. The geothermal flux is insignificant.

I think the key fact here which makes things counterintuitive is this. Even though conduction is very slow at _supplying_ heat, high temperatures are possible underground because conduction is the _only way to lose heat_. But as soon as you get above the surface, conduction becomes totally unimportant because radiation is possible.

-Rob

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1. Rob,
The heat flux is small when the temperature difference is low. Forget the sun, and set atmosphere to 0K. The flux at the top will not be 0.1 W/m^2. To think that is incredibly stupid.

Again, you’ve avoided answering what happens when you place a cold obect next to one at thermal equilibrium (i.e. internal heat flux = zero).

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1. Rob says:

As a rough estimate, the geothermal heat flux is k DeltaT/L, where DeltaT is the temperature difference across the lithosphere and L is the thickness. The current temperature difference is about 1300 K – 300 K = 1000 K. If the surface were at 0 K, the flux would increase by 30% to 0.13 W/m^2.

“Again, you’ve avoided answering what happens when you place a cold obect next to one at thermal equilibrium (i.e. internal heat flux = zero).”

I’m not quite sure what the question is. If the two objects were in thermal contact (either touching or via radiation), heat would begin flowing from the hotter to the colder object until equilibrium was established.

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2. Rob,
I really feel there is something wrong with you.

“If the surface were at 0 K, the flux would increase by 30% to 0.13 W/m^2.”

The surface is not FIXED at 0 K, silly.

Have you ever cooked food? Does the pan top stay at room temperature?

“heat would begin flowing from the hotter to the colder object until equilibrium was established.”
The first object is at thermal equilbrium, i.e it has uniform temperature, i.e. its conductive heat flux is zero.

And here you admit that 0 W/m^2 conductive heat flux will warm 2nd object to the same temperature.

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3. Rob says:

You said, “set atmosphere to 0K. The flux at the top will not be 0.1 W/m^2.” So I set the surface to 0K and calculated the flux as 0.13 W/m^2. I don’t know how else you expected it to be interpreted.

0.13 W/m^2 represents the maximum geothermal heat flux, any change to the surface temperature would result in a smaller flux.

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4. No, I said set it initially to 0K. You FIXED it at a constant temperature. Don’t hold it unphysically permanently fixed.

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5. Rob says:

Right, as opposed to the totally realistic and physical scenario where the atmosphere is 0 K only temporarily.

You never said “initially”, and in any case I’m not sure what point you’re trying to make here. No matter what temperature the surface is, the geothermal flux is negligible.

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6. You know exactly the point I’m making, but you are morally corrupt and so you play dumb.

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7. Rob says:

Zoe,

I have been nothing but straightforward and honest. Instead of providing a coherent reply to my arguments, you have called me morally bankrupt and insincere, and implied I have some agenda.

People who are actually interested in the scientific process welcome criticism and are not emotionally attached to their theories. The physicists I know propose and discard new ideas on a daily basis as part of their work. You should have done the same with your geothermal hypothesis long ago. This is not an attack against you, it just a statement of fact.

Your theory relies on arguments that lack any recognizable formalism of physics. You disregard correct arguments with flippant, nonsense replies like, “He’s equating a heat flux between two locations to an absolute energy flux equivalent at one location. Is that conservation of energy? No!” You invent new techniques like assigning direction to units, when sound mathematics gives direction to vectors. When your theory is in direct conflict with established knowledge, you ignore or misunderstand the latter. And on top of all that, when you’re offered corrections, you lash out with personal attacks and claims that the whole scientific community is wrong.

Placing your own hand-wavey arguments over mathematical concepts that have been known for _centuries_ is just hubris. I can’t stop you from doing so, but you’ll certainly be wrong.

– Rob

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8. Rob,
Heat flow is not a conserved quantity. Energy is. You still don’t know the difference between heat and energy.

You’re not enlightening the readers by clarifying your points, but leaving a mess of contradictions … and I think you know that, which is why you get this treatment from me.

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9. Rob says:

Since you are consistently mischaracterizing my argument, I will just repeat my very first post:

Heat is a form of energy, and it is perfectly valid to look at how much heat flux is crossing into some volume. Energy conservation can be written as a continuity equation ( https://en.wikipedia.org/wiki/Continuity_equation#Energy_and_heat ). Integrating over a volume, the rate of change of energy within the volume equals the surface integral of energy fluxes, including heat flux. If 390 W/m^2 is leaving the earth’s surface, and only 160 + 0.1 W/m^2 is entering, then it would be cooling rapidly. But this is not the case.

As you can see I have never said or implied that there is “conservation of heat flow”. However, heat being a form of energy transfer, it must be included for conservation of energy.

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10. I understand your argument Rob, but I don’t think you understand mine.

Imagine there is a cavern deep in the Earth, such that radiation is possible from bottom to top and top to bottom. The bottom emits 10000.1 W/m^2 to the top, and the top emits 10000 W/m^2 to the bottom. Obviously there is also conduction from below the floor , and it equals 0.1 W/m^2. And obviously the net radiation is also the same as that conduction.

Your job is to explain why it’s that hot at the bottom of this cavern. There’s two ways to do it, and one makes more sense than the other.

1) The bottom is that hot because of conductive heat flux (0.1) from below and because of radiative flux from above (10000).

This is the one you favor. The problem with it is that it leaves open the question of why it’s that hot at the top? Your answer to that is bottom radiation ( 10000.1 ) minus conduction ( 0.1 ) = 10000.

Do you see the problem with this? You have a circular dependency! What was the point of going through this exercise when you could have just said …

2) The bottom is that hot because it simply has a lot of energy! There’s enough Joules in that matter, and the mass and specific heat is such to create that temperature.T = Q/(mc). And this temperature will emit 10,000.1 W/m^2 !

This is a much better answer. The bottom (and more accurately: the deep “core”) is more than fully capable of heating the top. In fact it already did! And that’s where the 10,000 W/m^2 “Downwelling” IR came from.

Your explanation, OTH, is whack. It pretends that 0.1 W/m^2 can’t “produce” a 10000.1 W/m^2 bottom emergent radiation, and therefore needs 10000 W/m^2 to emerge from the top. Where did the top get it? the top has 0.1 W/m^2 flowing to the top top, so where did you get 10,000 W/m^2 flowing down? HMMM?

What is true here applies to the atmosphere. Geothermal heated the atmosphere a LONG TIME AGO, and that is why there is so little net flux from geo to atmo. You live at the bottom of that analogous cavern.

Your explanation only applies to NET flux, and doesn’t tell you ANYTHING about one way emergent radiation.

The 390 W/m^2 is simply geothermal emergent radiation (~330 W/m^2) plus ~60 W/m^2 from the sun. The rest from the sun got taken up by latent + sensible heat.

Mainstream climate scientists flip geothermal and make it backradiation from the atmosphere.

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11. Rob says:

Zoe,

Discussing physics with sentences instead of equations leaves far too much room for confusion. Concepts that can represented by a single symbol balloon into paragraphs. And words are ambiguous where equations are not. I agree with some points in your most recent post and disagree with others, but it would be much more productive to have this discussion in the language of math.

I urge you to make your next blog post a solution of the heat equation for the earth’s lithosphere. To simplify, feel free to just solve for the steady state, because the full solution to the heat equation is guaranteed to approach the unique steady state at long times anyway. And a simple 1D model should suffice (you could also solve for a sphere but the result will be very similar). Then you have this simple equation:

0 = d^2 T / dx^2

with boundary conditions

x=0:
T = 1300 K

x=150 km:
-k dT/dx = -240 W/m^2 + sigma*T^4

The first boundary condition says the temperature at the bottom of the lithosphere is 1300 K.
The second boundary condition results from energy continuity at the surface, 0 = dq/dx , where q is heat flux.

Once you have a solution, you can ask the following questions:

1) What is the resulting surface temperature in this model with no greenhouse effect?
2) How much does varying the temperature of the mantle (boundary condition 1) affect the surface temperature?

These answers will quantitatively tell you how important geothermal heat is.

I think this would be a really great blog post.

Best,
Rob

P.S. If you disagree with any of these equations, please explain which equation you would use instead.

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12. Rob,
The surface-air boundary are right next to each other. There’s conduction abd convection right there. Why are you using only radiation and from the middle of the troposphere?

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13. Rob says:

This simple model ignores the atmosphere entirely (conduction, convection, and greenhouse effect) to test whether geothermal and solar heat are enough as per your theory. The atmosphere also makes the physics significantly more complicated.

(Accidentally posted this in reply to the wrong comment below).

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14. Rob,
Your precious “greenhouse house” effect IS geothermal. Your atmosphere is not a raw source of energy. It got its energy from geothermal. Look at an energy budget diagram. There’s a circular dependency, and crazy people are trying to convince you it originates from the atmosphere.

How are you still not understanding this?

Without sun:
Geo HEAT Flux => Geo Upwelling – Atmo Downwelling

0.1 = 334.2 – 334.1

The atmosphere is just a continuation of geothermal!

Now the sun enters the picture. It delivers a range of radiation from ~340 to ~161 W/m^2 from top to bottom of the atmosphere. Of the 161 reaching the surface, only ~60 W/m^2 adds to surface temperature, as the rest is latent and sensible heat.

This 60 W/m^2 will add to the surface Temperature that geothermal brought (~4C by recent measure, 0C by old convention).

What’s the problem, Rob? What are you having difficulty understanding?

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15. Rob says:

Zoe, in physics we solve problems involving temperature and heat conduction with the heat equation. Can you show your argument works using the heat equation?

You said in the main article here that an atmosphere isn’t necessary to show the strong effect of geothermal heating, so it should be very simple to solve the heat equation as I wrote it out above.

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16. Rob,
You mixed geothermal heat flux with solar coming into ~middle of troposphere and one way radiation from the surface.

I already solved the problem for you with equations. I’ll repeat again, for NO sun:

Geothermal Heat Flux = NET radiation

k dT/dx = sigma*(T_surface^4 – T_lowest_atmosphere^4)

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17. Rob says:

Zoe,

From reading the article, it seemed to me that you said an atmosphere wasn’t necessary to see the effect of geothermal. For example, the diagram of your fictitious planet has no indication of an atmosphere, and you say “But did I say this fantasy planet even has an atmosphere? What if not?” and “The surface here is 1000°C because it’s 1010°C a hundred meters below the surface. Simple.”

So we should be able to make a simple model that uses the equations of physics (heat equation), doesn’t have an atmosphere, and shows the effect of geothermal. Am I understanding correctly?

If so, can you offer an appropriate boundary condition similar to what’s in your last post, but with no atmosphere term? And then we can use that to solve the heat equation. If necessary we could later make a more complicated model that includes an atmosphere.

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18. Rob,
Without the sun or atmosphere, and introducing a monitoring satellite:

k dT/dx = sigma(T_surface^4 – T_satellite^4 )

Wherever the satellite is passing by.

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19. Rob says:

Zoe,

Great, now we have enough to solve the heat equation. The satellite should generally have little effect on the earth’s temperature since it is very small, so I’ll ignore it to start. Then our equations are:

0 = d^2 T / dx^2
with boundary conditions
x=0: T = T_mantle
x=150 km: -k dT/dx = sigma*T_surface^4

The differential equation has general solution:
T(x) = a*x + b

Plugging in the x=0 BC,
T(0) = a*0+b = T_mantle
so b = T_mantle.

From the x=150 km BC,
-k * d/dx (a*x + T_mantle) = sigma * (a*150 km + T_mantle)^4
-k*a = sigma * (a*150 km + T_mantle)^4

Wolfram Alpha can solve this equation ( https://www.wolframalpha.com/input/?i=Solve%5B5.67*10%5E-8*%28a*150000%2B1300%29%5E4%2B3*a+%3D%3D+0%2Ca%5D ) giving a=-0.0085 Kelvin/m or a=-0.0088 Kelvin/m. The second one gives a negative surface temperature so it’s unphysical.

Finally, we have a surface temperature of T_surface=T(150 km)=T_mantle – 8.5 Kelvin/km * 150 km = 25 Kelvin.

That’s pretty cold. What if we double the mantle temperature to 2600 K? Then T_surface increases to only 35 Kelvin. Not a huge effect.

Clearly according to this model, geothermal heat is not enough to give a high surface temperature. Either this conclusion is correct, or we need to add something to the model. What do you think?

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20. Rob
“-k dT/dx = sigma*T_surface^4”

This is highly fallacious. You can not equate heat to one-way radiation. You need to use radiative heat transfer equation. OK, you dropped it because you think it’s zero. It would actually be 2.725K from the universe, still excluding the sun.

The big problem with your math is that you conserve thermal conductivity (k)!

k is not the same throughout. k is a product of kappa (thermal diffusivity), density, and heat capacity. Assuming all geo matter is the same, you still have to deal with the density gradient!

This means Earth could be at thermal equilibrium and still have a temperature gradient.

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21. Rob says:

Zoe,

Definitely it is important to consider incoming radiation. For now I’m ignoring the atmosphere because it’s much easier and we’re trying to isolate the effect of geothermal. But we can easily include blackbody radiation from the universe. Then boundary condition 2 becomes -k dT/dx = sigma*(T_surface^4 – (2.75 K)^4) . But this changes the surface temperature by less than a part in 1 million.

I agree that thermal conductivity isn’t homogeneous in the earth, so we should examine how sensitive the model is to different k. Note that even with constant k there is already a temperature gradient in the lithosphere of 8.5 K/km. As an extreme example, let’s see what happens if we use the thermal conductivity of copper: 400 W/mK, about 100x that of rock. Then the predicted temperature gradient is 8.1 K/km ( https://www.wolframalpha.com/input/?i=Solve%5B5.67*10%5E-8*%28a*150000%2B1300%29%5E4%2B400*a+%3D%3D+0%2Ca%5D ), giving a surface temperature of 1300-8.1*150 = 85 K. Even with this very high k, that’s still pretty cold.

What do you think? We could add other things to the model as well and see how it changes. Could even add a simple model of a radiating atmosphere (a layer model), then check how sensitive the result is to changes in the mantle temperature.

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22. Rob,
“thermal conductivity isn’t homogeneous in the earth, so we should examine how sensitive the model is to different k … ”

All you did was replace rock with copper. That’s irrelevant. You need to examine the k GRADIENT due to DENSITY gradient.

I’m glad you didn’t go in the direction of arguing that the surface would be 2.725K and then internal conductive flux changes to 0.0000031 W/m^2.

I would like to congratulate you for showing that a copper top layer would create 85K at the surface!

That’s the power of geothermal!

Now replace the uniform density copper with rock of decreasing density (thus k), and you get ~273K as geophysicists show.

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23. Rob says:

“Now replace the uniform density copper with rock of decreasing density (thus k), and you get ~273K as geophysicists show.”

Are you able to prove this statement using the heat equation? It is an extraordinary claim that, while a uniformly high thermal conductivity (400 W/m^2) gives a surface temperature of 85 K, a much _lower_ thermal conductivity (rock is ~3 W/m^2) could give a surface temperature of 273 K, just because k is nonuniform. As I’ve shown, a smaller thermal conductivity generally _decreases_ the surface temperature.

You would need to show this mathematically for it to be at all believable.

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24. Rob,
“As I’ve shown, a smaller thermal conductivity generally _decreases_ the surface temperature.”

That’s because you’re using the wrong formula. You need to use the one I provided. It’s in the top right of my diagram: “Geothermal Emission”.

As I’ve already stated multiple times: You CANNOT equate an internal heat flux to emergent radiation. The heat equation only applies to matter and radiation is NOT heat transfer, unless hotter matter radiates to cooler matter.

You get around this fact by treating space as matter and setting its temperature to zero so you can eliminate the second term in the radiative heat transfer equation, and pretend that there is actual heat flow from planet to NOTHING. But you can’t actually do that.

Here is the FULL radiative heat transfer equation for concentric spheres:

Notice this is heat transfer from matter to MATTER

Is space MATTER? I don’t think so!

What is the emissivity of space? (epsilon_2)

It is ZERO. Now what happens with this equation?

Setting epsilon_2 to zero causes the denominator to be infinity, and the result is Q(1-2) = 0.

ZERO!

There is ZERO heat transfer from a planet to space. There is however heat transfer from a planet to a satellite. A satellite will quickly come to equilibrium with the surface without draining any significant heat content from the planet.

Another way to see the obvious is the basic heat equation: Q = m * c * dT

Q planet = Q space

mp * cp * dTp = ms * cs * dTs

where p and s are planet and space.

So how much heat transfer is there between planet and space?

Pretty obvious … ms = 0 … dTp = 0

Space has no mass, therefore there is ZERO heat transfer from planet to space.

Therefore it is illegitimate to equate conductive heat flux with emergent radiation and allow it to determine surface T.

Your equation is therefore invalid, despite being very popular.

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25. Rob says:

Zoe,

The boundary condition I gave is correct. Here is a more detailed explanation. Please let me know exactly which part you find objectionable.

1) energy conservation says du/dt = -dq/dx where q is net energy flux. In steady state 0 = dq/dx. Thus q is constant. I.e. for two given planes q1=q2. So far this is just energy conservation: surely you can’t have a problem with this.

2) take the plane just below the surface. The only energy flux is conduction. So q1 = -k dT/dx. This is just Fourier’s law, there is no reasonable objection.

3) take the plane just above the surface. There is an outward energy flux due sigma*T_surface^4. There is an inward energy flux sigma*(2.7 K)^4. The net is q2 = sigma*(T_surface^4 – (2.7 K)^4). This is just the Stefan-Boltzmann law, which you cannot possibly disagree with.

4) Subbing into q1=q2, we get -k dT/dx = sigma*(T_surface^4 – (2.7 K)^4) . This is simple algebra.

Being as specific as possible, please tell me which step you disagree with. So far you have only said “you can’t equate internal heat fluxes with outgoing radiation”, but it’s not clear to me what you mean mathematically. Each step has one equation. Which equation do you disagree with: energy conservation, Fourier’s law, or the Stefan-Boltzmann law?

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26. “There is an outward energy flux due sigma*T_surface^4.”

To what matter?
Energy doesn’t flow to nothing. Did I not explain that already?

“The net is q2 = sigma*(T_surface^4 – (2.7 K)^4)”

It looks like you’re sending radiation back to the distant stars and galaxies that gave you 2.7K. But that is them to you. They’re hotter than you! You’re transfering NOTHING to them.

The only significant object the Earth transfers heat to is the moon: ~0.1 W/m^2 … but not our whole surface at the same time.

The universe is 5% atoms. Your view factor is not even 1 for that emission.

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27. Rob says:

I see, you diagree with the Stefan-Boltzmann law.

Well, there is a wealth of experimental and theoretical evidence for it. The Stefan-Boltzmann law does not have a condition about “energy flowing to matter”. _Any_ object emits thermal radiation from its surface, including objects surrounded by empty space.

You can even point an IR thermometer or spectrometer at the ground to measure the outgoing thermal radiation. It’s undeniable.

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28. Rob,
Potential radiation and heat transfer are not the same thing.

An IR thermometer is matter, silly.

You’re confusing an over simplification for lay people with what actually happens.

Get a heat transfer textbook.

I already gave you an image of the formula.

You can’t just willy nilly disregard surface areas, view factor or mutual emissivities.

Your latest formula is very strange. You’re setting T_surface to cosmic rqdiation and then setting internal conduction with it. So you get 0.0000031 W/m^2 for both fluxes. Why?

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29. Rob says:

“Potential radiation and heat transfer are not the same thing.”

Indeed, “potential radiation” is not a thing at all. You just made it up.

You say I’m oversimplifying, but in this case it really is *that simple* to disprove the hypothesis that geothermal heat contributes significantly to surface temperature.

You have tried many ways to complicate the issue and obscure this fact. First you said my calculation needed to include inhomogeneous k in the lithosphere. I showed that this was not important. Then you said the model needed to include radiation from the cosmic microwave background at 2.7 K. I proved that this is also a tiny effect. You’ve said that the model needs to include satellites and distant galaxies. And finally you’re straight up making up words like “potential radiation” and insisting they need to be included for an accurate calculation. It is completely clear that you are throwing out a bunch of nonsense to muddy the picture.

I’m sure you could think of a million more things I could include in the calculation, and I could spend all day calculating. But I’ve already done this for several of your requests and none have changed the result. Because ultimately all it takes is conservation of energy, Fourier’s law, and the Stefan-Boltzmann law to disprove your hypothesis.

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30. Rob,
You seem to be upset that I have to explain it you in many different ways. Before “potential radiation” I said “NET radiation”, and you still didn’t understand it.

Perhaps you need to brush up on the basics:

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

Now cosmic radiation is good to start, but you need to include the moon:
https://phzoe.com/2020/10/05/geothermal-to-the-moon/

“to disprove the hypothesis that geothermal heat contributes significantly to surface temperature.”

You’ve done no such thing. Read again.

Remember …

0.1 = 334.2 – 333.1

Geo Heat Flux = NET radiation

You keep pretending geo doesn’t heat the atmosphere … so you end up confusing initial heat transfer with steady state.

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31. Rob says:

I’m annoyed because you’re throwing out one red herring after another. Before you stated quite confidently,

“ The big problem with your math is that you conserve thermal conductivity (k)!”

I put in the time to show, mathematically, that actually this is not a problem at all.

And suddenly you forget about that. Now the real, actual problem is something else entirely.

Now it’s the moon that’s the key to geothermal heating? Give me a break. When one wild theory is disproved, you come up with another. Why would this theory be any different? I’m sure when I disprove this one you’ll say actually I need to account for the solar wind, then the orbit of Mars, then some other irrelevancy. Meanwhile your own equations purported to prove your geothermal hypothesis don’t account for any of these effects.

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32. Rob,
Everything goes together. Would you like me to write an article explaining it all from the beginning?

You asked to remove the atmosphere, so I did. Whether it’s there or not doesn’t change the fact the geothermal flux shows up all the way to the moon.

It’s pretty obvious your job was to purposely ignore everything and just repeat your talking points.

What you ignored was your concession. Would you like to try again? Or will you continue to act like a grand standing ignoramous?

Which of the following is the correct way to do heat transfer?

2) Geo Flux = NET radiation

Justify your answer with a textbook reference if you can’t do it righr.

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33. Rob says:

Zoe,

In answer to your question: conservation of energy in a steady state implies dq/dx = 0, or energy flux q1=q2 for any two planes. In the plane below the surface q is just conductive heat flux. In the plane above q is the net radiative heat flux (plus any other heat fluxes, i.e. latent and sensible heat). So your second option is correct.

Since geothermal conduction is quite small, to a very good approximation one can say net radiative heat flux (plus net latent and sensible heat) is zero. This is what what you see in the famous energy balance diagrams.

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34. Rob,
Your brain seems to have stopped working.

Small number -> Big Number – Almost-as Big number

Conductive fluxes become small as the cold side T reaches hot side T.

So what if net radiative heat flux is small? This tells you nothing about temperature.

The full equation is
k dT/dx = k(atmo) dT/dx + h(Tsurface – Tatmo) + sigma*(Tsurface^4 – T_atmo^4)

If atmo starts at zero kelvin, you will get a huge initial flux. Earth will cool very slightly as atmo heat content is small in comparison. As Earth cools and atmo heats … it almost reaches equilibrium .. only 0.1 W/m^2 off. This is the steady state.

You for some strange (ideological) reason can’t grasp this.

You’ve tried every permutation of wrong possible!

You’re given a geo top T of 0C, and a geo flux of 0.1, and you desperately want to change either or both … which is not reality.

Do you at least acknowledge that geo delivers 0C?

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35. Rob says:

“ Do you at least acknowledge that geo delivers 0C?”

Above I solved the heat equation with geothermal and no atmosphere. The resulting surface temperature was 25 Kelvin. Actually I used the exact boundary condition you just gave:

“ k dT/dx = k(atmo) dT/dx + h(Tsurface – Tatmo) + sigma*(Tsurface^4 – T_atmo^4)”

But with no atmosphere terms.

If you have an equation which has no atmosphere and solution T_surface=0C, I’d be interested to see it.

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36. Rob,
You abandoned science when you neglected heat transfer from matter to matter ONLY.

You believe that a vacuum is a heat sink.

Do you even know how a vacuum thermos works?

P.S. Without the sun, Earth would have 25K somewhere very high in the atmophere, but not for the reason you give.

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37. Rob says:

Zoe,

Do you have equation which shows T_surface=0C with no atmosphere, only geothermal?

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38. Rob,
It’s already ~0C from internal energy. Can you show me the real-time temperature drop by your “radiation” into a vacuum?

I ask again, do you know how a vacuum thermos works?

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39. Rob says:

“ It’s already ~0C from internal energy.”

Zoe, the internal energy would change without the Sun and atmosphere (only geothermal heating). Do you have an equation that calculates this change?

“Can you show me the real-time temperature drop by your “radiation” into a vacuum?”

Any hot object in space cools off by radiation, albeit slowly. For example, a comet warms enough to outgas when it passes near the sun. Later it cools back down by radiation.

“I ask again, do you know how a vacuum thermos works?”

Of course.

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40. “For example, a comet warms enough to outgas when it passes near the sun. Later it cools back down by radiation.”

You can’t use the sun. The fact that energy flows in from the sun, and then later stops is irrelevant. There was real time heating. You have to show the comet draining its own internal energy into a vacuum, without reference to any other body.

“the internal energy would change without the Sun and atmosphere (only geothermal heating). Do you have an equation that calculates this change?”

I gave you the equation. I was asking you to backup your claim that internal energy is drained into space. So tell me: how quickly temperature drops per second the surface matter doing the “radiating” into a vacuum?

“Of course”

OK, let’s see. Does the distance between the cylinder walls change how much heat leaves the inner wall?

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41. Rob says:

“You can’t use the sun. The fact that energy flows in from the sun, and then later stops is irrelevant. There was real time heating. You have to show the comet draining its own internal energy into a vacuum, without reference to any other body.”

Well it’s certainly not the sun cooling it back down is it?

Are you seriously arguing now that it’s impossible for an object to radiate energy to space?

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42. The sun induced extra molecular motion. That extra motion was only due to the sun. Once it disappears, the molecules go back to doing what they were doing before.

There is no cooling due to radiation to space. It’s all your imagination. The cooling happens because the process of heating was removed.

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43. Rob says:

Zoe,

You have some deep misunderstandings of physics. Cooling cannot happen without losing energy.

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44. And you can’t get warmed by the sun if you’re losing energy as fast as you’re getting it, genius.

Your answer to the thermos question is incorrect. You’re not qualified to do physics, but only mock those who do.

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45. Rob says:

“There is no cooling due to radiation to space. It’s all your imagination.”

If this is really what you believe, you need a lot of education in the basics, which would take more time than I care to spend. Best of luck.

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46. Rob,
Basics are basic. When you look at the details, you realize that experiments that allowed formulation of SB Law used a small black object inside a large cavity.

Is a cavity space or is it matter?

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2. gbaikie says:

“However, the small thermal conductivity means that very little heat power flows upward toward the surface. This is given by Fourier’s law, which shows approximately 0.1 W/m^2 flows upward through the lithosphere toward the surface. But above the surface, much larger heat fluxes are at play — about 240 W/m^2 come in from the sun, and 390 W/m^2 are emitted as thermal radiation. The geothermal flux is insignificant.”

One aspect of this is amount watts per square added by CO2 is also a small value- I don’t think anyone put it above 1 watt per square meter.
I have my disagreements with Zoe, but is it your view that adding 1 watt per square meter is also insignificant?
I would tend to agree that 1 watt added is rather insignificant, but my reasoning is we don’t have an exact number and that 240 watts being emitted is only somewhere close how much is emitted by Earth.
Another issue is if we imagine that Earth does exactly emit 240 watts, is it constant value over say, last million years.
And perhaps, to make it simpler, if we assume the sun energy reaching Earth over last million years is also constant over last million year, would the amount emitted remain at 240 watts for over last million years?

Or does Earth emit same 240 watts during glacial and interglacial periods?
I believe the correct answer to give in high school test is that during glacial period, Earth emits less and absorbs less sunlight.
And then, we need reasons why the earth recovers so than absorbs more energy in order to get out of the glaciation period- and some reasons are given.
Over a century age is was suggested changes in CO2 level explained why entered and exited from a glaciation period. But of course that was proven wrong- as ice cores show quite the opposite.
But then as is now, it is considered that small change could make a difference.

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3. Jarle says:

Rob,

So radiation carries away energy at the surface. Of course. But why does this contradict 0C being delivered to the surface by geothermal? What do you think the temperature profile from St. Kingas Chapel and up to the surface looks like?

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1. Rob says:

“So radiation carries away energy at the surface. Of course. But why does this contradict 0C being delivered to the surface by geothermal?”

Well, it doesn’t make a lot of sense physically to say that a temperature is delivered to the surface. Instead, a geothermal heat flux is delivered to the surface, and we can figure out what temperature that translates to.

If suddenly geothermal were the only heat source for the surface, then the surface would be receiving about 0.1 W/m^2. At the surface’s current temperature, it is radiating about 390 W/m^2. Obviously this is a net loss of energy, so the surface would begin cooling. It would continue to cool until radiative heat outflow balanced the geothermal inflow. According to the Stefan-Boltzmann law (0.1 W/m^2 = sigma*T^4), this happens at -237 Celsius.

Of course this calculation is approximate. For example, as the surface cooled it would draw a little more geothermal heat. But not by much.

“What do you think the temperature profile from St. Kingas Chapel and up to the surface looks like?”

It will continue to look like a linear gradient until a few meters below the surface. At very shallow depths the ground temperature is more complicated because it is affected by surface conditions, seasons, etc. People who install geothermal heat pumps know a lot about this ( https://www.builditsolar.com/Projects/Cooling/EarthTemperatures.htm ).

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7. Jarle says:

Rob,

Think in terms of superimposition. Imagine linear profile all the way up to the surface, and then surface conditions like weather and insolation can do what it will to alter the top few meters. Would this thinking be unphysical too?

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1. Rob says:

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2. gbaikie says:

I would say that sunlight only heats a small portion of a sphere and mostly alters the top few meters and our molten ball rock heats uniformly, and ocean a large slab adding uniform heat. And ocean surface has highest average temperature, that makes the Average Atmosphere temperature.

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8. gbaikie says:

“Geothermal gradient is the rate of temperature change with respect to increasing depth in Earth’s interior.
As a general rule, the crust temperature is rising with depth due to the heat flow from the much hotter mantle; away from tectonic plate boundaries, temperature rises in about 25–30 °C/km (72–87 °F/mi) of depth near the surface in most of the world.”

Say one away from tectonic plate boundary, and drill hole/make pit 100 meters in diameter and 3 km deep or down to rock which is 75 to 90 K warmer than surface.

Air has “thermal gradient” or what is called lapse rate of about 6.5 C per km.
Or if there was instead no heated rock from Geothermal heat, the air 3 km down would be 6.5 C times 3 = 19.5 C.
The non geothermal heated air at bottom 3 km pit would 19.5 C warmer than air at the surface.
What temperature would air at bottom be if included the geothermal heat {75 to 90 K warmer than surface}?
Well, the air lapse rate would cool it, if one had an open top to the pit.
What happens if one puts ceiling/roof on the top of the 3 km deep pit?
Hot air rises. But if have hot air 3 km up, it will have the lapse rate of 6.5 C per km, down.
So with good insulation at ceiling, one get something like 75 to 90 C + 19.5 C air temperature at the bottom of the pit.

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1. gbaikie says:

How does this apply to Earth’s global temperature?

If you simply measure {and incorrectly not include the lapse rate] all the air of the atmosphere,
and averaged [the incorrectly measured] air, the number would indicate a very cold global air temperature.
But if you include the lapse rate {and all other factors- which are fairly minor factors} and entire average air
of entire atmosphere is around 15 C.
Or if took air at 2000 meter high and “brought it” to 1000 meter elevation the 2000 meter air made more
dense would be same temperature as 1000 meter air.
Or if 2000 meter air at 0 C fell 1000 meters and become denser because it’s at lower elevation {the falling and added
density] the air would become 6.5 C air.
But it note that 15 C air, is pretty cold air, and if the 15 C is say 3000 meter higher 15 – 19.5 K = -3.5 C air- which
will freeze water. And frozen ice falling will require warmer air below it to warm it {or it cool lower air}. And liquid droplets caused to higher in to colder air, will “warm” the colder air, and result in warmer air at the surface.
Or clouds “moderate” surface air temperature and such moderation in cooler regions, could be called a warming effect. Or more uniformity of global air, is called global warming.

Now if larger areas of earth’s surface “control” global air temperature. {it seems reasonable to me}
Larger areas are the ocean surface, 70% vs 30% of land surfaces.
And all land surface average temperature is currently at about 10 C
And all ocean surface temperature is “somewhere” around 17 C.
It seems one can say, ocean surface temperature, controls global average surface temperature.
AND Ocean surface warms land surface air. AND land surfaces cool global average surface air temperatures.

And related to such obvious “facts”, the temperature of entire ocean {which is about 3.5 C] controls
global surface air temperature. Or why we are in an Ice Age.

{And how people living in an Ice Age, and can worry about global warming, is a amazing wonder of human stupidity}

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1. gbaikie says:

So, say a bit more about why 3.5 C ocean, matters {a lot}, and also mention again my disagreement with Zoe.

In simplest terms, a 3.5 Ocean matters, because a 3.5 ocean allows the formation of polar ice sea ice.
Polar sea ice is rare in Earth’s long history {and any scientific discuss with people who think the Earth is
only 6000 year old, has inherent “problems”- one can believe this [and I am not saying anyone here believes it]
I am simply saying it requires a certain unnecessary amount of mental gyrations}.
Anyhow in Earth long history, polar sea ice is quite rare. And being in an Ice Age, we might think it is common or even needed. And it’s even possible that during our Holocene period, most of the time of this interglacial period has been with ice free polar sea ice. But doing our Ice Age polar sea ice is quite common.
In the psychotic news media, it’s common to express the end of the world being related to not having polar sea ice,
and it is true that the polar bear species has had polar sea ice for large amount of time of it’s existence, but it also a fact of having zero polar sea ice for thousands of years. And if Europe is getting stuff from Japan or China, ice free is shorter shipping route.
Having polar sea ice is sort like have more land area- area which cools rather than warms, and having more area which warms in coldest part of the world, is globally warming. Liquid ocean is area which doesn’t get much sunlight, warms because surface water cools and falls and is replaced by warmer water below it. Or warming is less severe cold particularly in winter when a very small amount sunlight gets to the region. Or it possible that polar bear will not even need hibernate- and will not need to consume such massive amount of blubber in relatively short period time.
Anyhow, what needed for more ice free arctic region is a more warmer ocean. And doesn’t mean ocean surface can’t freeze a bit, or it will not snow. And more likely is having on average more snowing. But one will have a shortage of time when air temperature is below -20 C in the arctic ocean area region. If want a lot -20 C air {or colder}, one will have to go to higher elevation.
So having ice free arctic will require more heat {and some more heat loss to space, but it should kept in mind that arctic ocean is a small part of the entire world. Warming a small but cold region of world, doesn’t require much heat, but turns a frozen wasteland into a place with more life- particularly, more trees. And it could be less of a mosquito hellscape.

Now, Zoe says Geothermal heat in the 30% of the world which gets less geothermal heat {and has actually been measured, somewhat- so can have more confidence in this} is making the world warmer.
And I guess, if we were instead at Mars distance from the sun, it would not make much difference- as the sunlight purportedly, adds little warmth to Earth.

It seems like it is quite hard to prove this.
And there happens to be a threat to billions of people of the world because there is this insane/stupid cargo cult which imagines Earth could become something like Venus.
{Venus having twice as much sunlight as Earth- due to it’s distance from the Sun.}

And I would say geothermal energy has a much larger effect, because we living in an Ice Age.
And I would say does a lot to explain what is happening during our Ice Age.

Though if our ocean was instead of being an average of about 3.5 C, was warmer than 5 C, the present level geothermal heating would have less effect upon global average temperature.

And we both agree that geothermal heating would prevent Earth from becoming a Snowball Earth.

And I would say 300 million ago {or longer] in Earth history, we probably had more geothermal energy, and even less possibility of a Snowball Earth. And geothermal heat could be part of answer of what is called, the faint young sun paradox:

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1. Jarle says:

Flat-temperature-people don’t know much about history (Or they choose to ignore it): from the site http://www.miljomytene.no (translated by google translate)

In the middle of the 17th century, wind conditions in the Pacific Ocean changed. During this period, a crossing took one month longer than both earlier and later. More ships sank than in other periods. The cold thus led to more storms and storms.

The cold in Asia caused the monsoon belt, ie the tropical rain belt caused by a low pressure on both sides of the equator (the intertropical convergence zone), to be pushed about 80 miles to the south. This led to famine in both India and China, caused by drought.

In North America, it was the late 16th and early 17th centuries, both long and cold winters, and dry summers. The French explorer Samuel de Chaplain (1567-1635), who founded the city of Québec, describes that the ice on one of the great lakes could be crossed on foot in June 1608. We do not know of any such thing in recent times.

Peter Bruegel the elder: The hunters in the snow. Painted in 1565, during a cold period during the Little Ice Age.

In the period 1580-1620, both summers and winters were cold in Europe. It was not uncommon for it to snow in the Alps in the middle of summer, even far down in the valleys. In Crete, snow fell in the winters. On the highest mountains in Scotland, the snow did not melt in the summer, and glaciers established themselves.

We know well when the ice went up on the Baltic Sea in the coldest years during the Little Ice Age. It tells of how cold the winters had been. A number of winters in the late 16th century and a little into the next century, it must have been 5-6 degrees colder than what is common today.

In the years 1620/1621 and 1621/1622, the Bosphorus froze. People could cross the strait between Europe and Asia. In the extremely cold winter of 1683/1684, the sea froze along the south coast of England, and in northern France. Very cold, and often very wet springs and summers had a major impact on agriculture. Crops in Europe often became miserable. 1627 was bad, and in the ensuing extremely cold year neither grain nor fruit ripened. In the 1640s, farming in the Alps was extremely poor, and at the end of this decade, both wine and grain crops in France failed. It repeated itself in the 1650s. Night frost hit Poland in mid-summer in the years 1664, 1666 and 1667. In June and July 1675 it was so cold in France that people had to fire in their houses. East of Asia it was very cold in some years. The spring of 1616 was icy cold in Japan, and the following year heavy snowfall fell in the subtropical parts of southern China.

During the cold years, the monsoon rains in Asia were reduced. In Europe, summers became humid, especially in the north. Yes, often very wet. We know it from Hungary in the years 1630-1641. There were major floods in places where it was usually quite dry. The years 1640-1643 were the wettest we know from southern Spain, and in 1649 northwestern Europe was hit by such heavy rainfall that the crops were destroyed.

With cold and long winters, as well as late and cold summers, glaciers grew all over the world. Farmers in the Alps noticed it well. The same thing happened in Norway; farms were vacated, and farmers asked to be allowed tithes.

The Swedish economic historian Gustav Utterström (1911-1985) was 50 years ahead of his time. He was the first to understand the link between a cold climate and the great economic problems in Europe. He was strongly opposed by the thinkers of the time, but his ideas eventually prevailed. He understood that although the average temperature in Europe during the coldest period did not fall very much (about 2 degrees colder than today), it was sufficient to reduce the crops around the continent. The cold weather shortened the growing season by three to four weeks, and lowered the height limit for crops by about 150 meters. Occasionally there was night frost, which ruined the crops. But not only lower temperatures reduced crops. With the cold came much larger rainfall, especially in northern Europe, which was harmful to crops.

With lower grain production, food prices increased. Especially the poor people starved. And hunger led to a greater risk of getting sick. Wars that often followed the bad times meant that armies acted as spreaders of infection, and major epidemics therefore broke out.

It is also part of history that the cold years brought about witch hunts. They became most extensive in the mountains and northern areas of Europe, ie places that were marginal to agriculture. And they were most numerous in the coldest and wettest years. In the 1620s, snow could fall in the summers in Europe, and it happened that the ice froze on the waters even in summer. It was precisely these years that the witch trials were at their worst. Several thousand women were executed during these years, accused of causing the bad times. Both Martin Luther, Jean Calvin and Huldrych Zwingly called for witch-burning, but the Catholic Church was also behind the crimes.

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9. Jarle says:

From my recent post: 80miles means 800km

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10. gbaikie says:

“As I’ve already stated multiple times: You CANNOT equate an internal heat flux to emergent radiation. The heat equation only applies to matter and radiation is NOT heat transfer, unless hotter matter radiates to cooler matter.”

But you can equate internal flux to how much it warms an ocean.
And it matters a little bit how cold the ocean is.
And I assume geothermal energy is absorbed by the ocean {the heat can’t “fly thru it”- or if any energy could travel at or near the speed of light- one could communicate with subs in deep water. Through sonar travels at speed of sound.}
In terms of a land surface, the heat would radiate {at or less speed of light] and conduct heat to air molecules and/or evaporate any H20 in the ground {and Earth surface/ground has water table in which any geothermal energy would go thru}. And btw if imagined there was lot geothermal energy, water would brought up from water table and plants would be getting water from it- or soil would not get below the 5% water content- which causes plants to die. Or one not need to irrigate crops.
Which brings up question why does water always evaporate. Or why doesn’t a empty cup {not rained on} fill up with water. There is a lot drying going on. Or ground will dry out if not rained on and part of it draining into the water table or down hill and rest evaporates into atmosphere. But if there was a high value connected to geothermal energy, water up and it would excess the amount evaporated to atmosphere.

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1. gbaikie says:

Though not saying that geothermal energy does not cause water to go up. Or I would say Mars lack of geothermal energy allow water to go down. Or I think there is a lot water in the Mars crust and mantle. And Earth’s mantle is drier than the Moon.
Or Mars is very dry, the Moon is far drier, and Earth mantle is the driest. But no idea how wet Moon mantle/interior is, but probably wetter than Earth’s mantle.

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2. Water starts to evaporate at 0C at sea level.
Geothermal delivers ~0C. So what do you expect? A very tiny amount of evaporation.

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11. gbaikie says:

–You believe that a vacuum is a heat sink.

Do you even know how a vacuum thermos works?

I ask again, do you know how a vacuum thermos works?

A thermos works mainly because H20 has high specific heat.
Water can’t lose a lot heat from evaporation. The warmer it is the higher the partial pressure.
https://en.wikipedia.org/wiki/Vapour_pressure_of_water
Water, 20 C. Vapor pressure: 0.0231 atm {0.33957 psi}
40 C. Vapor pressure: 0.0728 atm [1.07016 psi}

So, thermos have lid which seals. Don’t have lid on thermos, the coffee cools fairly fast.
In terms of vacuum, they have partial vacuum or just air space or foam insulation. Having partial vacuum
is better.
Incandescent lightbulbs have partial vacuums, also.
Partial Vacuum will transfer heat via conduction/convection. And perfect vacuums don’t have any conduction/convection {though stuff can evaporate into a partial or perfect vacuum {cool from that, Apollo cooled
it’s spacesuits by evaporating ice in their pack- to keep astronauts at cooler more comfortable temperature}.
[Other than perhaps boots or gloves, one doesn’t heating of spacesuit in space- as human body generates heat}.

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1. Thanks, gbaikie, but that was too much info.
My only concern is the distance between inner and outer wall.

I don’t care about real vacuum thermos’.

Just imagine two concentric cylinders of infinite height.

What do textbooks say about heat transfer between inside to outside? And what does Rob say …

According to Rob, the distance between matter doesn’t matter. The emissivity of receptor doesn’t matter. The view factor doesn’t matter.

He believes the inside emits according to SB Law, and that’s all. All textbooks are wrong.

How could that be? Either radiative HEAT TRANSFER is determined by all those things, or it isn’t. Both can’t be true at the same time.

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12. gbaikie says:

–Zoe Phin
January 15, 2021 at 7:50 AM
Water starts to evaporate at 0C at sea level.
Geothermal delivers ~0C. So what do you expect? A very tiny amount of evaporation.–

One reason water pipes are put under ground is to prevent the water from freezing- which occurs at 0 C.
One will having building codes which has different depth depending “how cold it gets in region”
I will google, Montana depth of water pipes
“Make sure the supply line from the well to the house is below the frost line — at least 5 feet deep in most areas. This will prevent line blockage and rupture caused by water freezing in the line during cold weather.”
Let’s try Oregon:
“C404. 10 Minimum burial depth. Underground piping systems shall be installed a minimum depth of 18 inches (305 mm) below grade, except as provided for in Section C404. 10.1”
Georgia:
“Top of pipe shall be buried not less than 1′ below the frost line (GA. Is 6”, so the minimum is 18”) (10.4. 2.) In areas where piping may be subject to mechanical damage,”
So frostline is 6″ but they want 18″, anyhow.
Let’s try Alaska {would guess they can’t have rule as it’s too friggin cold though some areas could be warm enough.
How deep is the frost line in Anchorage Alaska?
“42 inches
They state that the minimum frostline depth is 42 inches for warm foundations (i.e. bearing soils are maintained above freezing), and 60 inches for cold foundations (i.e. bearing soils are subjected to freezing).”
So would i say, got town called Anchorage because ground is warm enough. But you are crazy enough to want lay pipe tens of meters deep, I suppose one could have a town almost anywhere.
Polar region is pretty small region and most land region at depth the ground is generally over 0 C.
As in link in other post. at 0 C, vapor pressure is 0.0060 atm {0.0882 psi}.
And in vacuum of space at -150 C ice very very slowly evaporates- for practical purposes it doesn’t evaporate but -100 C ice for “practical purposes” does evaporate. Or Antarctica, ice evaporates but it also grows from sublimation, or in theory, an empty cup could have ice added to it. Or Antarctica air is very dry, but if the air become less dry, it could add to an empty cup. Or basically, why one needs a frost free freezer. Hmm, I guess you say if ice evaporates
why do need frost free freezer {cause you open the door}. Maybe better thing to say is, what is freezer burn:
“Question
What is “freezer burn?”
The outcome of frozen food losing its moisture as a result of poor wrapping.”
And ” These water molecules prefer the most hospitable environment- the coldest place in your freezer. The molecules migrate from the steak to the coldest place they can find,”
https://www.loc.gov/everyday-mysteries/food-and-nutrition/item/what-is-freezer-burn/
Or the frozen steak, evaporates and the water vapor sublimates somewhere colder- unless it’s sealed, well.

Liked by 1 person

13. Jarle says:

Zoe,

If you arm a mouse trap and then send it off to space, energy has left earth. It will not transfer its energy to another body before it hits one, and thereby releases the spring containing the potential energy. Are you saying that energetic photons leaving earth aren’t carrying away energy?

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1. gbaikie says:

Hmm.
It seems Zoe claims Earth is a star, and for an object to be warm, it has to be a star.
And stars don’t warm other stars.
I think might related to how cold the Moon is.
And I wonder if ever tried to determine the average temperature of Mercury- and if she did
this would “confirm” this idea.
Or I am sure what average temperature of Mercury is, but I think it’s like the Moon- cold.

Or many imagine Mercury is too hot for it to be habitable, but roughly I would say, only small portion of
Mercury is hot, it’s mostly quite cold. Or hotter than Moon, but one could claim the Moon is cold.

I say geothermal energy of Earth, is mostly about baseline temperature, which is overlaid with heat from
the Sun. And seems to me, Zoe is claiming the baseline temperature from geothermal energy is higher
than what I think it is.

Another way, to look at it, is greenhouse effect theory claims the Sun is only heating Earth to -18 C and greenhouse gases add 33 C to it’s average temperature. So in terms of what mean baseline, -18 C temperature is “baseline”.
And others think this baseline is even colder than -18 C.
I would baseline of Earth with ocean is about 5 C, but if remove the Earth’s ocean, Earth gets quite cold.
And I think ocean is a “warming effect” due to sunlight and geothermal heat.

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2. Gravitational collapse heating is linear, while radiative cooling is to the fourth power.

How would any body have temperature if its supposed to cool faster than it can warm?

Because space is not a coolant, but colder matter is.

Planck’s Law can’t be derived from light-as-a-particle theory.

The geometry of the cavity is important to which waves can form.

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1. gbaikie says:

If point is the cargo cult idea {or if like pseudo science} that most cooling occurs in upper atmosphere and/or gas can heated by sunlight, I have no argument about that. And Ozone is about a chemical process. Like calling it a greenhouse gas is like calling the Sulfuric acid of Venus clouds, a greenhouse gas. Or the clouds earth, “greenhouse gases”.
Earth warms and cools the atmosphere from a surface {ocean surface, land surface- and probably from cloudy “surfaces”}.

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3. Energy is matter in motion. Excess motion can only flow to where there’s less motion.

It is intra and inter-molecular motion that generates “photons”, and it would seem odd for electrons to slow moving because they moved.

It would seem odd to treat space as the most viscous fluid imaginable – capable of conduction to the 4th power.

And yet is this not what they do?

They do not say space conducts directly , and yet mathematically they treat it as such – to the 4th power!

I’m highly skeptical.

I think waves are only generated when there’s a difference in electromagnetic fields.

EM radiation is two orthogonal transverse waves traveling at the speed of light. I think there is something much faster traveling directly (not wavy) that allows negotiation to take place, so that “photons” will only be traveling from hot to cold.

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1. Jarle says:

The basis of the negotiation will be obsolete when the photon finally arives. Either the body is no longer there, or its temperature has changed in course of the the journey.

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1. Instant spooky action at a distance.

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1. Jarle says:

What is relevant in the video with regards to this discussion is the statement that in terms of heat, everything is connected to everything. Didn’t make a note of the minute mark. Apparently this was realized a long time ago.

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14. Jarle says:

New science reveals that 50% of the ordinary matter in the universe consists of ultra thin clouds of hydrogen gass residing between the galaxies and between the galaxy clusters. So thin that there are only a few atoms distributed over several cubic meters. Yet the temperature of these atoms is several million degrees celsius (plasma). How can these clouds retain such a high temperature 14.5 billion years after the birth of the universe if the theory of radiative heat loss to space itself is correct….

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1. Indeed, good point, Jarle. Although I can’t tell if they are actively heated by space electric currents ??? In which case it’s not a closed system.

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1. Jarle says:

Since 50% of the ordinary matter consists of structures that can be observed directly, and 50% consists of these clouds, it doesn’t leave room for any more ions. Perhaps dark matter ions. Or perhaps they got it wrong regarding the share of hydrogen clouds. One thing is certain, science evolves (unless hindered by the Iron triangle, as Richard Lindzen calls it)

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2. gbaikie says:

Gas temperature is velocity gas molecules travel.
Earth orbit has velocity of 30 km/sec [66,960 mph}. If had enough gas molecules to hit each other at 1 AU distance from Sun they would collide at high velocity {and be hot}.
At our galactic distance, Sol is traveling at 230 km/sec if enough gas at that distance to hit each other, then they would even hotter.
Hydrogen wall:
“Hydrogen wall and heliosheath Ly α absorption toward nearby stars: Possible constraints on the heliospheric interface plasma flow”
“The heliospheric interface is the region of the solar wind interaction with LIC. The heliopause is a contact discontinuity, which separates the plasma wind from interstellar plasmas. The termination shock decelerates the supersonic solar wind. “..
“and Mach number, respectively. For the inflowing partially ionized interstellar gas, we assume a velocity of 25.6 km s−1 and a temperature of 7000 K. These values are consistent with in situ observations of interstellar helium”
https://agupubs.onlinelibrary.wiley.com/doi/full/10.1029/2002JA009394
What is the hydrogen wall:
“The sun moves through the galaxy encases in a bubble formed by its own solar wind. … That hydrogen wall is the outer boundary of our home system, the place where our sun’s bubble of solar wind ends and where a mass of interstellar matter too small to bust through that wind builds up, pressing inward.”

So beyond Pluto one has Hydrogen Wall and gases are hot. And we not in a particularly “energic” part of Galaxy.
Gas glows hot when collide at high velocity- but space is not hot {or cold} due to the lower density of gas/plasma involved- it glows the glow would not warm anything. Or it’s same thing with thermosphere though our thermosphere is quite dense compared to space [even compared to our Moon’s “atmosphere”- which has better vacuum then can be easily made on Earth]
https://en.wikipedia.org/wiki/Thermosphere
“The highly attenuated gas in this layer can reach 2,500 °C (4,530 °F) during the day. Despite the high temperature, an observer or object will experience cold temperatures in the thermosphere, because the extremely low density of the gas (practically a hard vacuum) is insufficient for the molecules to conduct heat.”

But even in our dense surface air, air does not warm anything by radiant energy- or the surface air transfers heat via convection/conduction/ evaporative heat transfers. Unless one deal with intense radiation such as from flame or IR heat lamp {which any distance, unless huge area, is not very hot}. Or put you hand side way {not above, where hot gases is rising] of gas heating element of a stove.

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15. Pebbles says:

A remark: Your quote says thermal equilibrium is constant temperature and then you talk about uniform temperature. But this is not the same and one has to be careful with the term thermal equilibrium. In a isolated system (no external heat source or heat sink) thermal equilibrium is reached with uniform temperature. However, if a constant heat source and constant heat sink is added then thermal equilibrium is reached if the temperature is constant but might be non-uniform. Take, for example, a rod and heat one end with a torch and put the other end in a flow of coolant like liquid nitrogen (to keep the sink at a constant temperature). After a while the temperature distribution on the rod will stabilise but be non-uniform. So you have a non-zero heat flux (ΔT is non-zero) although the temperature in the rod is constant. Or take a wire that you heat on one side with a lighter and and the other end in a water stream from your tap.

I am not saying that CHF = CSR, but so is not CHF = 0.

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1. By “constant temperature” I meant the same temperature throughout.

Adding a “constant heat source”, and more importantly a “constant heat sink” would not produce a constant temperature indeed.

I’m not putting a rod into an a liquid nitrogen powered heat sink.

I’m using no heat sink, and allowing constant heat source to do its thing, which is to make the temperature same throughout (assuming density gradient is the same).

Don’t play with words.

“So you have a non-zero heat flux (ΔT is non-zero) although the temperature in the rod is constant.”

No. I didn’t mean constant as in final.

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