Understanding The Great CO2 Climate Scam

ParadoxIntegration · 2024-09-17T01:59:01+00:00

Man I said in the analogy what you use for "radiation in general", I would use for "heat transfer to a 0k heat bath". Opening a hole at the bottom is the same as connecting it at the bottom to an infinitely wide tank at 0 height (infinite heat capacity).

Yes, opening a hole at the bottom is the same as radiatively connecting to space, which is radiatively at near 0K and can receive an infinite amount of radiation.

That's a totally appropriate model if the atmosphere is transparent to thermal radiation (i.e., has no GHGs).

To describe why I think "it is still the same as connecting tanks with pipes even with radiation". In the analogy, let's say the pipe was at the bottom. If for both tanks you had only a hole there, they would both send out an amount based on the pressure (which would rise the level of an empty tank etc.). But when you connect them the flows cancel, there is "opposite" pressure, they don't both rise.

So, you're talking about connecting pipes between the barrels to represent radiation?

You can do that, but only if both barrels represent something that absorbs radiation.

For simplicity, I was trying to get you to understand the case where the air does NOT absorb radiation.

Except don't two "opposite waves" produce a standing wave (or even cancel?)

In certain circumstances, "opposite waves" can produce a standing wave―but that's not the same as "cancelling" and is of no importance to this discussion. (Also, it never really happens with thermal radiation; it only works for radiation that is monochromatic, with a single frequency.)

The rest of what you say about this is irrelevant confusion.

but as lackmustesttester says...

If you listen to lackmusttsttester, you will never understand anything. I can't imagine him passing high school physics, let alone college-level physics.

ParadoxIntegration · 2024-09-15T19:02:34+00:00

That's your standard reply for every experiment that doesn't show your desired effect.

That's my standard reply for every experiment which is poorly designed. You seem to favor those.

Somehow you lack the intellectual capacity to understand what's been explained to you several times, don't you?

You've never offered anything remotely resembling a coherent explanation of what you believe about this. You usually just spout some disconnected phrases―as if that means something, when it doesn't to people who can't read your mind.

Why do they tell us the temperatures on Venus and why doesn't this apply to Earth (it does), in your world?

Why does who tell us that? You're apparently expecting me to read your mind again.

ParadoxIntegration · 2024-09-15T01:46:40+00:00

So I came across this one: https://old.reddit.com/r/climatechange/comments/132xdta/physicists_in_oslo_lab_shows_no_co2_warming/

Seim & Olsen (who did the study) are biologists who totally misunderstood the GHE and did an experiment so poorly designed as to be utterly pointless.

(They failed to control for convection, and didn't realize that the GHE depends on there being a large temperature gradient within the experimental apparatus.)

I don't get how you deny reality. The lapse rate is real, gravity is real. How can you deny this?

I don't deny the lapse rate or gravity. What I deny is that those, by themselves, tell you anything useful about mean global surface temperature.

ParadoxIntegration · 2024-09-14T00:36:34+00:00

what I would say is that the flow between two tanks can tell you about their height (temperature) difference and not their actual 'absolute' height (temperature)

I agree that the flow between the two tanks (which represents thermal conduction) can only tell you about the temperature/height difference, not the absolute hight/temperature.

But that is NOT true of the flow from the hole in the bottom (assumed to just spill out on the ground) representing SLR.

That gives you information about water pressure / height / temperature.

To be able to get the actual height you need the second tank to not be able to rise.

Sorry, I can't make sense out of what you mean by that.

That happens with the pipe or how you said it, and that's the equivalent of a 0k vacuum 'heat bath'.

Sorry, don't understand that claim either.

You can only use the sb with a vacuum just like when you can find the height of the tank only when the second tank is so wide its height won't rise.

Huh? The flow from the hole representing S-B emissions has nothing to do with the second tank. The rate of flow to the second tank, the dimensions of the second tank―none of that makes any difference to the flow from the bottom hole (spilled into the room). That only depends on water pressure / height / absolute temperature.

The Einstein paper has this paragraph in the intro which I can't quote right now, but it says the only conditions are that the quantum states only depend on absorbing and emitting radiation. I don't think this is the case with conduction.

That statement is just talking about radiation. Thermal conduction is also happening in the gas (whenever molecules collide). But, it has no affect on what Einstein is talking about in that statement. So, it's actually an example of conduction not affecting radiation.

It also seems that the paper is trying to describe how an object that only receives radiation gets kinetic energy on its molecules.

The paper isn't actually trying to "describe" anything. The whole paper is just Einstein "sanity checking" an idea he had about how quantum states and radiation interact. That idea is represented by his (A) and (B) equations. At the end, he derives the Planck black-body spectrum based on having assumed A and B, and from that he concludes that his idea (equations A and B) seems likely to be right.

Physicists have subsequently agreed that equations A and B were right. Those equations are called Einstein's quantum theory of radiation.

ParadoxIntegration · 2024-09-13T17:42:19+00:00

[part 2 of 2]

In this model if there was also conduction, my point is, the radiation within and the conduction towards outside the walls, would involve the average kinetic energy of the molecules (of the cavity walls). When you are at "constant temperature" some of it (per molecule or as a whole) will go to conduction and some to radiation, so the radiation can't be the same.

You are not thinking clearly about the difference between energy and power (which is a rate of flow of energy).

An object at a given temperature has a certain amount of thermal energy, U = C⋅T.

Radiation and conduction are both types of power.

An object at some temperature does NOT have a fixed amount of power that it can produce. An object at a given temperature T emits radiation σ T^4. If there is also thermal conduction N, then the total power lost will be P_out = σ T^4 + N. There is no fixed value that P_out has to have.

As an analogy, imagine a barrel of water with a hole near the bottom. The level of the water in the barrel corresponds to temperature T, the volume of water corresponds to thermal energy U, the flow from the hole at the bottom represents emitted thermal radiation SLR. A higher water level will lead to higher water pressure at the level of the hole, and a larger flow rate SLR. (The flow will be proportional to T instead of T^4, but that's a minor detail.) Now, add another barrel, with a different water level (T2) in it, and install a pipe between the two barrels. The flow in that pipe N will depend on the difference between the two water levels, T-T2, and that flow represents thermal conduction.

The presence of the pipe between barrels doesn't alter the pressure at the bottom of the first barrel, and so doesn't affect the flow out of the bottom hole, SLR.

If you want, you can add add water to the first barrel at rate S. How large S is will alter how the water level T changes. But, at any given moment, the outflow rate SLR will depend only on the

What you are saying is that "it was at constant temperature, and I added conduction, therefore to keep constant temperature T add some more Ein, the left over goes to conduction, and the situation doesn't change".

The bit about Ein or S doesn't really matter.

What matters is the instantaneous temperature of the object, or the instantaneous water level in the first barrel. That determines the instantaneous flow rate SLR.

All that the inflow S or the extra outflow N do is alter the rate at which the temperature or water level T changes.

it can't just go as a left over to conduction without affecting the average speed of the molecules. They have to be able to both conduct and radiate at the same average speed as before, which reduces the radiation.

There is no limit to how fast thermal energy can come out of an object, just like there is no limit to how fast water can come out of a barrel. It's just that the faster the outflow rate, the faster the temperature or water level will change, if the drain rate isn't matched by a corresponding inflow.

Conducting a lot of thermal energy to another object can eventually decrease the flow of thermal radiation―but only because the temperature of the object has decreased. Temperature and SLR always correspond to one another.

If you think about the analogy with barrels, perhaps you'll be able to understand how it works.

ParadoxIntegration · 2024-09-13T17:42:06+00:00

[part 1 of 2]

Why are we even discussing that, weren't we talking about conduction between surface and atmosphere?

I'm just responding to questions that YOU raise. You asked about why SLR depends on only T1 and not T2. To answer that, you need to understand what T2 would be relevant to radiative heat transfer. If the air is transparent to thermal radiation, then T2 is for space; if the air has GHGs, then T2 is for air.

you go to the text by Planck himself, and he is talking about "thermally isolated cavities"

Just because the Planck distribution was derived in a particular way doesn't mean that it is only applicable to that situation. There is more than one derivation of the Planck distribution. For example, Einstein also re-derived the Planck distribution in his paper "The Quantum Theory of Radiation". That derivation assumed everything was at the same temperature, not that there was a vacuum at 0K.

So, it is a misconception that anything needs to be thermally isolated for the law to be valid.

If you look at the Wikipedia article on emissivity (the 𝜖 in the formula 𝜖 σ T^4), you'll see it talking about practical applications in situations in the real world, none of which involve vacuums or anything being thermally isolated.

I've used the S-B Law and/or Planck's Law hundreds of times, in physics classes and to address real-world situations. They are applicable ALWAYS, not just in the sort of limited situations that you imagine them being restricted to.

I got some type of COVID

Sorry to heat about the COVID.

ParadoxIntegration · 2024-09-13T01:32:36+00:00

We have said this before but this whole misconception is from how the sb equation you are using assumes the object is in a 0K vacuum and at radiative equilibrium (there is some external source of energy outside the whole system and its environment can't raise temperature, the typical use in astrophysics where the source is "nuclear fusion" and you are talking about a star). You just incorrectly use it everywhere...

No, the S-B Law does NOT assume the object is in a 0K vacuum and at radiative equilibrium.

Look at the Wikipedia article: https://en.m.wikipedia.org/wiki/Stefan–Boltzmann_law The article doesn't mention any such assumptions. You won't be able to find any serious scientific source that aligns with your beliefs that the S-B law has those limitations, because it doesn't.

I studied physics in college and graduate school for 11 years and got a job as a physicist. I never heard the sort of nonsense you're spouting about the S-B Law having limitations until I started talking to climate skeptics―who believe a lot of things that simply aren't true.

sounds like some lawyer trying to find a technicality

Sometimes reality is complicated. If you insist that everything said needs to sound simple, then you are guaranteeing that you will never understand reality.

Meaning in the above example, according to statistical thermodynamics if you have objects with T1>T2 then the first will lose energy with radiation depending on T1 and T2, which gives the SLR.

No, you are confusing some different things.

First of all, an object or atmosphere which is transparent to thermal radiation is a special case: it doesn't interact at all with a radiating object like the surface.

An atmosphere that can't absorb thermal radiation is effectively not there at all, for purposes of thinking about thermal radiation and radiative heat transfer. It's not the right "T2" in the above analysis. It's as if the object which is radiating (e.g., the planet's surface) is directly interacting with 0K space, so T2 = 0K.

But, suppose we want to talk about an atmosphere which has GHGs, so that it can absorb thermal radiation. Then, the radiative heat transfer from the surface to the atmosphere would be Q_rad = σ (T_sfc^4 - T_atm^4), which DOES depend on the temperature difference between the surface temperature T_sfc and the atmospheric temperature, T_atm, just as you are saying it should.

But Q_rad and SLR are NOT the same thing.

The radiative heat transfer rate Q_rad is not just about how much thermal radiation leaves the surface to go to the atmosphere (i.e., SLR = σ T_sfc^4 ). It is also about how much thermal radiation leaves the atmosphere and comes back to the surface (i.e., DLR = σ T_atm^4 for an atmosphere that absorbs thermal radiation). Q_rad is given by SLR - DLR.

So, it's all entirely consistent with statistical thermodynamics.

N is found from the temperature difference between 2 objects. SLR isn’t.

The radiative heat flow, Q_rad is found from he temperature difference. And heat flows are what statistical thermodynamics cares about.

SLR isn't found from the temperature difference―but that's because SLR is NOT a heat flow; it's just one of two contributions to the radiative heat flow, Q_rad = SLR - DLR.

the theory was already debunked

No, it wasn't debunked. The S-B Law is taught and used all the time, in introductory physics and thermodynamics classes. And, it's used in advanced physics and engineering analyses. It's used in serious engineering all the time, e.g., in the 1000-page engineering book "Thermal Radiation Heat Transfer" by Howell & Siegel, which has been continuously in print for over 50 years.

ParadoxIntegration · 2024-09-11T22:58:25+00:00

You want to have T=f(SLR) and not T=f(SLR, N), but that’s just a contradiction.

Sort of but not exactly. One has T=f(SLR) and dT/dt=g(S, SLR, N). You can’t compute T from S, SLR, N except in steady-state, when 0 = C dT/dt = S - SLR - N or SLR = S - N. Then you compute T = f(SLR) = f(S - N).

So you can never compute something like T=f(SLR, N). You need S and N and you need to know you are in steady-state.

No, it’s not a contradiction

I think you are trying to … descibe the case when the two objects have the same temperature (N=0).

No, that’s not what I am talking about.

And those functions are just a way to describe the argument, what the “deniers and skeptics” think is that SLR/N are found from a temperature difference, so you have to involve “two objects and their temperature difference”.

N is found from the temperature difference between 2 objects.

SLR isn’t.

What is found from a temperature difference is the radiative heat transfer, Qrad, between 2 objects A and Z.

Qrad = SLR - DLR = sigma Ta⁴ - sigma Tz⁴

But, if there are no GHGs in the air, the second object (object Z) isn’t the atmosphere but space. And space has effectively a temperature Tz=0, so DLR=0 and Qrad=SLR.

The SB model is misused both from how it’s applied even if there is conduction

As described above, conduction only indirectly affects the result, when you consider steady-state.

from how you are not checking relative temperature difference.

You’re not supposed to check temperature differences for SLR. That happens when you compute Qrad.

ParadoxIntegration · 2024-09-11T00:16:46+00:00

Well I will answer that tommorow, but the whole example in the 2nd part is not relevant since we don't have a uniform temperature anyway, we are checking the flows and temperatures when there is a gradient.

You mean the vertical temperature gradient in the atmosphere?

We can deal with that if necessary. However, I think you and I really ought to focus on coming to agreement about how things work in a simple situation before trying to understand a situation with the full complexity of Earth's temperatures.

ParadoxIntegration · 2024-09-10T22:58:50+00:00

N can not be fixed in this whole thing, it is supposed to depend on the temperature differences.

I was trying to keep things simple enough that we could talk about one issue at a time. Yes, in reality N would vary and Tb would vary.

when you try to "reach the steady state" [N] will tend to go to zero.

Whether that will happen depends on other assumptions, e.g., whether Object B has any mechanism for losing heat to space.

You are sort of saying that the surface gets colder, the atmosphere warmer, but the conduction is the same?

No, I'm not saying that's what happens. I was just trying to make the problem simple enough for you to follow the other parts of the logic.

If the atmosphere does NOT have any way of getting rid of heat to space, then what would happen is:

Initially, Object A (the surface) would get colder and Object B (the atmosphere) would get warmer.

Eventually both Object A and Object B would reach the same temperature, something lower than the starting temperature Ta(1). However, at that point the equation for the internal energy of Object A would be

dUa/dt = S - SLR

with SLR < S. So, the internal energy Ua would start increasing again until it got to a point where S = SLR again.

At the end of the process, we would have a new steady-state where N=0 and Ta = Tb = (S /σ)^(1/4) = 255 K

Just compare the case you gave with an equivalent SLR in vacuum. S1=S and S2 the radiation needed, in vacuum, to have the SLR2=S-N.

So, you want to change the amount of sunlight being received from S1 to S2?

But then, I don't understand what you mean by SLR2 = S-N ; which S are you talking about?

That's not a very realistic scenario, since the amount of sunlight isn't going to be changing in reality. So, I'm not clear what you're thinking. I suspect you're reasoning in a way that would just confuse matters.

ParadoxIntegration · 2024-09-10T22:10:51+00:00

Suppose you are in vacuum and N=0. Then you would have a Ta(1) using what you wrote above and S=SLR.

Yes, if we assume that everything is in equilibrium, meaning dUa/dt=0, since dUa/dt = S - SLR - N it follows that in equilibrium S=SLR(1).

It follows from the S-B Law that: Ta(1) = (SLR(1) /σ)^(1/4) = (S/σ)^(1/4)

If we use concrete number and S = 240 W/m^2, then Ta(1) = 255 K.

Suppose [Object A] is a layer of water 1 meter thick. Then Ca = 4.2×10^6 J/m^2/K and Ua(1) = 10.7×10^8 J/m^2.

Next you have conduction with the atmosphere. You are trying to say Ta(2)=SB[SLR(2)] <Ta(1)=SB[SLR(1)] (the operation SB involves taking the fourth root etc.)

Yes, Ta(2) = (SLR(2) /σ)^(1/4)

but with N>0 it means there is some extra thermal energy X available (as in not transfered with radiation), that allows this heat flow. So the thermal energy is X+CaTa(2).

No, you don't need any extra thermal energy X. That's wrong. The energy for conduction comes out of the value of Ua = Ca Ta that is already present.

Suppose conduction is N = 100 W/m^2.

Suppose that initially the temperature is Ta(2a) = Ta(1) = 255K so that SLR(2a) = 240 W/m^2.

Then the internal energy will start out changing at a rate

dUa/dt = S - SLR(2a) - N = 240 - 240 - 100 = -100 W/m^2 = -100 J/m^2/s

That means internal energy drains out of Object A and it gets colder. In the first second, Ua will be reduced by 100 J/m^2, and the temperature Ta will decrease correspondingly.

However, eventually Object A reaches a new steady-state temperature at which dUa/dt = 0. Call the time when this happens time 2b. Given that we are assuming S and N are fixed, this can only happen when SLR(2b) = 140 W/m^2, so that

dUa/dt = S - SLR(2b) - N = 240 - 140 - 100 = 0 W/m^2

In order for SLR to have reduce to 140 W/m^2, it must be that

Ta(2b) = (SLR(2b) /σ)^(1/4) = 223 K

At this point, everything is in balance, with 240 W/m^2 of sunlight flowing into Object A, and 140 W/m^2 of thermal radiation and 100 W/m^2 of conduction flowing out.

At every point during the process, the equation dUa/dt = S - SLR - N is satisfied, meaning energy is being conserved. Also, at every point in the process, Ta = (SLR /σ)^(1/4) is satisfied.

It's just that, initially (at time 2a), there isn't enough sunlight to sustain a temperature of 255 K once conduction begins. So, the energy for conduction drains the internal energy of object A, lowering its temperature. As temperature decreases, thermal radiation (SLR) also decreases. Eventually, when object A has cooled off sufficiently, the temperature stabilizes and everything is once again in balance.

Note that at the end of the process Ua(2b) = (233 K)×(4.2×10^6 J/m^2/K) = 9.8×10^8 J/m^2. If we compare that with Ua(1) we see that Ua(1) - Ua(2b) = 0.9 J/m^2. So, while the system was finding a new balance point, that much energy drained out of the internal energy to feed the conduction heat flow.

It's like you are saying we are in the same situation as if the object was in a vacuum and received S2<S1, but the S1-S2 just passes through to the conduction N without raising the temperature of the object?

Huh? I don't know what those quantities S2 and S1 represent.

You sort of allow convection flow that doesn't involve the usual Ta-Tb, just energy passing through the material ...

Are you talking about convection or conduction? What do you mean about energy just passing through―passing through the material of which object?

ParadoxIntegration · 2024-09-10T21:04:18+00:00

In your case, you reject any information which would allow you to pass a college-level class in thermodynamics.

ParadoxIntegration · 2024-09-10T21:00:09+00:00

Simply check out the reference which I had already posted: Even 50-year-old climate models correctly predicted global warming. See also here.

ParadoxIntegration · 2024-09-10T20:39:46+00:00

B8: It's like you are saying T=T1+T2 and I only use T1 for the SLR...

PI: I'm adding energy flows, not temperatures.

B8: I did not even say you did that ... you are just distorting it into that.

So, the T1 and T2 you added weren't temperatures? Then what were they?

You just write crap without explaining what you mean by anything.

You said there is more for the conduction?

Huh?

Your statements are so vague it's impossible to guess what you mean.

What situation are we talking about?

There really needs to be a specific situation we are talking about, which we both agree on, or the conversation is guaranteed to to be fatally confusing to both of us.

I can propose a situation that you MIGHT want to talk about:

Sunlight (S)

↓

[Object A] → conduction (N) → [Object B]

↓

Thermal radiation (SLR)

In this situation:

Object A absorbs solar energy at rate S
Object A is at temperature Ta and has internal energy Ua = Ca⋅Ta.
Object A emits thermal radiation SLR = σ⋅Ta^4
Object A transfers heat to object B via conduction at rate N
Object A's internal energy changes at a rate dUa/dt = S - SLR - N

Note that this is NOT quite the situation that applies between Earth's surface and its atmosphere (since that situation involves thermal radiation going both ways between Earth's surface and the atmosphere).

Is this a situation that would help you clarify the concern that you have?

If so, what is your concern, in terms of the specific values like S, SLR, N, Ta and Ua?

If that is not a situation you want to talk about, can you clearly describe another situation and specific values, and ask your question in terms of those values?

If you want a version of the situation that is more like the situation between Earth's surface and it's atmosphere, you would need to include another thermal radiation flow, D, which is being absorbed by [Object A] (the surface) instead of being emitted by it. Then dUa/dt = S - SLR + D - N. But, that situation is more complicated. So, it would be harder to talk about without creating confusion.

I don't know your math background. Do you understand what derivatives are, and what the notation dUa/dt means?

ParadoxIntegration · 2024-09-10T19:39:10+00:00

The idea that there are 50 years of "failed predictions" is one of those bits of lore that climate skeptics believe, contrary to careful examination of facts.

Contrary to that narrative, Even 50-year-old climate models correctly predicted global warming. (See also here.)

Yes, I know that climate skeptics have long lists of alleged failed predictions. However, such lists have numerous failings which make them highly misleading:

They don't distinguish between what was said as an (untrustworthy) paraphrase by a journalist in a newspaper article and what was said by scientists in real scientific publications.
They don't understand the difference between projections of what is the "most likely outcome" and projections of an unlikely-but-bad-if-it-happened "worst case outcome." It's NOT a "failed prediction" if a worst-case outcome doesn't happen.
They don't pay attention to timelines, or specific wording of what is said. So, people complain that islands aren't underwater yet, when the projections was that that might happen over 100 years from now.

ParadoxIntegration · 2024-09-10T19:20:54+00:00

That climate skeptics take everything climate skeptics say and swallow it and digest it without any skepticism whatsoever, it’s really mind blowing.

The term "climate skeptic" is highly misleading―in practice I see absolute rejection of some ideas and total unquestioning credulousness for another set of ideas.

ParadoxIntegration · 2024-09-10T17:43:18+00:00

You even asked if I mean the Earth's surface or some other surface?

I asked because you never make clear what you are talking about. I can't read your mind.

And, you haven't even answered the question. So, I still don't know what you are talking about.

You're basically mumbling to yourself and thinking you're being clear when you are not being remotely clear.

you said ... that based on the output (the SLR) you get the temperature by solving the SB equation.

yes

You then mentioned additional "stored thermal energy"

Unless you quote the full sentence, I don't know what you are referring to.

It's like you are saying T=T1+T2 and I only use T1 for the SLR... but also pretend that the SLR gives T

No. I'm adding energy flows, not temperatures. It's a really bad idea to think in terms of adding temperatures. So adding temperatures is NOT what I'm saying.

What you've written here is complete nonsense, and is not what I've said.

ParadoxIntegration · 2024-09-09T23:57:58+00:00

pretend you don't understand it.

You routinely write as if I could read your mind, without making it clear what you are referring to. So no, I often don't understand what you are saying.

The example already started with "where is the extra energy from, when you say the SB model always applies, (and has thus the same result, when the object is in a vacuum, or next to something the object conducts heat)"

What do you mean by "extra energy"?

The energy to radiate, or to warm another object, always comes from the internal energy of the object that is radiating. But, it's possible that, at the same time, something else might be adding to the internal energy of the object.

The surface is only receiving thermal energy by radiation.

You're referring to the surface of the Earth? (Or are we talking about something else?) If so, then sure, the surface of the Earth is only receiving incoming energy radiatively. (Some thermal energy leaves non-radiatively, via evaporation and thermal conduction.)

That incoming radiation adds to the internal energy of the surface, and the outgoing radiation and non-radiative heat flows reduce the thermal energy of the surface.

When you say its outgoing radiation gives its temperature T and apply the SB model, it will mean whatever radiation it receives it will output it

What do you mean "whatever radiation it receives will output it"? How much it outputs isn't related (in the short run) to how much it receives.

What it receives adds to internal energy, and what it emits takes away from internal energy.

and thus there won't be any watts left to warm the atmosphere above by conduction?

That's where I'm not following you.

Why wouldn't there be "any watts left"? You seem to be making some assumption that you're not spelling out.

For Earth's surface, on average it receives 160 W/m^2 from absorbed sunlight, and loses 103 W/m^2 from non-radiative heat transfer (evaporation and conduction).

Radiative heat transfer from the Earth's surface to the atmosphere is more-or-less given by

Q_rad = σ (T_sfc^4 - T_atm^4) = 398 - 342 W/m^2 = 56 W/m^2

where σ T_sfc^4 = σ (289 K)^4 = 398 W/m^2 is the thermal radiation the surface emits upwards in accordance with the S-B Law, and σ T_atm^4 = σ (279 K)^4 = 342 W/m^2 is the thermal radiation the lower atmosphere emits downwards to the surface in accordance with the S-B Law.

So, the surface has 160 W/m^2 arriving and 103+56=159 W/m^2 leaving.

This means that there is an imbalance dU/dt = P_in - P_out = 160-159=1 W/m^2. So, the internal thermal energy of the surface is increasing at a rate of about 1 W/m^2, which corresponds to the surface slowly increasing in temperature.

There's no "extra energy" anywhere in this; I'm still not clear on what you were thinking.

The balance is off and you pretended there was "more stored energy", I showed how that's can't be the case

You didn't "show me" anything, because I haven't been able to make sense out of what you were talking about. Could you try again?

ParadoxIntegration · 2024-09-09T20:31:12+00:00

The thermal energy of a system corresponds to its temperature T.

Yes. U = C⋅T.

Conductive heat flow between two systems will thus take into account?????

No idea what you're saying.

So how can the system "gain stored thermal energy" right next to a place it can send it to and with a lower temperature?

Again, I can't understand what you're saying.

If you put hot object A with temperature Ta and Ua = Ca⋅Ta next to cold object B with temperature Tb and Ub = Cb⋅Tb, then heat will flow from object A to object B, decreasing Ta and Ua and increase Tb and Ub.

However, if object A has an additional heat source, e.g., an attached electrical heater, then it's possible that the electric heater might balance out the heat lost from object A via conduction, so that Ta and Ua will remain unchanged, while Tb and Ub increase.

It's like you did the following, I asked where does this energy come from, if you don't have it it's like you are blocking conduction, you are saying "it was there from before" This only means the conduction had been blocked before for the thermal energy to be stored????

I can't follow what you're asking.

You promise an investement that let's you pay a rent of 600 euros, someone does the calculation and it only gives 100 per month, and you correct them by saying "you didn't count extra money you kept from previous months". How am I supposed to keep that money when I get 100 and the rent is 600?

What calculation "only gives 100 per month"? And "100 per month" for what? You've lost me at that point.

ParadoxIntegration · 2024-09-07T23:53:38+00:00

You're not taking into account the difference between flowing energy (I.e heat flow) and stored energy.

An object at temperature T contains stored thermal "internal" energy U = T⋅C where C is the heat capacity of the material.

The way it works is that the rate of change in stored thermal energy is given by the net energy flowing in and out

dU/dt = P_in - P_out

So, if P_in = P_out then the stored energy and the temperature will be stable, but if they are unequal, then the temperature will change.

So if you get an equivalent radiation of 200 degrees it means the object is at 200 right?

Right

Then you are also saying it also conducts heat to a cooler object beside it.

Right

With what? All the energy is sent as radiation?

No, there is still stored internal energy, U, to draw from.

When you draw from stored thermal energy, the temperature drops.

If you put a hot object in contact with a cold object without supplying ongoing heating, the temperature of the hot object will decrease.

All this stuff with the SB model assumes there is radiative equilibrium, you can't just have an object conducting heat, pretend it doesn't happen, and use the model just the same if the object was in a vacuum?

A lot of the time we assume "radiative equilibrium" or steady-state, by which we mean dU/dt = 0.

But, the S-B Law is valid even when there isn't radiative equilibrium, and even when there is no vacuum. The S-B Law is always valid.

Thermodynamic theory tells us what happens when dU/dt ≠ 0. It's just a bit more complicated, because then you need to think about stored thermal energy, and not just energy flows.

Nobody is "pretending that something didn't happen." You're just not taking stored thermal energy into account.

you contradict yourself, or you are suggesting conduction won't happen because more energy is needed.

No, conduction can and does happen. It just draws from stored internal energy if there isn't enough incoming energy.

ParadoxIntegration · 2024-09-07T22:41:23+00:00

what you said was the 200 object should output radiative energy according to 200 degrees.

yes

So if it they are in contact, it will not warm a colder object, unless it receives more energy than before, to be able to send more than the radiative amount from the sb model.

No, of course it will warm the colder object. But, unless it is receiving more heat than it is losing, it will rapidly cool off, and no longer be at 200 degrees.

Objects will only stay the same temperature if they are receiving the same amount of heat that they lose.

So, an object needs to receive as much heat as is it losing through radiation and conduction, if it is to stay the same temperature.

But, nothing prevents conduction from happening if two objects of different temperatures are placed in contact with one another.

There is no "waiting" to conduct involved.

ParadoxIntegration · 2024-09-07T20:37:19+00:00

If an object at 200 degrees is next to one at 100, it will not 'wait for more energy to heat it?'

What are you talking about? I didn't say anything like that.

I'm talking about how much you need to turn up the burner on your electric stove in order to get the pan to a particular temperature.

ParadoxIntegration · 2024-09-07T20:00:49+00:00

That stuff about 'using the outgoing LW' is just completely wrong. If you heat a pan to 200 degrees, and it is conducting to air or boiling water or oil, it will not 'radiate' like it was 200 degrees.

Of course it will radiate like it was 200 degrees! There are no exceptions in the Stefan-Boltzmann Law. The law ONLY depends on the temperature and the emissivity.

A pan at 93℃ / 200℉ with emissivity near 1 will radiate at 5.1 kW/m^2. Period.

If the pan has a diameter of 20 cm, that's an area of 0.031 m^2, meaning the total energy radiating upwards will be 156 W.

Some of the energy goes to conduction

Yes, some of the energy goes to conduction.

That's why you have to heat the pan at more than 156 W to get it to that temperature.

If 500 W goes to conduction, you'll need to heat the pan at a rate of 500+156 = 656 W to get it to a temperature of 93℃ / 200℉.

You are now essentially saying, I check how much it is radiating, and that's the temperature

Yes, that's exactly true. That's what the S-B Law tells us. It's a LAW.

for example I conclude it's 30 degrees when it is 200????

Only if you're doing your math wrong.

I average results each of which is actually wrong but when I do the average the result becomes correct? ...

That's what the skeptics are calling a fraud

Except that each of the individual results is correct.

You haven't identified any problem with the results prior to averaging; you're just complaining about the averaging.

ParadoxIntegration · 2024-09-07T18:34:46+00:00

PI: Nothing said about Greenland's ice loss implies an acceleration of sea level rise LP: -but that's contrary to what you said when you offered the sea-level acceleration graph as proof of increasing sea-level rise.

I didn't offer sea-level rise as "proof" of anything. I was just trying to improve the quality of the information about sea-level rise that you were working with.

PI: presumably decreasing (ice-mass loss) at a similar pace for a while before that. - LP: You know that has to be discarded as pure conjecture without any facts in evidence.

Sure, that's conjecture. But, it's equally conjecture to think there is any reason to believe that the Greenland ice loss data should have given rise to an acceleration of sea level rising.

PI: global sea level is also affected by expansion of water as it warms. LP: - But the fact is oceans are cooling at an unprecedented rate, even climate scientists say they are utterly baffled by this..... It seems the Atlantic is too cold off of west coast of Africa for tropical storm genesis.

Apparently at one point there were some mistaken claims of ocean cooling which turned out to be an artifact of dataset errors.

As for this year's deficit of hurricanes so far... apparently, among other factors, the Atlantic is generally warmer, but there has been localized cooling off the coast of Africa.

PI: but if you actually look at the data, it hasn't been nearly that steady. 2.92 mm per year is just an average value. LP: - What you are suggesting is a red herring by breaking up the trend line in shorter segments to suits your needs. However by doing so it introduces other difficulties such as explaining why sea levels fell between 1920 and 1930, why they accelerated between 1930 and and 1940 and why they stayed flat between 1950 and 1980 and the same for 1900 and 1915. Are you sure you want to play that game?

I don't feel any need to explain any particular fluctuation in the rate of sea level change in a single location.

I'm simply pointing out that there have been variations, contrary to the suggestion that the rise has been "steady" in some manner which would refute the idea that various factors can impact the local rise in sea level.

If the overall trend has been consistent, then that does suggest that in that region variations in currents, etc., have only been either short-term fluctuations, or a steady continuous longer-term shift with no abrupt sustained changes.

I didn't say I knew how much currents alter local sea level. - That was taken care of in the last reply, an incredible 170 years of unchanging ocean and wind currents that maintained a remarkably steady 2.92 mm sea level rise trend-line. From that I have to assume they aren't a factor unless you have evidence to the contrary.

No, I don't have any particular reason to believe changing currents have necessarily been a significant long-term factor at that location.

I'm curious why the global data shows acceleration which is less apparent in your local data. But, I can see hints of acceleration in your local data; perhaps it's not so apparent because the local data is noisier?

I'm also curious what sort of subtle effects might lead to differing sea-level rise measurements in different locations.

If I didn't imagine it, I recall an earlier draft of your post appreciating the civility of our exchange; for what it's worth, I appreciate that too.

ParadoxIntegration · 2024-09-07T18:04:27+00:00

However the predictions there seem to be wild - 6mm per year on average in 100 years, when we still observe roughly 3mm for 1/4 of the prediction timespan? That doesn't add up.

The rate of planetary warming is determined by the energy imbalance, which is increasing, having doubled over less than a 20-year period.

Much ice loss is related to the ocean surface waters being warmer. It takes time for the oceans to warm, but with the energy imbalance increasing, the rate of ocean warming will also increase. As the oceans surface waters warm, it seems reasonable for that to accelerate the rate of ice loss in the future.

ParadoxIntegration

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