Over at the religious page scienceofdoom.com you can find this post where the priest is preaching about how he misunderstands science so badly that he disproves his own theory. I am going to answer his post here, because I like freedom of speech, and he doesn´t, which he admits in the post.
He starts with blabbing about the radiative transfer formula for radiation emitted at the surface and how it passes through the atmosphere:
Iλ(0) = Iλ(τm)e-τm + ∫ Bλ(T)e-τ dτ
He explains it like this:
”The intensity at the top of atmosphere equals.. The surface radiation attenuated by the transmittance of the atmosphere, plus.. The sum of all the contributions of atmospheric radiation – each contribution attenuated by the transmittance from that location to the top of atmosphere.”
This is all fine. But the problem is he thinks he has calculated a greenhouse effect, even though he himself writes ”attenuation”. He must not know what that means.
He then shows this spectrum and points out how good the match is:
And he is correct. It is a beautiful calculation with accuracy that almost gives you goosebumps. Now comes the funny part, he thinks it is a spectrum that proves his theory!!!
Look at the left scale, it says: Intensity+units. This is intensity of heat, and the higher up the graph goes, the higher the intensity(hotter)
Now look at the action of co2 in wavelengths 20-13. What happens to the intensity?
A deep drop in intensity, that is what happens. This means that the action of co2 in the atmosphere, is to decrease intensity/heat emission. That is not a greenhouse effect, that is the effect you get when you spray dry ice(co2) in air. It cools.
But the greenhouse-believer think that heat can hide, so they claim that the invisible ghost-heat stays in the system and accumulates.
Non-religious people know from logic that dry ice doesn´t make hot surfaces hotter, just colder.
Then we get this nice little picture, adding to the display of ignorance.
The picture shows, and the text describes, how heat transfers to surroundings of a body. What we know is that the sb-law tells us that at equal temperature there is no transfer of heat between surroundings and solid. When discussing gh-theory, the defenders like to add ”net”, which is pathetic, because the sb-law only calculates heat transfer. Nothing changes from ”net”, because ”net” is heat. The rest is not heat and cannot heat anything.
Do you notice anything missing in the picture? Yeah, a sun. Another proof of ignorance delivered by the priest himself.
Again, we know that at equal temperature there is NO transfer of heat. 0.
In the gh-theory there is the idea that if one body is at lower temperature, it starts to transfer the negative difference, which is not ”net” and therefore not heat.
Anyone with knowledge about heat transfer and thermodynamics knows that only work and heat can increase temperature. Co2 is not producing heat, even greenhouse believers know this. Co2 absorbs heat. The question is then, does co2 do work on the system? Of course not.
Knowing this, can co2 raise the temperature of it´s own heat source? Absolutely not!
Kids understand this, but not blanket-people.
I somewhat miss the opportunity on this blog to directly find the latest comments – or maybe I’m just not seeing it. But I feel sometimes comments might simply drwon in the depths of the thread, for it has become quite long. So I’m starting a new head comment from time to time.
This one regards to some conceptions the commenter Bryan has come up with:
”1. If the container is closed then convection is suppressed.
Convection is a very powerful means of heat transfer.
2. If Argon a non IR active gas shows the same or greater temperature rise than CO2 then the radiative effect is non existent or marginal
3. If greenhouse theory predicts a rise of no more than 0.3K for that size of enclosure then your ‘experiment’ is bogus because it contradicts its own theory.”
To 1.: ” If the container is closed then convection is suppressed.”
We should highlight, that convection is supressed EQUALLY in all containers of experiments I have shown. So, yes, of course this will cut of a big contribution to heat transfer that occurs normally within our atmosphere. But this is identical for all systems – if they show differences nonetheless, those differences must be determined by something they have not in common. Supressed convection they DO HAVE in common.
Not in common they have:
– the density of contained gases
– their total mass ( those with denser gases have more mass )
– the ability of their contents to absorb radiation at certain wavelengths
I can easily agree that those experiments suffer from one main problem: The sorrounding of the closed systems is not vacuum, and therefore mainly convection transferres heat away from the boundaries. One must assume the convective transport to be the same at the boundaries of each system, otherwise the experiment fails.
Much better would be to have closed systems made of some long wave transparent material filled with some identical black bodies but different gases and be placed in vacuum. Not that easy to do in standard labs and with restricted economical power.
Anyway, we have to assure everybody knows that suppressing of convection in some locality only, does not change the sum of energy within the system represented by earth. It is just hindering energy distribution – what accumulates in one place ( let it be a real greenhouse ), is missing in some other place. With it’s environment, which is space, earth cannot interchange energy convectively. It can only do so by radiation.
And that leads by pure logic to some conclusion: It does not matter, if some experiment that should prove ( or disprove ) the temperature dependence of systems from their containment of long wave active gases supresses convection or does not so. The conditions for compared systems must be the same for each system, despite their variance in gases.
Since the GHE claims to be existent due to a massive difference in rate of energy transport ( convection much slower than radiation ), a purely radiative accessable environment would be the most important thing if someone wants to represent something near to real GHE.
Anyway, the differences needs to be explained. Anybody?
To 2.: ”2. If Argon a non IR active gas shows the same or greater temperature rise than CO2 then the radiative effect is non existent or marginal”
Since the system was designed open, the entire experiment fails from first principles. Before equilibrium temperature could be detected, the measurement crushes due to convective losses OUT OF THE SYSTEM. Redo the experiment under identical closed conditions.
To 3.: ”If greenhouse theory predicts a rise of no more than 0.3K for that size of enclosure then your ‘experiment’ is bogus because it contradicts its own theory.”
Greenhouse theory works with the lapse rate, which is not present here. How do you come to 0,3K? Anyway, what DOES cause the difference?
GillaGilla
I just came now to read all the new comments.
It has shown to be without sense to give you sources, for you even question the basic equations of physics to be something different than they are.
– You mix up mass with weight.
– You mix up accelaration with force.
– You mix up Einsteins E with thermodynamic E.
– You demand every unit could be transferred to every other unit.
– You claim a general mass dependence of temperature.
– You claim a general time independence of averaging of thermodynamic states.
Plus many things I forgot here. Each and every of this is wrong. You want references? Newton, Einstein, the SI system of units, the equations themselves, every singel textbook of physics – you can choose which of those references is strong enough.
Maybe try something on yourself: Wrap your PC CPU in some mass of room temperature wool. According to you it will not heat up, for an additional heat absorber was placed.
GillaGilla
”You want references? ”
Yes, I want links and citations. Your words are worth nothing.
”Wrap your PC CPU in some mass of room temperature wool. According to you it will not heat up, for an additional heat absorber was placed.”
You prove my point by not knowing how insulation works.
https://en.wikipedia.org/wiki/Thermal_insulation
”Heat flow is an inevitable consequence of contact between objects of different temperature. Thermal insulation provides a region of insulation in which thermal conduction is reduced or thermal radiation is reflected rather than absorbed by the lower-temperature body.”
You see, insulation prevents absorption of heat in the surroundings of a body. It is the opposite of the gh-effect, which INCREASES absorption in the surroundings. I am very surprised that practically all alarmist blanket people don´t know this basic fact.
To retain heat, you need to decrease absorption of heat in the surroundings. Not add potent heat absorbers like dry ice.
Dry ice is used in fire extuingishers partly because it cools very efficiently by massive heat absorption. Why would it have an opposite effect in an atmosphere?
GillaGillad av 1 person
”You want references?”
Yeah, we gave you references. Wikipedia, and I quoted directly out of the physics textbook sitting two feet to my right.
Man, you don’t even got your high school physics right.
GillaGilla
I gave you a very clear reference, explaining insulation in a way that is exactly contradicting you claims of the atmosphere`s relation to the solid planet. My reference is a very strong fact which proves that an atmosphere at low temperatre has an opposite effect t o the principles that is involved in thermal insulation.
What is your response to that? You claimed insulative effects of cold air, with an analogy of insulation and a CPU. The proven and applied physics, says that you are wrong.
Absorption of heat in the surroundings of a warm body, is exactly what insulation aims to eliminate. Because it cools. Reducing absorption=retaining heat. Increasing absorption= increased cooling.
GillaGilla
”You see, insulation prevents absorption of heat in the surroundings of a body.”
”Absorption of heat in the surroundings of a warm body, is exactly what insulation aims to eliminate.”
That is all perfectly correct.
”Reducing absorption=retaining heat. Increasing absorption= increased cooling.”
Not wrong, but imcomplete.
Cooling is not only caused by absorption of heat, it can also be consequence of the radiation of energy to the nothingness of space. Thus reducing that radiation also equals retaining heat.
”It is the opposite of the gh-effect, which INCREASES absorption in the surroundings.”
That is wrong and there is reference for that in your own text.
The sorrounding to which the system earth-atmosphere cools is space. To space radiation is REDUCED at certain wavelengths, as the figure YOU show in your article demonstrates nicely. That is insulation of earth against space, according to the definition you have given yourself.
You have not understood, that radiation to space is directly from a planet’s surface, if no long-wave active gases at all are in the atmosphere. With such gases radiation occurs from the atmosphere rather than the surface. Simple logic is enough to get meaning from this fact.
GillaGilla
”Cooling is not only caused by absorption of heat, it can also be consequence of the radiation of energy to the nothingness of space. Thus reducing that radiation also equals retaining heat.”
Correct. The problem is, the atmosphere doesn´t reduce the rate of energy flowing from the surface, it adds to it.
In any point in time, the surface has to sustain emission from both the atmosphere by transfer of heat, and emission according to its temperature.
Without an atmosphere, the surface would only need energy to emit according to its temperature. The solid earth sphere, supplies energy for emission from both itself and the atmosphere, simultaneously. It has to power both bodies to have constant emission. The atmosphere is not a barrier, or obstacle, for emission from the surface. The emission doesn´t depend on the external atmosphere, it only depends on the internal state.
You cannot constrain emission of heat from a cold body by adding cold fluids which absorbs heat.
Adding cold air, is equal to adding a second heat sink, which adds further absorption to the absorbing infinite heat sink of space.
”The sorrounding to which the system earth-atmosphere cools is space.”
Of course. Now I want you to argue that the atmosphere doesn´t surround the solid earth.
If it surrounds earth, it is part of the surroundings.
”To space radiation is REDUCED at certain wavelengths”
Yes, but not from the surface. The surface shows no reduction of heat emission from co2. The atmosphere, on the other hand, shows reduction of emitted heat. Reduction of heat emission=lower temperature.
The effect of the atmosphere is lower temperature.
The surface emission depends only on the internal state.
Show me a reference, with citations, which proves your claim that cold fluids cause increasing temperature by reduction of heat emission in the surroundings.
”You have not understood, that radiation to space is directly from a planet’s surface, if no long-wave active gases at all are in the atmosphere. With such gases radiation occurs from the atmosphere rather than the surface.”
Oh, yes! I understand that perfectly.
Without an atmosphere, there is only surface emission from two dimensions. With an atmosphere, the solid surface muste provide the energy for both surface emission and emission from the atmosphere, simultaneously.
The surface emits, independently of the atmosphere, according to the internal state. Simultaneously, it also transfers to the atmosphere, according to its emission and the relation to differences in temperature of the surrounding air.
The atmosphere emits 244W/m^2. Surface emission is 383W/m^2. According to the inverse square law, the source needs to transfer 4*244=976W/m^2.
383+976=1361W/m^2. Exactly balanced to the heat source, the sun.
Will you counter this by showing how the atmosphere transfers 333W/m^2 to the surface at a temperature of 255K, in line with the energy budget by Trenberth?
Give one single example of another observed physical process, where the addition of a low temperature heat absorber, causes an increasing power density of the heat flow which heats the absorber. If your claims are valid, this principle must be universal and confirmed by observations.
GillaGillad av 1 person
In your example you state that the surface would emit at 333 if at 255K. However, instead it emits at 383, which equates to a surface temperature of 286K. What is the reasoning behind the surface temperature being warmer in your mind?
”Give one single example of another observed physical process, where the addition of a low temperature heat absorber, causes an increasing power density of the heat flow which heats the absorber. If your claims are valid, this principle must be universal and confirmed by observations.”
Your example of a 255K surface temperature now being at 286K seems like a great example of the surface temperature warming with the addition of an atmosphere to it.
GillaGilla
”Give one single example of another observed physical process, where the addition of a low temperature heat absorber, causes an increasing power density of the heat flow which heats the absorber. If your claims are valid, this principle must be universal and confirmed by observations.”
Maybe it would be wise to define precisely what the universal principle is, that I claim to exist. It is this:
A system heated by any constant power supply will over time go into steady state, where input of energy into the system equals output of energy from the system to the environment. From this point of time on temperature is unchanging.
( Let the time being needed to reach such a steady state be ”t” and the temperature in this state be ”T”. )
If the system is modified in such a way that the rate with which energy is transferred through the system to the boundary surface with the environment changes, ”t” as well as ”T” will have different values compared to the former case.
Would you agree so far?
Example for that is the famous pot on a plate.
Let three identical pots be heated on three identical plates with identical power each. Let the system be the pot + it’s content.
Do we both agree that this setting will lead to different steady state temperatures?
Now let the contents of the pots be 1) air, 2) water, 3) glycerol. Let the plate have enough power to bring each liquid to boiling point. In which of the cases will the bottom of the pot be warmest in steady state? And where is the mass of the system the greatest?
I claim bottom pot plate temperatures to be water<glycerol<air. The masses are definitely air<water<glycerol. The time "t" to come to steady state I claim to be air<water<glycerol as well.
It's an easy to perform everyday experiment if you have access to a lab. I have.
What is your statement to this?
GillaGilla
LIT,
”The atmosphere emits 244W/m^2. Surface emission is 383W/m^2. According to the inverse square law, the source needs to transfer 4*244=976W/m^2.
383+976=1361W/m^2. Exactly balanced to the heat source, the sun.”
You are again mixing laws and principals incorrectly. Does inverse square law apply to radiative energy transfer? Yes. Can it be used for transfer through the atmosphere? No. Let me explain this first before I point out the clearing error in your math.
Inverse square, when used with radiative transfer, is only applicable if the radiation is traveling through vacuum or a medium that it is transparent to. Since the atmosphere is not transparent to longwave radiation, inverse square doesn’t apply. We use inverse square to calculate the reduction in density when radiation can travel uninhibited. This isn’t the case for the atmosphere.
The error in math comes from the ” 4*244″. Working the math backwards means that 244 = x/4, with x being the originating source of power. Nothing wrong with this and the inverse square law if the source of power is 2 meters away. So if you think that the surface, which is the source of radiation for the atmosphere, is only 2 meters from the top of where this emission of 244 happens ten we have a whole lot of issues.
GillaGilla
ChrisA
Your recently linked video’s over at SOD are examples of stupid or careless experiments.
This linked PDF was written by scientists who believe in the Greenhouse Effect .
However they point out that the experiments ( linked in video) are bogus and they worry that such mistakes will weaken the case for belief in the greenhouse effect.
The experimental presentations by Greenhouse Effect advocates are often full of such mistakes
Klicka för att komma åt Wagoner%20AJP%202010.pdf
GillaGilla
Bryan,
It’s good to see you supporting that the greenhouse effect isn’t bogus.
Replication in a class room would be very difficult given the scale of the atmosphere. Shrinking it down to size means that the values we would be trying to measure would be very minute. Thankfully we have readings from the actual planet and atmosphere. We have satellites orbiting the planet measuring incoming and outgoing LW and SW radiation independently at the top of atmosphere (ToA) as well as equipment on the ground measuring incoming and outgoing LW and SW radiation. So we have a planet sized lab experiment going on. And what has been shown is that the equations used to model what the ToA radiation would look like given the makeup of our atmosphere matches what we measure, very closely. What has also been shown is that these same equations that predict ToA flux work very well to describe the back radiation that again, matches to surface measurements.
I haven’t watched the videos that ChrisA linked so I’m not going to speak specifically to the validity or content of them. But dismissal of experimentation can’t be considered a valuable retort to GHE given that we have been measuring incoming and outgoing radiation, by wavelength, at the surface and ToA for years and have observed exactly what the models for a GHE caused by radiative feedback would predict.
GillaGilla
Bryan ( and Brad a bit ),
the experiments showed how identical containers which are heated identically by some lamp ( or the sun ) come to different temperatures according to their containment. CO2 led to higher temperatures than normal air.
This experiments are claimed to demonstrate the real GHE by the experimenters.
However, I have not shown them for the purpose of GHE verification ( for even I think they are not that well suitable for this ).
I have shown them to counter a general statement of lifeisthermal, who claimed the addition of CO2 ( as well as the addition of mass ), will act as an additional heat sink and thus leads to more cooling. Since CO2 is denser than air, even more mass is in CO2 rich containers compared to those filled with air. This was stated as general and not restricted to radiation cases only.
This experiments might say nothing about GHE, but this general concept given by LIT they prove wrong.
I myself have therefore given an example of three pots on identical heating plates with different contents, for such a setup will not be mixed up with GHE.
GillaGilla
Bryan,
I just spent a few minutes to watch both of these videos. I honestly expected one of the videos to be a demonstration similar to the experiment in the pdf that you linked. Neither of them was similar to that experiment. So I’m curious, what are the problems that you see with the experiments Chris shared? What are the reasons for the higher temperature in the co2 environments vs the control environments?
GillaGilla
Brad,
according to the paper suppression of convection contributes very much more to reduction of cooling than absorption of thermal radiation.
The paper Bryan linked is not even able to show the steady state condition, which alone is of interest for the discussion here. Fig.2 breaks down due to losses of CO2/Ar before the steady state everybody needs to know is reached. Reason: The system was not sealed. It is so stupid.
I showed sealed setups, because if you want to show a difference which is not caused by supression of convection, the convective conditions must be the same for each system.
Anyway, lifeisthermal demanded this: ”Give one single example of another observed physical process, where the addition of a low temperature heat absorber, causes an increasing power density of the heat flow which heats the absorber. If your claims are valid, this principle must be universal and confirmed by observations.”
Well, ALL systems showed ( my videos as well as the pdf delivered by Bryan ) this, what should not be. They also prove the misuse of E = mc^2 to be wrong, for CO2 is denser than air and thus mass is increased at identical volumes.
According to lifeisthermal this must lead to decline of E. We observe in fact rising temperature.
GillaGilla
There has been so much going on in this thread. Maybe a summary of the basic misconceptions might help to find things concentrated at one place.
1.) Stefan-Boltzmann Law is about heat transfer.
Not true. Heat transfer in vacuum can be derived from this law only if bodies with known temperatures radiate to each other. In this case heat transfer is defined by the part of energy that the warmer body emits more than the colder body emits.
2.) Systems with energy input from a source have a matching output of the same quantity at any point in time.
Not true. If that was the case, neither cooling ( excess output ), nor warming ( excess input ) of systems fed by a source were possible. A pot of water getting to boil on a heat plate contradicts this.
3.) Gravity does work.
Not true. Gravity alone is simply a potential with the units of acceleration. There is no text in physic’s literature stating that gravity would do any work.
4.) Gravity gives birth to thermal resistance.
Completely new. No known physician has ever written about this. It is not clear at all what that should be. At least the units for thermal resistivity and gravity are impossible to match.
5.) If mass is added to a system that receives constant input by an energy source, internal energy will be distributed over more mass and thus system temperature fall. This can easily be seen by E = mc^2. If energy E is held constant and mass rises, c^2 must decrease.
Not true. E is no temperature related energy at all. c is one of nature’s constants and therefore unchangeable.
Furthermore the internal energy is not constant, but developes time dependent into a steady state. Here the misconception of 2.) also take it’s toll.
6.) Input from sun is equilibrated with the core of earth.
Not true. The distance from surface to core combined with the limited heat conductivity through earth make this impossible. Furthermore the energy of sun arriving at earth has a flux density far below the one at the core of earth.
If someone asks for references: Nearly every textbook about basic physics. Or nearly every webpage about basic physics. Or the papers of the scientists who developed those laws. Or the very formulas of the laws, for they tell explicitly due to their units what they are, and what they are not.
This list can be discussed, corrected or something added. Brave people like Ben and Brad have tried to point out how the neglecting of GHE has led to a neglecting of entire modern physics.
Maybe silent byreaders will find some corrections here in concentrated form, for it is otherwise dissociated over the entire thread.
GillaGilla
”Heat transfer in vacuum can be derived from this law only if bodies with known temperatures radiate to each other. In this case heat transfer is defined by the part of energy that the warmer body emits more than the colder body emits.”
And the earth is in vacuum, so it obeys the s-b law for radiative transfer of heat, to- and from- the system. I use it to find effective emission with TSI and surface temperature, otherwise I only use it for emissive power in a point where temperature is measured. Or, more precisely, where we set an average temperature from measurements, like the surface mean temperature.
For the large part, I use it to find the heat flow in a point, and then I stack the mean emission from different parts to find a balance to TSI. Because the source must be balanced to the whole system in any given point in time. The surface, plus the atmosphere, is fed by solar heat. TSI must supply a total amount of heat that support constant emission from both the surface and the atmosphere at a given point in time. Of course there is fluctuations, like seasons, in heating from the sun. But here we are dealing with averages which includes all variations, so the system is treated as instantaneous.
”Not true. If that was the case, neither cooling ( excess output ), nor warming ( excess input ) of systems fed by a source were possible. A pot of water getting to boil on a heat plate contradicts this.”
As I wrote, when dealing with averages, which the greenhouse effect does, the system should be treated as instantaneous. In the case of boiling water, with constant input, you deal with the system by averaging the input over time into a steady state of heat flow through the water, which will behave as instantaneous.
”Not true. Gravity alone is simply a potential with the units of acceleration. There is no text in physic’s literature stating that gravity would do any work.”
The definition of work is displacement of mass. Gravity displaces mass with a constant acceleration of 9.8m/s, or, constantly pulls on static mass with 9.8Nm per second with an equal opposing force keeping it in place.
If a rock disconnects from a mountain from erosion, and tumbles down some distance to end up stopped by the ground making a dent in the soil, is there not any work done? Where does the energy come from, if not displacement of mass?
”Completely new. No known physician has ever written about this.”
I know. It´s awesome that they left it behind on the road, for me to find. You know, new discoveries are like that, nobody has written or thought about them.
” It is not clear at all what that should be. At least the units for thermal resistivity and gravity are impossible to match.”
Not at all. We know that light is affected by gravity from gravitational lensing. There is no difference between light and heat, they are just different parts of the spectrum and have different intensity, where visible(light) and not visible(the rest) a re relative to the human eye. If gravity is proven to affect light, we can safely assume that it affects heat. The units for thermal resistance, stress and pressure is Nm^2, and squaring g results in that unit, we are definately dealing with thermal/light resistance.
”Not true. E is no temperature related energy at all. c is one of nature’s constants and therefore unchangeable.”
E is ”energy”. Temperature is, by taking it to the fourth power, a measure of power. Power is a definition of intensity of energy flow, so power is energy. And also, the definition of temperature, is the ”average kinetic ENERGY of a volume of mass”. So temperature is most definately energy related. Remember that Einstein based his investigations on thermodynamics, so he most likely refers to the concept heat/light/temperature when he writes about E.
”Furthermore the internal energy is not constant, but developes time dependent into a steady state.”
Nothing depends on time. Time is what we use to define the evolution of action. In the end, the steady state eliminates any influence of time. The final state of any system, is determined by the energy in heat and forces acting on the system. When considering a steady state, time is irrelevant. Because any change inside the system, is limited in its effects by the constant heat flow and the constant forces. Co2 adds neither constant heat or constant force.
”The distance from surface to core combined with the limited heat conductivity through earth make this impossible. Furthermore the energy of sun arriving at earth has a flux density far below the one at the core of earth.”
What? Distance makes interaction between bodies impossible?
The difference in energy of the heat flow from the sun and the core, is exactly my point. They equilibrate at the surface, as two sources of emission in balance. The very small addition of heat from internal generation is the proof. Earth adds only ~90mW/m^2 to the heat flow, that is very, very close to equilibrium. So, at the surface, according to the s-b law for transfer, there is no ”net” transfer from TSI. The solid earth emits equally to what the sun provides, +90mW/m^2. Whatever processes that generate heat inside earth, they are equal to what the sun provides, at the surface.
But, in the atmosphere, we have additional mass emitting heat at 244W/m^2, and the atmosphere has no internal generation of heat. So, at the same point in time, where internal generation produces surface emission at 383W/m^2, there is also heat transferred at the rate
1360.8-383=4*244W/m^2, that supplies the heat emitted by the atmosphere.
” neglecting of GHE has led to a neglecting of entire modern physics.”
Impossible, since there is no reference anywhere to experimental data showing how the addition of heat absorbers to a constant heat flow, can increase the density of power of the heat flow. I challenge you, to find a reference where the addition of co2 to the surroundings of a heat source causes increased power of the heat source.
GillaGilla
Quick response in regards to your gravity and rock example. Gravity will have accelerated the rock at 9.8 m/s^2, that isn’t the same as 9.8 Nm of work. To find work we need the force first, which is acceleration x mass. And then work is force x displacement. So work done on the rock would be 9.8m/s^2 x mass of rock displacement of rock.
Seriously, pick up a textbook.
GillaGilla
If a rock is accelerated by a constant force of 9.8Nm, that is equal to addition of energy to that rock, of 9.8Nm per second. This means that a force is acting on the rock at a constant level of 9.8Nm each second. If the rock has a bottom surface area of 1m^2 and weighs 1kg, the force acting on that rock is 95.7Nm^2 if using the surface acceleration given by Nasa. Slightly more, if using the mean value of gravity at ~9.8Nm/s.
9.8Nm of force is acting on mass above the earth surface. The effect on the fluid air, will be 96N/m^2 of acceleration towards the surface, which needs a minimum source power, according to the inverse square law, at 4*96W/m^2, equal to 4*g^2Nm^2.
Exactly, which part of this is hard for you to understand?
”So work done on the rock would be 9.8m/s^2 x mass of rock displacement of rock.”
The work done in a point at the center of mass, is 9.8/Nm each second. The work done on a square meter of mass is 96Nm^2 each second.
The squared unit of time, s^2, is only there to tell you that the acceleration caused by the work done in the last second, is added to the acceleration caused in the next second. Which means, time adds nothing, because adding time dilutes the energy density. But adding energy or force, concentrates the power per unit time.
Time is a measure of inflation where a constant flow of energy(for example TSI) dilates in space. The rate of expansion is dependent on the intensity of power, and the level of expansion is dependent on the intensity of power and the amount of time passed.
Add an amount of time to a density of power, and you get a rate of energy flow. Increase the amount of time, and you get dilution of energy.
Adding co2 to a system does not add power to the heat source, and it is not a force acting on the system. It adds mass, or possibly volume, to the flow of energy. Neither an increase of mass or volume can increase the power from a source. Only an increasing density of energy or addition of a force, can increase the power of a heat source.
GillaGilla
You need to read. Gravity is an acceleration. At the surface that is 9.8 m/s^2.
We can’t begin to have a conversation about force until you have a mass to accelerate. Once we have a mass with known kg then we can talk about how many N of force there are because force= mass x acceleration (N= kg x m/s^2, see how the units turn into something meaningful). Until then, we can only talk about the acceleration of gravity being 9.8 m/s^2. Elementary physics.
GillaGilla
You need to stop it now. Are you drunk?
The FORCE is constant. It is 9.8m/s, or 9.8Nm. It is independent of mass, because the acceleration is 9.8m/s no matter what the mass is.
Gravitational CONSTANT. Big G: 6.67*10^-11. g=9.8
The word CONSTANT should be a clue. It means that the force is constant.
GillaGilla
Lifeisthermal,
2 objects with mass m1 and m2 are separated by a radius of r when measured. Please let me know with specific equations what is the force of gravity between them, the acceleration of gravity on either object, you pick which one, and the equation for the total gravitational potential of the system. In normal physics, these are 3 equations that look rather similar. I’m curious what your equations are for it though.
Force =?
Acceleration (m1 or m2) = ?
Total gravitational potential=?
GillaGilla
Brad and ChrisA
ChrisA said
”Bryan,
It’s good to see you supporting that the greenhouse effect isn’t bogus.”
No I did not
I just pointed out this this embarrassment of an experiment is all too common with GHE advocates.
Shallow science and ‘ failed experiments’ give the game away.
So IR gases send some radiation back to the source and it is absorbed – end of theory- that as far as I know is the science behind GHE
This then leads to a 33K increase in surface temperatures.
Some people are easily fooled .
Compare a real experimenter like R W Wood (see earlier link) and how a simple experiment falsified the greenhouse theory
GillaGilla
Bryan,
So why the differences in temperature in the videos Chris shared?
Also, is this the Woods experiment you are referring to?
http://clim.stanford.edu/WoodExpt/
GillaGilla
Bryan,
Here is another reference to the Woods experiment from someone a bit more prominent on the denier side of agw.
http://www.drroyspencer.com/2013/08/revisiting-woods-1909-greenhouse-box-experiment-part-i/
GillaGilla
Brad
For the slow learner I suppose I should break down what is wrong with your experiment and any of these 3 points sinks its validity
1. If the container is closed then convection is suppressed.
Convection is a very powerful means of heat transfer.
2. If Argon a non IR active gas shows the same or greater temperature rise than CO2 then the radiative effect is non existent or marginal
3. If greenhouse theory predicts a rise of no more than 0.3K for that size of enclosure then your ‘experiment’ is bogus because it contradicts its own theory.
Roy Spencers experiment and W Pratts attempts should also be subject to the 3 tests above.
GillaGilla
Brad and ChrisA
I had a look at the Roy Spencer link
Here is a comment I made at the time
http://www.drroyspencer.com/2013/08/revisiting-woods-1909-greenhouse-box-experiment-part-ii-first-results/#comment-86921
The results of a multi year experiment to find out the best material for real agricultural greenhouses.
Does an IR active covering give warmer internal temperatures than a non IR active material.
What is really worrying is that that link which worked perfectly for 3 or 4 years now has been suppressed as you will find out.
Again it is typical that experiments and papers that contradict the GH effect are removed.
GillaGilla
Bryan,
The links I provide speak very clearly to the issues with the Woods experiment. You act as if he carried out some great experiment that disapproves GHE but it wasn’t. Both boxes were covered with glass. How can one expect to get to different temperatures if they are both covered with glass?
Second, the point of the paper you linked was that the cause of the heat increase in these open container experiments want radiative feedback. Rather it was a dense layer of co2 crafting a natural layer that delayed convection. This was demonstrated as the cause when argon was used as well. The experiments that Chris shared both made sure convection wasn’t a variable by using sealed containers.
I still fail to see how the post you shared invalidates that experiments Chris shared.
I do understand the point of the paper and agree that we should all be careful in jumping to conclusions about causality of an experiment.
In regards to the link in your comment on Spencer’s blog, would be interesting to see the information. I’m curious what their measured difference in incoming radiation was between an IR active covering vs non since three ir active is going to reduce incoming radiation.
GillaGilla
It seems to come down to one single point of confusion:
1) Thermal energy never transfers (net) from a colder object to a warmer object. Or, practically never – Maxwell’s demons still apply here.
2) Adding insulation to a hotter body will cause it to lose heat more slowly. If the body was at steady-state equilibrium with its surroundings; if thermal energy was generated at the same rate as it was lost; then the rate of loss will decrease, the thermal energy in the body will increase, and the temperature of the body will rise. This is what happens when more thermal energy is generated than lost; the temperature rises.
The 1LOT is about #1. The greenhouse effect is about #2.
No one is saying that heat will transfer (net) from a colder body to a warmer one. We’re saying that adding radiative insulation will decrease the loss of heat from the warmer body, so the temperature will rise.
GillaGilla
”Adding insulation to a hotter body will cause it to lose heat more slowly”
No. Adding insulation reduces absorption of heat in the surroundings.
Absorption of heat from a hot body, in the surroundings, makes it cooler.
Take a coin, heat it to a glow(at least 798K), drop it in a bucket of water.
What happens?
You added a heat absorber to the hot coin. Did it get hotter?
If not, why do you expect a planet to behave the opposite way?
Arrhenius was a complete retard when it comes to heat. Maybe he can be excused from lack of data of atmospheric temperature.
GillaGilla
(me) ”Adding insulation to a hotter body will cause it to lose heat more slowly”
(lifeisthermal) No. Adding insulation reduces absorption of heat in the surroundings. Absorption of heat from a hot body, in the surroundings, makes it cooler.”
Yes, adding insulation reduces the transferrance of thermal energy from the hotter body to the colder one. Which means the colder one absorbs less, sure, but it also means that the hotter body loses heat more slowly. These are equivalent. Insulation slows the rate of thermal energy transfer.
”Arrhenius was a complete retard when it comes to heat.”
His work has stood over a century of challenge from physicists.
His work is standard, accepted research by 99% of the world’s physical scientists. Thermodynamics, materials science, statistical mechanics, engineers… we all use his work. I’ve got 3 modern textbooks on my shelf that cite his work or the work derived from his work.
You should write up your rebuttal of his work. I’m sure your Nobel Prize awaits. =D
I dunno man, at some point, when you’re arguing with century-old science that’s underlies a huge range of modern technologies… maybe you kinda have to ask yourself if maybe you’re the one misunderstanding it, rather than everyone else.
GillaGilla
Lifeisthermal,
2 objects with mass m1 and m2 are separated by a radius of r when measured. Please let me know with specific equations what is the force of gravity between them, the acceleration of gravity on either object, you pick which one, and the equation for the total gravitational potential of the system. In normal physics, these are 3 equations that look rather similar. I’m curious what your equations are for it though.
Force =?
Acceleration (m1 or m2) = ?
Total gravitational potential=?
GillaGilla
I would really like to have clarified what the author lifeisthermal thinks about this:
What happens, if a body and a heat source are together enclosed by some reflective material?
Will the body that is heated by the source be heated to a higher temperature than without the enclosure, remain the same, or cool down?
The enclosure is of course colder than either the heat source, or the body.
Let’s even get more into detail: Let the body be an living, sleeping mouse, the heat source a warmed gel pack and the enclosure a reflective foil at only room temperature.
Will the body core temperature of the mouse be higher with gel pack + foil than with a gel pack only, or be the same or lower?
Thermodynamics should answer this I hope?
Since colder objects can never rise the temperature of warmer ones, we should see no change, right? But correct me if I’m wrong.
GillaGilla
Why do you want to know what happens when a heat source is enclosed with reflective material? How does that relate to adding a heat absorber to a heat flow? Co2 is not reflective, it doesn´t act equal to reflection, it is not in an enclosure.
Why do you include a sleeping mouse? Why do you include gel and foil?
Not a single thing of what you wrote, is relevant to how a cold fluid acts when it´s heated by a hotter heat source.
GillaGilla
Dear lifeisthermal,
my comment was about a very basic question about first principles of thermodynamics, namely this:
Will something that reduces energy output of a system containing a constant heat source cause warming of the system beyond the temperature that would be there without such a ”reducer”? Yes/No? ( The reducer is colder than the rest of the system ).
That is very general and can be abstracted to three body systems like sun-earth-atmosphere, or gel – mouse – foil. Or immersion heater – tea – cup. Or metabolism – human body – space suit. Or so many else.
In all that cases the mechanisms differ, but the underlying principle remains unchanged: Energy input is held constant into a system, while energy output is varied.
Now many commenters claim ( and so do I ),that this will lead to higher internal energy if output is reduced due to insulation, and thus higher temperatures. You claimed otherwise.
Correct or not?
GillaGilla
”Will something that reduces energy output of a system containing a constant heat source cause warming of the system beyond the temperature that would be there without such a ”reducer”? Yes/No? ( The reducer is colder than the rest of the system ).”
If the heat flow is reduced from a system, it means that the system just got colder. Increased heat flow, means hotter.
I think you are mistaking a radiant barrier for reduction of heat flow by addition of cold fluid absorption. A barrier reduces the transfer to conduction, that is a whole other story. When a cold fluid, which cools a heat source, gets colder, it doesn´t make the heat source hotter.
”Energy input is held constant into a system, while energy output is varied.”
The energy input into a pot on the boiler plate is constant. Pour water into it and it gets colder from absorption of heat. The output is reduced and nothing gets hotter, only colder. How can you not understand this?
GillaGilla
Correction to above comment:
Not ”energy output” is varied of course.
Energy transfer speed to the outer limit of the observed system is varied by different insulators, while input equals output in steady state.
That way it’s correct.
GillaGilla
And the insulators have a thermal resistance. Thermal resistance is measured in Nm^2 and gravity provides a thermal resistance of g^2. The insulators(bad word for heat absorbers) are there because of the heat flow and resistance of gravity. An atmosphere is excitement of the surface, it is carried by the heat flow. It is caused by heat, not a cause of heat.
GillaGilla
”Thermal resistance is measured in Nm^2”
Wat. No. Thermal resistivity is measured in inverse thermal conductance; so, (m*K/W). This is the measure of how difficult it for thermal energy to pass through a material. Note that the units are completely different from yours.
…just checking, but when was the last time you sat down with a modern college physics textbook and worked through some problems? Your approach to even classical problems is a bit, uhh, unconventional.
GillaGilla
I will return with the reference for thermal resistance. But, if you are not completely retarded, you can understand that a unit for pressure, will have an effect of thermal resistance. More mass per volume, means more resistance for the heat flow.
Even without the reference for thermal resistance, using gravity as pressure with units Nm^2 is enough to realize the relationship to emissive power.
Pressure:
https://sv.wikipedia.org/wiki/Pascal_(enhet)
GillaGilla
”If the heat flow is reduced from a system, it means that the system just got colder. Increased heat flow, means hotter.”
Did you mix up the terms ”from” and ”to” in the statement above? I hope so. Heat flow ”from” a system means that heat leaves that system. You stated above an increase of that flow away means hotter. Of course the opposite is true: The faster heat leaves a system, the more the system cools. Less heat flows mean more energy staying as internal energy in a system, thus less cooling.
That is the principle of insulators: They do absorb heat as every other matter, but with a different ( slower ) rate.
If a constant heat supply is transfered with different rates, the 1st law only holds if slower rates go together with higher temperatures at the source for the flux has to be constant. Faster rates will go together with lower temperatures for the same reason. That is consequence of 1st law.
If you fill a pot on an oven with water you sped up(!!!) the rate drastically, for water conducts and convects heat much better than air per volume element. No wonder you end up colder. Compared with air water is not an insulator. Air is the insulator compared to water. And which pot gets hotter, the one with air, or the one with water? Yeah…
This example shows that slow down of heat loss leads to higher temperatures, if a heat source gives constant input.
What happens if you cover the CPU of your PC with, let’s say, wool? Will it get colder, for the wool additionally absorbs heat? ( Wool is a very weak heat conductor ).
You didn’t answer the question to the mouses. Might occure absurd to you, but I told you the underlying principle. Can you answer this question with your knowledge of physics, or not?
”gravity provides a thermal resistance of g^2”
Please show me the paper or the textbook where this comes from!
GillaGilla
” Heat flow ”from” a system means that heat leaves that system. You stated above an increase of that flow away means hotter.”
The heat flow is independently related to the internal state. A change in heat flow is caused by a change in the internal state. Problem?
”the 1st law only holds”
The first law always holds. Give a reference for situations where it doesn´t, or shut up.
” if slower rates go together with higher temperatures at the source for the flux has to be constant. Faster rates will go together with lower temperatures for the same reason. That is consequence of 1st law.”
Reference, please.
”If you fill a pot on an oven with water you sped up(!!!) the rate drastically, for water conducts and convects heat much better than air per volume element. No wonder you end up colder. Compared with air water is not an insulator. Air is the insulator compared to water. And which pot gets hotter, the one with air, or the one with water? Yeah…”
Good, good. You understood how water interacts as a fluid with a heat source.
Now, compare a body in vacuum emitting 400W/m^2 from a two-dimensional surface, to a body in vacuum with a cold layer of fluid air, emitting 400W/m^2(solid surface)+244W/m^2(cold fluid air) from a two-dimensional surface AND a threedimensional volume of fluid.
Which body needs more power. The one with only a surface emitting 400W/m^2, or the one with a surface and a fluid emitting 400W/m^2+244W/m^2.
If you remove the fluid from the double layered body with fluid and solid, what happens? Will the solid have to emit 400+244=644W/m^2 from the surface?
And, what is the effect of adding a volume of cold fluid to a hot body? Does the hot body get hotter?
If it does, you need a reference.
”Please show me the paper or the textbook where this comes from!”
Do you want a reference for heat/light being affected by gravity? Take a look at the nobel prize this year.
Do you want a reference for thermal resistance, stress and pressure being measured in Nm^2? Google it.
Come on!
This is like talking to two-year olds. You have a brain. Use it.
GillaGilla
”If the heat flow is reduced from a system, it means that the system just got colder. Increased heat flow, means hotter.”
Again, wat.
When a system sheds thermal energy, it *loses* heat. Which means the temperature will be lower than otherwise.
It’s a matter of how much heat is being gained (generated or absorbed) versus how much is being shed. When the more is being generated/absorbed than is being shed, the temperature goes up. And when it’s the other way, when more is being shed, the temperature goes down.
Over time, if the input is held constant, a system will head towards ”steady-state equilibrium”, where the output will eventually decrease to match, as the temperature drops. This outflow is governed by the Stefan-Boltzmann Law. See, if the temperature is too high, the radiated energy will be greater than the energy absorbed or generated, and the system will lose thermal energy. And with that, the temperature decreases. And because the temperature decreases, so does the amount of emitted energy (by the S-B Law). This continues until the system balances out.
So, no, when heat flow is reduced from a system, it does *not* mean the system just got colder. It’s just the opposite: reducing the heat flow from a system will make it warmer, if all else remains the same.
There are other ways to reduce the heat flow out from a system besides cooling it off, y’know. And then *this* results in heating, which increases the temperature, and then the radiated energy will eventually increase to balance. But in the meantime, the decreased outward energy flow happens because the system is heating.
GillaGilla
”When a system sheds thermal energy, it *loses* heat. Which means the temperature will be lower than otherwise.”
Remember, -constant heat flow-. The emission is independently related to the the internal state, the temperature of the emitter. Reduced heat flow means that the emitting body got colder.
”See, if the temperature is too high, the radiated energy will be greater than the energy absorbed or generated, and the system will lose thermal energy.”
Too high for what?
The radiated energy is dependent on the internal state. The absorption of energy is dependent on the radiated energy. The ”loss” of energy depends on the internal density of energy, the internal state.
”And with that, the temperature decreases. And because the temperature decreases, so does the amount of emitted energy (by the S-B Law). This continues until the system balances out.”
The emission decreases as an effect of decreasing temperature. The temperature doesn´t decrease as an effect of decreasing emission of radiation. I repeat, the emission depends on the internal state only, this is not a questioned fact. It doesn´t say: the internal state depends on emission. Obviously, because that would be retarded.
”So, no, when heat flow is reduced from a system, it does *not* mean the system just got colder. It’s just the opposite: reducing the heat flow from a system will make it warmer, if all else remains the same.”
Reference please.
Prevost stated; the emission from a body depends on the internal state only. It has not been questioned.
You say: the internal state depends on the emission.
What you claim is the opposite of very old undisputed physics. You have to support your claims, that cooling of a heat absorbing fluid causes increased power of the heat source heating the fluid.
My claims, that the heat source controls emission of the fluid, that the internal state is independently the cause of emission, is grounded in already proven physics.
”There are other ways to reduce the heat flow out from a system besides cooling it off, y’know.”
Interesting. Please tell me more. And don´t forget, describe how they support the claim that adding dry ice to a cold fluid heated by a warm solid body, causes increasing output of power from the heat source. By creation of energy, like in the greenhouse theory.
I guess that energy-corporations worldwide is interested in how they can add co2 to their energy production to increase the output power of their power plants.
GillaGilla
Ben,
Is it semantically correct to say that an object radiates heat? I feel like lifeisthermal will attach himself to use of the word heat when you say a system loses heat. When I was in physics, heat was always talked about the sum of the kinetic energy of the molecules and temperature the average. The links to hyper physics make it sound like use of heat in that sense is no longer accepted. Heat is the net transfer from a colder object to a warmer, which makes conversations with LIT incredibly difficult since he will focus solely on the misuse of word, like heat, while completely abusing century old equations, slap whatever units he feels appropriate on to values, apply laws from one field of physics to others where they don’t belong (using radiative heat transfer equations for conductive heating). But how dare you misuse the word heat.
And I’m honestly asking. I’ve always use the word heat as you did, but is it correct to do so? An analogy, if an object is in a vacuum by itself, it radiates. But if there isn’t another object for that radiation to interact with, is it heat? I’ve resorted to using internal energy when talking about radiation and leaving heat for ugly the net transfer in the positive direction, from colder to warmer object
GillaGilla
Ben,
Good to know I’m not completely crazy in my usage of the words then. Saw a similar conversation on scienceofdoom in regards to use of equilibrium instead of steady state. So trying to use those terms properly as well as since I have often referred to the earth being in equilibrium with the sun. It’s all context.
GillaGilla
”Saw a similar conversation on scienceofdoom in regards to use of equilibrium instead of steady state.”
Oh yeah, hahaha, that’s definitely another one where we do it. ”True” equilibrium is thermodynamic equilibrium, but we use the word ”equilibrium” to also describe steady-state eq., pseudo-steady-state eq., etc. Just because it’s too damn wordy to say ”pseudo-steady-state equilibrium”.
The context is again obvious; we’re talking about when they system is in balance with regards to X, where usually we’ve been talking about the flows or balance of X for the last 10 minutes, so the context is incredibly clear. It could be thermal flows, elastic stresses, electric fields, etc., etc. Practically anything.
GillaGilla
Lifeisthermal,
Absorption of radiation is not dependent on emission. It is dependant on the absorption of the surface. Net energy transfer is then the difference in amount absorbed and amount emitted. If this amount is negative, the surface temperature will drop until absorbed= emitted. If the number is positive, that means more is being absorbed than emitted, the surface will warm until absorbed= emitted. Do you agree?
GillaGilla
Logistical,
Units units units!
”Even without the reference for thermal resistance, using gravity as pressure with units Nm^2 is enough to realize the relationship to emissive power.”
Unit for pressure is N/m^2, not Nm^2. No reason to have a discussion on thermodynamics when basic units of calculations are thrown around haphazardly.
GillaGilla
That depends. At the receiving end, you have N which can be used on a m^2, as N/m^2. Tracing it to the source, a uniform force of a Newton in a point, must have equal force in the next point beside it, and the next, and the next. So then, you have N*m^2. With a source power of 4*xNm^2.
Gravity is a uniform force. And from our point of view, being affected by it, it has a constant effect of 9.8^2Nm^2. From the source point of view, it has an effect of 4*g^2N/m^2, declining according to the inverse square law.
P.S. Thank you for your questions, they help me to question myself and my concepts. I appreciate it very much.
Now, to save your soul, we just need you to abandon the idea that one can create energy by adding dry ice to a heat source. Then you will be a free and good man. Stop obeying the oppressive forces of the guilt-spreading priests in the doomsday-cult of IPCC. It is just a power-tool, trying to replace the christian concept of original sin. A way to compensate for the loss of religious blame-giving, where they try to make us believe that we should feel guilty for just being born. The church says: you are a sinner, repent by paying us. IPCC says, you are a sinner for using the comforts of living in civilization, pay us.
You recognize these charlatans by their arguments, that you are too stupid to know your own best, and you should give your money to them, and they will save you.
Never listen to anyone that uses that kind of arguments.
GillaGilla
You really need to open a textbook, let at least Wikipedia gravity is not a constant force. It is proportional to the mass of the 2 objects being pulled together. At a set distance from the center of mass what is constant is the acceleration. For our distance from the center of mass of the earth, that is 9.8m/s^2. Do you understand this? Not Newton’s, not Newton’s/m^2, not Newton’s m^2. It’s 9.8 m/s^2. Do you understand?
Gravity is not a uniform force. If it were lighter objects would fall quicker than heavier objects. Again, no need in discussing anything more complicated than this if the elementary basics of gravity can’t be understood by you.
GillaGilla
” It is proportional to the mass of the 2 objects being pulled together.”
Ok. Why then, does two objects of different mass, accelerate equally towards the ground? If resistance of air is eliminated.
In vacuum, gravity-acceleration is independent of mass. An apple falls as fast as a feather. A bowling ball falls as fast as a piece of styrofoam.
If gravity is proportional to mass, why does gravity accelerate a 1000-tonnes boulder, as it does a goose-feather pillow?
Gravity is not dependent on mass, it is constant. Mass varies, gravity doesn´t.
”Not Newton’s, not Newton’s/m^2, not Newton’s m^2. It’s 9.8 m/s^2. Do you understand?
There is no difference between the force in Nm, or m/s^2, or the energy needed to provide such a force in W/m or W/m^2. Do you understand?
”If it were lighter objects would fall quicker than heavier objects.”
Wow. How dare you enter a discussion of physics with such low level of knowledge?
Just watch how gravity accelerates a feather and a bowling ball at the same rate in vacuum:
I mean, really?
Is this the level of climate scientist physics?
You don`t even know the physics of gravity?
Let me know, was this embarassing to you? That you didn´t know that gravity is a force independent of the mass it it attracts?
GillaGilla
Because the force of gravity is based on both masses. Force of gravity is Gm1m2/r^2. m1 and m2 are the 2 masses in a gravitational field, r is radius separating the center of mass, G is gravitational constant, and no, that isn’t 9.8m/s^2. Look it up if you need to understand
If the force was only based on the pulling object then the acceleration would differ between objects with different mass. Acceleration is constant based on radius from center of mass. It is Gm1/r^2.
GillaGilla
So, why is a feather accelerated at equal rate by gravity, as a bowling ball?
If gravity depends on mass, the bowling ball would fall faster.
GillaGilla
Because at a given radius the acceleration is constant, so they both accelerate at the same rate. Get it?
That means though that there is a stronger force on the ball since it has a larger mass. Because acceleration= force/mass. So see, when the mass gets larger, the force has to grow for the acceleration to stay the same. Grade school stuff.
GillaGilla
The force is ALWAYS 9.8Nm. The WEIGHT/MASS varies. So, THE FORCE is CONSTANT, while the MASS VARIES.
This means that the FORCE is independent of VARIATIONS of mass.
Regardless of what the mass is, the accelerating force is constant.
If the force is dependent of mass, then the force would vary with mass. It doesn´t. It is 9.8Nm or 6.67*10^-11Nm^2, independently of mass.
GillaGilla
”That means though that there is a stronger force on the ball since it has a larger mass.”
This is so completely retarded. The force is 9.8Nm. Newton means per kg, The force is equal per weight. Weight/volume is mass,
The FORCE is 9.8Nm, no matter what the mass/weight is. It is NOT stronger on the ball. The ball is DENSER, in a field of constant force.
Kindergarten stuff.
GillaGilla
You show your complete lack of understanding based on the fact the you keep talking about a force being Nm. That is a unit for work. Unit for force is N, or kg m/s^2. See it shift there in the units, force is mass x acceleration. If the force stays constant and the mass changes, then the acceleration changes. That’s what would happen if force of gravity was a constant 9.8N
But that doesn’t happen because the acceleration is constant. That’s why 2 objects fall at the same rate.
Can you tell me the equation for force?
GillaGilla
”That is a unit for work. ”
It is a unit for a force. Or work.
Any unit can be converted into another. It is a key to understanding the relationships in physics, try it.
”If the force stays constant and the mass changes, then the acceleration changes.”
You fucking imbecill!
The acceleration is the derivative of the energy-addition per unit time. Which is equal to the force which adds energy to the body.
Shame on you, trying to fool me into believing that you were average-intelligent.
You are at the mental state as a piece of wood.
GillaGilla
From hyper physics,
Please tell me what the force of gravity is based on?
http://www.hyperphysics.de/hyperphysics/hbase/hframe.html
GillaGilla
I have already told you. I have given you equations that show exact relationships including the force of gravity.
So, instead, YOU show ME what gravity is based on. And why the measured force of gravity should not be accounted for as a thermodynamic force.
GillaGilla
”The FORCE is constant. It is 9.8m/s, or 9.8Nm. It is independent of mass, because the acceleration is 9.8m/s no matter what the mass is.”
Oh jeebus.
Yes, the *acceleration* is essentially the same, regardless of the mass. The *force* is not. Force = mass X acceleration.
It takes twice as much force to accelerate a 10kg bowling ball by a given amount as it does for a 5kg bowling ball. And the gravitational force acting on the 10kg bowling ball is twice that of the gravitational force acting on a 5kg ball, so you get the *same acceleration*.
LIT: please, please, please go take a physics course. You’re confusing acceleration and force. Please recognize that you need to get back and understand the basics, first.
GillaGilla
Force and work are different things in physics. You don’t just get to say that it is a unit for one of the other. Unit of force is N. Unit of work is Nm.
I’ve typed the equation in your blog as well as provided you a link for the compassion for the force of gravity. Once more though:
Force of gravity is G*m1*m2/r^2
Acceleration of gravity is G*m1/r^2
Notice how the formula for the acceleration is only dependant on 1 mass, not 2. Because everyone on earth feels the same acceleration of gravity. But the force required for that acceleration varies. That’s why when 2 different people step on a scale it shows a different weight. Weight is a measure of force, or mass x acceleration.
GillaGilla
”Weight/volume is mass,”
This is also really really really wrong.
Weight is a measure of force in the scientific community, where weight = mass X normal gravitational acceleration. In laypeople’s terms, weight is used interchangeably with mass, which is not great; it conflates two different units, but whatever.
But in no system does weight/volume = mass. Density is mass/volume. And weight/volume is… not really anything.
In SI units:
mass is (kg)
weight = Newtons = mass*acceleration, or (kg * m / s^2)
weight/volume is Newtons/m^3, or (kg / (s^2 * m^2) )
Note that (kg) and (kg / (s^2*m^2)) are two different units.
LifeIsThermal, you are making fundamental mistakes here about the relationship between different units. Take. a. physics. class.
GillaGilla
”Weight is a measure of force in the scientific community, where weight = mass X ”
”weight refers to the force exerted on an object by gravity.”
https://en.wikipedia.org/wiki/Mass_versus_weight
So, weight is the force of gravity acting on a mass x.
Density:
”mass per unit volume”
https://en.wikipedia.org/wiki/Density
So density of mass is equal to density of energy in E=mc^2.
And a force is equal to energy, in f=ma, where the energy needed for a force f, is equal to =ma.
And your problem is?
GillaGilla
This is why you weigh different amounts on the Earth and on the Moon.
Your *mass* is the same on each, but the gravitational *acceleration* is different between the two. Gravity is different on the moon than on the Earth.
Since weight is force, and Force = mass * acceleration, the change in acceleration between the Earth and Moon also means a change in force or weight.
GillaGilla
”Your *mass* is the same ”
What is the mass of one dm^3 of water on the moon? What is the mass of one dm^3 on earth?
GillaGilla
”Mass is both a property of a physical body and a measure of its resistance to acceleration”
https://en.wikipedia.org/wiki/Mass
So, mass is relative to the force acting on it. The force is NOT relative to the mass.
GillaGilla
”So, mass is relative to the force acting on it. The force is NOT relative to the mass.”
No. The mass of a body is the mass of a body, regardless of what force acts on it.
However. The force of the Earth’s gravitational pull on a body (yours, mine, a feather, a bowling ball) *depends* on the mass of the body. More mass => more force. This gives the same acceleration for all bodies:
a = F/m
The force scales with the mass.
The gravitational pull exerted on a 10kg ball is twice that of an adjacent 5kg ball. Since the force is twice as much, and the mass is twice as much, and (2F/2m) = (F/m), the acceleration of both balls would be the same if you dropped them.
GillaGilla
”The mass of a body is the mass of a body, regardless of what force acts on it.”
Ok. Define mass.
”No. The mass of a body is the mass of a body, regardless of what force acts on it.”
Ok. So, if a force in a field is constant, like G, or g, then mass behaves relative to the force?
”*depends* on the mass of the body.”
,
But still, on earth, the force is 9.8Nm per kg. Independent of mass. The force is always the same on earth. But the mass varies.
Show me a reference where a body have a force greater than 9.8Nm/s on earth. If you can´t. you are wrong.
As you say, a=F/m. So 9.8=F/m. Whatever the mass, m, is, the acceleration is a. A is the constant. Not m.
GillaGilla
LIT,
From the Wikipedia link on mass
”An object on the Moon would weigh less than it does on Earth because of the lower gravity, but it would still have the same mass. This is because weight is a force, while mass is the property that (along with gravity) determines the strength of this force.”
Elementary school stuff man.
GillaGilla
”This is because weight is a force”
You are not aware of the trap I am luring you into.
If weight is a force, the addition of co2 to a volume of mass, is an increase of force on earth.
Are you saying that increasing the amount of co2 in the cold fluid atmosphere is increasing the force of gravity that does work on the solid earth?
You keep claiming that mass is the cause of gravitational force, so increasing the amount of co2 in a given volume of air, must increase the mass of that parcel of air, If the mass increases in the fluid, the gravitational force must increase, according to you. Then, the force of air must increase, gravitationally.
Are you saying that co2 increases the gravitation of earth?
GillaGilla
— ”Are you saying that co2 increases the gravitation of earth?”
The CO2 came from the Earth. You’re not adding mass to the Earth as a whole, because the O2 in the CO2 came from the air, and the carbon came from the ground. It’s just moving some carbon atoms from the ground to the air; a change in elevation, which has a very, very, *very* small effect on gravitation. This effect is completely negligible.
Start with understanding units first, I think.
GillaGilla
” It’s just moving some carbon atoms from the ground to the air; a change in elevation, which has a very, very, *very* small effect on gravitation. This effect is completely negligible.”
If all parts of co2 comes from earth, air and ground, it cannot have ANY effect on gravity. Even if you think that the force of gravity originates in mass. Because mass is then constant.
So, when m is constant, and c^2 is constant, how can the energy E increase?
Can you give an example of observations where the addition of any kind of mass, co2 or whatever, increases the energy of a system?
Even better: an example of addition of mass, where temperature increases without any increase in power from the heat source?
GillaGilla
LIT,
What is your weight in Newton’s? My mass is 65kg
That means I have a weight, or force against the ground of 637 Newton’s.
Your getting closer with the units.Your comment about Nm/kg is almost the baseline unit for acceleration. Drop the m and you’re there. Acceleration is N/kg
GillaGilla
”My mass is 65kg”
The concept of Kg, is relative to earth. Kg is a product of g=9.8Nm, you are aware of this, right?
65kg would not be not be your mass on another body.
Your mass is relative to earth gravity, I am sure that you understand this.
The concept of kilogram, is a product of the gravitational force on earth, I hope you understand this.
Relativity is a concept of independence, where you need to distance yourself from surroundings.
”Acceleration is N/kg”
No. Acceleration is N/time. Nm/s. Supplied with energy at W/s.
The acceleration is always 9.8Nm. The energy of a given mass accelerated is F=ma, F=kg*9.8, on earth.
Read carefully: The force on mass is CONSTANT. Always 9.8Nm, 9.8m/s, regardless of mass, on earth. G=constant, g=constant, independent of mass.
G=g/(m1*m2)/r^2
In the above equation, G and g never change. They are the same, even if m1 and m2 changes. g/G(m1*m2)/r^2
So, m1 and m2 can change, as well as the distance between them. But G and g will be constant.
The distance to the sun/heat is relative to G, and the distance to the center of mass/heat on earth is relative to g. But g doesn´t vary with kg, and G doesn´t vary with kg. Because G varies with area in N/m^2, and g varies with distance in N/m.
The energy of force per unit time, depends on the density or energy in the source, independent of mass. Because G*time*area is constant each second, while mass can be at any value. Exactly like the heat at any distance from the sun, is W*time*area. So, heat and work is independent of mass, because it is a derivative of area per unit time.
Whatever the mass of a body in the suns electromagnetic field, the heat is constant, and the force is constant.
The force and heat, is present, independently of any mass.
GillaGilla
— ”The concept of Kg, is relative to earth. Kg is a product of g=9.8Nm, you are aware of this, right?
65kg would not be not be your mass on another body.”
Incorrect. Your mass would be the same on the moon; your weight changes. See:
”An object on the Moon would weigh less than it does on Earth because of the lower gravity, but it would still have the same mass.”
https://en.wikipedia.org/wiki/Mass
–
— ”Acceleration is N/time”
No. Acceleration is (m/s^2). Think about it; this makes sense. Speed is meters per second. Acceleration, or the rate of change in speed, is meters per second per second. If you go from standing still at t=0, and at t=2 second, you’re running at 4 meters per second, then your average acceleration during that time was 2 meters per second, per second. E.g., each second, you increased your speed by 2 meters per second.
https://en.wikipedia.org/wiki/Acceleration
And, fine, I’ll grab my undergraduate physics textbook off the shelf[1]:
”Because the dimensions of velocity are L/T (Length/Time) and the dimension of time is T, acceleration is meters per second squared (m/s^2).”
–
— ”In the [G=g/(m1*m2)/r^2] equation, G and g never change. They are the same, even if m1 and m2 changes. G/g=(m1*m2)/r^2”
No, again, completely wrong. The equation should have an ”F” where the ”g” is, e.g., [F = G*(m1*m2)/r^2]. This is force, not acceleration.
And G is fixed as the gravitational constant across the universe (as far as we can tell). But the gravitational force between two bodies, ”F”, depends on the mass of the first body (m1), the mass of the other body (m2), and the distance between them (r).
Alright, I should bow out. You can pick up any textbook on physics from high school to graduate school, and they all contradict you on these points. If you’re curious why people disagree with you on climate change, it’s because of stuff like this. You haven’t even learned the basics.
[1] Physics for Scientists and Engineers, 6th Edition, by Serway and Jewett, p. 31
GillaGilla
2″Your mass”
The concept of mass is a product of the weight and volume. Mass is the density of matter in an object. It cannot be determined without a reference to the force of gravity acting on the object.
In the vacuum of space, it is impossible to determine the mass of an object, if not in relation to G or g. So, mass, is relative to earth, or the sun.
”No. Acceleration is (m/s^2)”
And m/s is equal to N/m, or W/s, at a mass of 1kg. The force of gravitational acceleration on earth is 9.8Nm per second. 1Nm is 1W/s. The square, s^2, is only there to define the derivative. Over time, the average force of 9.8m/s, adds up to a force of 9.8m*s^2. From an altitude, the force on impact of an object, is 9.8*kg*s*m. The derivative cleans out the mass and time to get the independent force.
”And G is fixed as the gravitational constant across the universe ”
I would never dare to go that far. We only have evidence of the force in the solar system. It might be universal, but it is not wise to use subjective assumptions from the solar system as a universal truth.
” The equation should have an ”F” where the ”g” is, e.g., [F = G*(m1*m2)/r^2]. This is force, not acceleration.”
Ok. So, g=G*(m1*m2/r^2. Which is equal to g/G=(m1*m2)/r^2
As you can see, both masses and distance is a function of gravity, both g and G.
The force is g/G, and both masses is a function of gravity. Or, gravity is a function of mass and distance.
Considering the facts of the first law, that force and heat is a function of energy, and the fact that gravity is a force, I think that it is more wise to regard heat and gravity as the source of mass, instead of mass being the source of force and heat. Because the first law doesn´t include mass.
”they all contradict you on these points”
No, no, no. What they contradict is this:
Adding dry ice to a heat source makes it hotter.
GillaGilla
Oh no, it’s a trap!
GillaGilla
Lol.
Humour is a sign of intelligence. Good for you.
You are not a lost case.
GillaGilla
I’ve given the links, if I could post a screenshot I would.
Force of gravity is Gm1m2/r^2
g (which represents the accelerating of gravity) is GM(earth)*r^2
Type these numbers into a calculator and see what you get:
6.674*10^-11 * 5.972*10^24 / ((6.37*10^6)^2)
What do you get? If you can type this stuff into a calculator and tell me what the number represents with units maybe we can continue a conversation.
GillaGilla
”6.674*10^-11” is the force, in units Nm^2, independent of mass in kg, and time in s(^2).
At earth´s location, no matter the value of mass in kg:s, the force is constant at ”6.674*10^-11”.
At any location above the surface of earth, independent of mass, the force is (6.674*10^-11)*(mEarth-mAnyobject)/distance to center(r)^2.
So, the measured force, G, and the measured distance, is fixed. Whatever distance, at this constant level of force, any mass will experience the same effect.
So, the mass of an object, at a distance from the heat source, is a product of a constant force at a certain distance.
The force is constant. It doesn´t matter what the mass is, it will still experience a constant force at the level of G.
This means: G is independent of m at a certain distance, just like heat flow from the sun is independent of m at a certain distance.
It also means: the temperature, which is a measure of the average kinetic energy, the net of all forces and heat in a point of measurement, is a product of heat and G. Because g is determined by the distance, r, to the sun, from the force G at this location. Combined with the heat flow, TSI, at this location.
The temperature of any mass, at a certain distance from the sun, is a product of the heat flow from the sun and the gravitational force from the sun.
GillaGilla
”So, when m is constant, and c^2 is constant, how can the energy E increase?”
It’s been said over and over that E and e are not equivalent. Both are measures of energy but they shouldn’t be focused with each other and can’t simply be exchanged for one another. Your misuse of them only further illustrates your complete lack of understanding and comprehension.
GillaGilla
At any location above the surface of earth, independent of mass, the force is (6.674*10^-11)*(mEarth-mAnyobject)/distance to center(r)^2.
You haven’t answered my question. Type the numbers in and tell me what you get. Also in your equation above you say that the force isn’t dependant on the mass of the second object. Yet in your formula you are using the mass of the second object. Why is that?
GillaGilla
I guess that you are aware of that
g=G*(m(earth)*m(object)/r^2(distance between center and object, squared)
The numbers doesn´t matter, it is the relationship between them that matters. The relativity.
G is constant. Mass is relative.
C is constant, distance is relative.
Energy is constant, space is relative.
GillaGilla
A mistake to include mass. g is a result of G. Without G there would be no g. But m can vary.
G exists independent of m. But m is entirely dependent on G. The product of G, the work done, is exactly relative to dU(TSI).
Two constant observed values of balanced force and power, and a non-interacting mass.
What I mean is, heat flow is independent of mass, that is known. And apparently, the heat flow has a precise relation to the force of gravity. It is not strange that the force has an exact relation to mass, and the same is true also for heat. But this doesn´t mean that gravity and mass is cause and effect. Actually, it is more likely that a force is a result of power in a heat flow. The indepence of heat flow, and the exact relation to the force of gravity, implies that gravity is independent of mass as well. Relative to mass, yes. Caused by mass, no.
This reasoning comes from the results of calculations. I did not expect to find such relations when investigating the balance of the heat flow. But seeing the precise relationships made me think about the internal relationships. Which is in line with known laws and principles for heat and forces.
I am not saying that I am right. I am just saying that there are equalities are precise, and they don´t contradict physics. When finding such exact relationships it would be stupid to ignore them. The precise agreement with observation deserves a closer look, and further consideration to confirm or refute them.
My toy-model accurately calculate average emission at the surface. This is what the gh-model failed to do. This fact alone makes it very interesting and also superior in performance compared to the gh-model which has no support in observation or litterature.
There is not a single paper or observation which supports a claim that adding co2 to a heat source, will increase the power of the heat source.
Even if it would act as insulation, insulation can not increase power of the heat source.
GillaGilla
”I guess that you are aware of that
g=G*(m(earth)*m(object)/r^2(distance between center and object, squared)
The numbers doesn´t matter, it is the relationship between them that matters. The relativity.
G is constant.”
You are completely wrong.
G is a constant, you are right about that. Unfortunately that’s about the only thing you are right about. By the way, G is a universal constant. You would use it to calculate the acceleration of or force of gravity between any two objects, not just the earth.
g is shorthand to refer to the acceleration of gravity on the surface of earth. It is found by doing this calculation:
6.674*10^-11 * 5.972*10^24 / ((6.37*10^6)^2)
G * mass of Earth / (radius to the surface ^2)
The units for this calculation are m/s^2. If you key in the above number into a calculator you will get approx 9.82 m/s^2
You can see by checking any number of websites, including (hyperphysics http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html), Wikipedia (https://en.wikipedia.org/wiki/Gravitational_constant), or any other site that prides itself on factual information, that what I have represented here is accurate and correct.
Also, for more information on mass, check hyperphysics here: hyperphysics: http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
This is all very trivial stuff but shows that any conversation about more complicated matters need not be addressed. So if you can start to actually communicate in a way that demonstrates you understand what mass is and it’s units (g or kg), what acceleration is and it’s proper units (m/s^2), proper units for force (N or kgm/s^2), proper application and use of work (Nm), proper use and units for pressure (N/m^2), and so many other things I’m done. I hope for the sake of your future readers you leave all of these conversations in place so that they can research this info on their own.
The thought has occurred to me that maybe there is some sort of wording that is getting lost in translation. I hope for your sake that is the case. Or you are just being an internet troll arguing for the sake of arguing. Because if it isn’t an issue of translation, and you are really serious about all of this and think that you are right, then it’s frankly a very sad case. You need to take some very basic classes before diving into anything deeper than a puddle when it comes to physics.
GillaGilla
The point is, that I show g is relative to TSI and emission of heat from earth. The emission of heat includes mass, because it flows through it, and it is independently related to the internal state of mass. Calculating g from mass and radius with G, does not contradict a relationship between heat flow and gravitational contraction of mass.
Knowing that gravity is a force which does work, and all forces needs a power source driving it, a relation to the flow of energy is more direct and follows the known physical relationships defined in the laws of thermodynamics. To not include it in a model of energy distribution on a planet, and regard it as mystical with unknown origin, takes out a large part of the total energy. A force, which is measured, needs an exact amount of power to drive it. The only known source of power is heat flow through the system. It is rational to balance heat and force. It is not rational to claim that cold fluid with increasing amounts of dry ice makes its own heat source hotter. It is not rational to think that mass is magically producing energy in the shape of a force called gravity. Energy equal to 96W/m^2 is constantly added by gravity. It must have a source of constantly flowing energy to drive it. The source must, according to the inverse square law, have a power of 383W/m^2.
GillaGilla
Your punt is completely invalid and wrong. The formula for little g is G*mass of earth/waits off earth squared.
Where in that formula did you see tsi?
Where in that formula do you see radiative emission from the surface?
In physics, when we say, for example, watts is based on work done and time, you would expect to see work and time in the equation. There may be some other variables there too, but work and time should be. Consequently, the equation for watts is:
Force*distance/time or just work/time. Do you see how’s that works.
And again, radiation isn’t a force, it’s an acceleration.
There have been too many links that I have provided to wiki pages and hyper physics pages that explain all of this.
GillaGilla
”Your punt is completely invalid and wrong. The formula for little g is G*mass of earth/waits off earth squared.”
Punt? Wait?
But sure, the known way to calculate g is by its relation to the density and volume of mass, into an equality between constant force of G in relation to a threedimensional volume of mass. It is acting on a square meter surface and we determine the power density in units N/m^2. In trelation to the volume of mass, it is relative to radius and density. The density is dependent on temperature/heat flow. Its action on the surroundings is measured as a force in a point, unlike G. But it can easily be converted to its strength on a square meter.
The behaviour and character of gravity, is opposite to heat. It expands and decline like heat. It travels as fast as heat. It is observed in relation to heat, universally. Heat and force is proven to be closely related. Based on observation and established physics, gravity and heat is closely related. This is the most rational conclusion possible. Free from subjective asumptions. Based on proven and applied physics.
An exact relationship independent of anything else, between force(gravity) and power of the heat source, is not a surprise. It is expected.
”Where in that formula did you see tsi?”
In the formula for a ball with a shell with uniform charge:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesph.html
Which is a result of geometry, force and flow of energy.
Your equation is the relation between force and the mass it acts on. My equation is the relation between force and power of the source(s).
”Force*distance/time or just work/time.”
The force acting per unit time, tells you what density of power needed to constantly drive the force.
To have constant force, constantly adding energy as work, you need constant flow of power.
Apparently, the force of gravity is exactly balanced to the heating power of the solid planet. It is not really something that should be surprising, since the most solid physics known to man, the laws of thermodynamics, is about that exact relationship. dU=Q+/-W
”And again, radiation isn’t a force, it’s an acceleration.”
Radiation is flow of em-energy, into or out of mass. But fine, call it acceleration, it is a definition which applies also to gravity.
It seems like a supportive argument of my calculations, not your blanket-theory where dry ice creates energy by cooling a heat source.
A reference to this statement would be useful. Cite the part of interest, and provide a link.
GillaGilla
My last comment should have said gravity is an acceleration, not radiation.
Lifeisthermal,
2 objects with mass m1 and m2 are separated by a radius of r when measured. Please let me know with specific equations what is the force of gravity between them, the acceleration of gravity on either object, you pick which one, and the equation for the total gravitational potential of the system. In normal physics, these are 3 equations that look rather similar. I’m curious what your equations are for it though.
Force =?
Acceleration (m1 or m2) = ?
Total gravitational potential=?
GillaGilla
It seems to come down to one single point of confusion:
1) Thermal energy never transfers (net) from a colder object to a warmer object. Or, practically never – Maxwell’s demons still apply here.
2) Adding insulation to a hotter body will cause it to lose heat more slowly. If the body was at steady-state equilibrium with its surroundings; if thermal energy was generated at the same rate as it was lost; then the rate of loss will decrease, the thermal energy in the body will increase, and the temperature of the body will rise. This is what happens when more thermal energy is generated than lost; the temperature rises.
The 1LOT is about #1. The greenhouse effect is about #2.
No one is saying that heat will transfer (net) from a colder body to a warmer one. We’re saying that adding radiative insulation will decrease the loss of heat from the warmer body, so the temperature will rise.
GillaGilla
Since you like to play games with animals, try this:
Put a mouse in a cage hanging at some level above ground, at some distance from an open air fire. Take a water-hose and shower the mouse with water(like rain, vapor or oceans). Then spray dry ice on the mouse with a fire-extuinguisher.
What happens to the temperature of the mouse? And why do you need analogies for ice, water, cold air and dry ice, in the shape of foil and gel-packs?
You are describing air, water and co2. You don´t need analogies, the action of these substances on heat sources is not unknown.
GillaGilla
Side question but not OT:
The cooling garment of space suits is according to NASA engineers vital for cooling against envoirenment ( if space walk is done in the sun ) as well as cooling away the heat produced by the metabolism.
How is the latter case possible? The colder space suit should be unable to heat the human body further than its normal 37C.
Anyway the astronauts are about to get a heat shock within a suit that is colder than them if not cooled actively.
Why?
GillaGilla
The 4k is the contribution from the core. Your numbers are incorrect for it’s contribution to surface temperature.
GillaGilla
Aah, you mean 4000 Kelvin? As far as I know, the core is closer to 6000K, maybe more. You have to think in terms of equilibrium, since the surface has only an addition of 90mW/m^2 from internal processes it has to mean it is in equilibrium. No addition to the flow means equilibrium, but it still emits 383W/m^2. So the surface emission comes from a subsurface layer with at minimum 1530W/m^2 heat flow. 1530W/m^2 is equal to a force of 16g^2.
GillaGilla
No, as in 4 kelvin. You are giving the volume of the earth way too much conductivity to think that the core warms us as much as you say.
And again your units. Force=mass×acceleration. This means g^2 is (kg^2m^2)/s^4. Watts/ m^2 which is kg m^2/m^2s^3 Or simplified kg/s^3. I realize these units may not make sense to think about in these terms, but the simple fact is that if your equation that you want to use to demonstrate causation doesn’t end with the proper units given as an answer, then there is no causation or correlation, it’s only circumstantial, and thus irrelevant for use as determining the physical properties in question.
GillaGilla
The core temperature is balanced to solar heat, of course. The whole system has to be balanced, not just the surface and the atmosphere. I don´t know why you people ignore 99% of the earth volume when you play in the greenhouse. Try that with a steam engine and see what you get, it
doesn´t work that way.
g=9.8N/m, N/m^2 is equal to W/m^2. G is in N/m^3. So I don´t think there is a problem here, even if you would like to find one.
Gravity acts with a force of 9.8N per meter, I set the dimensions equal to the heat flow by squaring it. Nothing strange with that, that is how it should be done if you want to compare the power of the heat source and the force in a heat engine. The first law, dU=W+Q includes no mass, but is universal. No problem.
One of my big points: I do nothing strange, I add nothing new, I use only old 19th century, proven and widely applied physics.
GillaGilla
There is still even more confusion I have not adressed in my previous comment.
”The change is in m, mass. When the second body is removed, there is less mass, m. But you still have the same energy E. So m decreases, but E is constant, so the energy per unit mass must increase.”
That is, simply by logic, not true. In space incident energy on surfaces determines if energy is added to a body, for radiation is the only transfer mechanism of energy there.
If you reduce mass within the radiation field of a star by removing a body you are also removing a surface. Thus radiation input in the unit of W/m² is reduced and your statement is falsified by logic.
E is NOT constant. It is is determined by surface area. Removing a body results therefore in less mass AND less E.
”Q is the TSI, viewed from a solar perspective. But from a surface perspective, Q is effective emission, the emitted heat. From the surface perspective, TSI is dU, because it is the added energy, the change in energy inside the system.”
The laws of themordynamics are always to be applied for entire systems. They are laws about the sum of all parts of a system. If you split a system into parts the laws are not allowed to be applied on those single parts. You do this, anyway.
You can not split the sun from your view and still come to valid claims that are solely designed for complete systems.
”To be clear: if you add heat absorbers to a constant, limited heat flow, you get less energy per molecule. Less energy per molecule means lower temperature.”
In this statement you forget time.
Systems accumulate heat within different times to steady state, if different amounts of bodies and thus different masses are fed with energy.
But we only look at steady states, so the energies per molecule are not simply related to the number of heat absorbers, but crucially to the time needed to rech steady state.
GillaGilla
”That is, simply by logic, not true. In space incident energy on surfaces determines if energy is added to a body, for radiation is the only transfer mechanism of energy there.”
You like painting yourself into a corner, don´t you?
The sb-equation for radiative transfer is known to hold in vacuum. And there is no doubt that it proves that the added energy depends on the temperature of the absorber. The higher the temperature of the colder body, the less energy is transferred. So, proven and applied laws of physics contradict what you say. That is really not a good position to argue from, don´t you agree?
”If you reduce mass within the radiation field of a star by removing a body you are also removing a surface. Thus radiation input in the unit of W/m² is reduced and your statement is falsified by logic.
E is NOT constant. It is is determined by surface area. Removing a body results therefore in less mass AND less E.”
This is ignorance. In vacuum at our location, E=TSI+G, independent of any surfaces. It is a matter of density of force and power, the energy that is absorbed is per unit mass/surface, total energy is not relevant for the density of force and heat flow. My calculations show that a massless, non-interacting cavity of any size, would have the same temperature as earth at this position. The same goes for Mars and Venus. So, shut up and calculate, you are only pooping words, no substance.
This is fact: you claim that adding heat-absorbing mass to a constant heat flow, without adding any heat, increases the average energy per molecule, and that is pure stupidt y.
”In this statement you forget time.
Systems accumulate heat within different times to steady state, if different amounts of bodies and thus different masses are fed with energy.
But we only look at steady states, so the energies per molecule are not simply related to the number of heat absorbers, but crucially to the time needed to rech steady state.”
Again, shut up and calculate. Time is not energy, energy doesn´t increase from waiting. You need constant input of energy, and if the heat-absorbing mass is increased, if not the power of the heat source is increased, the average energy per molecule will decrease.
A pot of water on a boiler-plate in equilibrium, add more cold water. Does it matter how long you wait? Will it get hotter than it was before you added more water? Don´t be silly.
Heat is instantaneous, time has nothing to do with heat. I have the numbers to back it up, what do you have? My calculations show exactly that: time should not be included to find balance. In a steady state, all quantities must be treated parallel and instantaneous. That is logic. Show me your calculation of how time creates energy.
GillaGilla
”My calculations show that a massless, non-interacting cavity of any size, would have the same temperature as earth at this position. ”
would love to see these calculations. The G represents gravity? And since gravity is based on mass I’m curious how this works since your object is massless. Also, temperature is a function of internal energy, density, and specific heat of a substance. So how does a massless object have a temperate?
GillaGilla
No mass, just empty dots. From the results I would say mass results from heat and gravity. Which would be logical since there was no mass at the moment of big bang, if there was a big bang (I have another idea, similar but very different). Heat and mass on this planet has the same center, what is cause and what is effect? But I try to not make too many assumptions, because that is how we got in this mess. The shell theorem and heat transfer combined, that is enough to say.
https://lifeisthermal.wordpress.com/2017/04/13/just-numbers-no-blankets-2/
GillaGilla
”It’s correct to add ”net” regardless of whether you’re discussing gh-theory or not. Unfortunately, when discussing heat transfer, everyone already knows that we’re talking about ”net” so we usually just drop it.”
I disagree, because if there is no difference in temperature there is no transfer of heat at all. This we agree on, right?
With differing temperatures, there is heat transferred, this is something we agree on as well, right?
The heat is what the blanket-animals like to name ”net”, and they imply that there is some other energy in transfer, that is not included in ”net”. Since the only things that can raise temperature of a body is heat and work, and heat is ”net”, then the ”not-net” energy must be work, right? And we know it´s not, so you are wrong. Co2 does no work, neither does the entire greenhouse effect, so there is only ”net” and that is heat. The equation only calculates heat, and the colder body only draws energy, it adds nothing unless it´s hotter.
GillaGilla
Sure. Both gross and net.
Emission by SB Law is ”gross” emission; the emission only depends on the state of the body doing the emitting, and does not depend on any other body.
If you put two bodies with the same temperature side-by-side, they’ll both emit by the S-B law. But the net energy transfer between them is nil. All bodies emit by S-B Law, regardless of their neighbors and total energy change.
Whereas, net energy change is defined by the difference of incoming and outgoing energy.
So: two bodies, both at the same initial non-zero temperature. Both emitting at that temperature, as defined by S-B Law. One of them emits to an empty vaccuum and gets no radiation back. The other one is receiving radiation back from an adjacent colder body at half the temperature.
Which one loses energy faster? Not a trick question.
GillaGilla
” Both gross and net.”
Good try!
But, No! Not gross.
Show me a non-greenhouse calculation of the ”gross” heat. If there is a thermodynamic definition of ”gross”, you will not have a problem finding it. Good Luck!
GillaGilla
It’s correct to add ”net” regardless of whether you’re discussing gh-theory or not. Unfortunately, when discussing heat transfer, everyone already knows that we’re talking about ”net” so we usually just drop it.
Think about how thermal energy is transferred within a body of gas: conduction and radiation. Molecules collide inelastically; transferring energy from one molecule to another. Molecules radiate, according to their spectra and the S-B Law, and radiation is absorbed. This is how thermal energy moves around within a body of gas.
Now consider a second body of gas, adjacent to and with the same temperature as the first. Will the molecules between the first and second bodies still transfer thermal energy? Certainly, yes. The inelastic collisions still happen, as does radiation and absorption. So heat transfer is still happening — it’s just that there’s no net heat transfer. The heat transfer between the two bodies will happen equally in both directions (or, very close to equally, at least).
GillaGilla
Ben Winchester says
”It’s correct to add ”net” regardless of whether you’re discussing gh-theory or not. Unfortunately, when discussing heat transfer, everyone already knows that we’re talking about ”net” so we usually just drop it. ”
Perhaps you did not get a chance to read my link
‘The word heat is better reserved to describe the process of transfer of energy from a high temperature object to a lower temperature one. ‘
http://www.hyperphysics.de/hyperphysics/hbase/thermo/heat.html
I have several physics textbooks all saying the same thing in different ways
Heat energy transfer only occurs spontaneously from a higher to a lower temperature.
The whole warmist alarm is based on a colder atmosphere being able to HEAT a warmer planet.
This erroneous theory contradicts basic thermodynamics.
GillaGilla
There is no contradiction in the idea that the atmosphere can act as an insulator, helping to reduce the rate of radiation loss to space. If you don’t believe me, check the equations on hyperphysics.
http://www.hyperphysics.de/hyperphysics/hbase/thermo/stefan.html#c2
See, the net rate of internal energy loss is found by taking into account the object and it’s surroundings. So why should surroundings matter, according to your comment, if they are cooler then the direction of heat transfer will be out to the environment. You are correct that the direction of heat will be from higher to lower temperature, but the rate at which that happens will vary based on the temperature difference. Maximum rate of heat loss would be for an object to radiate into a vacuum. As a system earth is doing this. However, the surface is not emitting into the vacuum, it is emitting to the atmosphere. So to calculate the rate of heat transfer, or internal energy lost or gained, we need to account for the temperature of the atmosphere. While the atmosphere might be cold, it’s still not vacuum. There is a whole lot more to it than my simplistic explanation but no need to get into that without agreement.
The explanation for why the temperature of the surroundings matter from the perspective of the hotter object is that the colder object/ surroundings are emitting radiation as well. If I’m understanding the terminology better, hyperphysics is saying that heat only happens in a positive direction. Which makes sense when we use it to state the object a is heating object b. Doesn’t sound right to say object b is negatively heating object a. Regardless of the semantics though, the net energy transfer can be positive or negative, depending on the frame of reference. ObjectA, the one with a higher temp, is losing internal energy, that is that its net transfer is negative. ObjectB is gaining, since it is colder, internal energy, thus it’s net transfer is positive. The kicker is that ObjectA isn’t losing internal energy as fast as it would if ObjectB wasn’t there.
GillaGilla
No, no, no. The net rate of TRANSFER is what you calculate with sb(T1^4-T2^4). Not the net rate of energy loss. Both bodies still emit T^4, but they ALSO transfer according to the difference. You must stack them. Transfer rate AND T^4. Then you see that the addition of surroundings means that more energy is emitted by two bodies than one. The heat source must continously supply both bodies with heat instantaneously. Just look at my calculations, that is exactly what I have done, and I got perfect results. In the greenhouse there is only failed predictions.
The mistake you make is to think that T1 is depending on T2(colder), like it is some kind of barrier. It´s not, it is an added heat sink that draws energy from T1, making it colder. Without the external atmospheric shell, the surface would emit 1/2TSI/(4/3)=510W/m^2. With the added heat sink it emits 383W/m^2.
This is important: Prevost, a pioneer in thermodynamics, stated that the emission from a body depends on the internal state only. ONLY. The internal state is what we measure as temperature, because temperature is the sum of all energy and forces in a volume, the ”net”.
You are trying to argue that the solid earth´s emission depends on the EXTERNAL state of the atmosphere. Are you trying to say that Prevost was wrong? If so, then you are on very thin ice, it has not been contradicted before. The problem is, that nobody in climate (change) science seems to have read any thermodynamics. If they had, they would have found the answers already, and there would be no greenhouse-theory.
GillaGilla
The mistake you make is to think that T1 is depending on T2(colder), like it is some kind of barrier. It´s not, it is an added heat sink that draws energy from T1, making it colder. Without the external atmospheric shell, the surface would emit 1/2TSI/(4/3)=510W/m^2. With the added heat sink it emits 383W/m^2.
This is fundamentally inaccurate. T1 will emit radiation at a rate proportional to its temperature, regardless of whether there is another object or not. Adding an object that is colder into the system doesn’t increase it. Sb law states the maximum amount that a body can emit. When there is another object in the system we take in to account that t1 is now receiving radiation from t2 as well, thus t1-t2. Adding a second object, all be it a colder one, still reduces that rate of internal energy loss of t1.
GillaGilla
Between two bodies, yes. But radiation is emitted regardless, from both bodies, the colder and the warmer. It’s just that the warmer one emits more energy, and the colder one emits less, so the net energy transfer is from the warmer one to the colder one.
It’s not like the colder body ”knows” that there’s a warmer body out there, so it stops emitting in that direction. No, it just emits less than the warmer one does, and that’s why the net transfer of energy is from the warm to the cold.
Notably, the site that you keep linking for descriptions of the physics also explains the greenhouse effect:
http://www.hyperphysics.de/hyperphysics/hbase/thermo/grnhse.html#c1
GillaGilla
”Between two bodies, yes. But radiation is emitted regardless, from both bodies, the colder and the warmer. It’s just that the warmer one emits more energy, and the colder one emits less, so the net energy transfer is from the warmer one to the colder one.”
Here is the problem: you might be right, you might be wrong.
BUT!!!
The equations used for heat flow and heat transfer, does only include the transfer from hot to cold. They don´t include any energy outside of the transfer from hot to cold.
So, it doesn´t matter if you (and/or me), thinks that there is any transfer except the transfer from hot to cold. Because the calculations (that include a law of nature, the s-b law), clearly shows that if there is any other transfer than ”net”, it doesn´t have any effect on temperature/heat flow.
It doesn´t matter if you say that dry ice transfer a million gross watts per square meter, from the atmosphere, because the s-b equation shows that it is irrelevant for heat transfer and temperature. Heat is old physics. We know how it works. What you say, that a cold fluid is involved in a transfer of ”gross” energy that increases surface temperature of the solid earth, is contradictive to known, proven and applied physics, and it can`t be found outside of the gh-theory.
To educate yourself, try to find a reference outside gh-theory where the addition of co2 to a heat flow/heat source, causes an increase in power output of the heat source.
GillaGilla