Guest Post by Willis Eschenbach
Short answer? Of course not, that would violate the Second Law of Thermodynamics … BUT it can leave the hot object warmer than it would be if the cold object weren’t there. Let me explain why this is so.
Let me start by introducing the ideas of individual flows and net flows. Suppose I owe you twenty-five dollars. I run into you, but all I have is a hundred dollar bill. You say no problem, you have seventy-five in cash. I give you the hundred, you give me the seventy-five, and the debt is paid.
Now, there are two equally valid ways to describe that transaction. One way looks at both of the individual flows, and the other way just looks at the net flow. Here they are:
Figure 1. Net flows and individual flows. The individual flows are from me to you, $100, and from you to me, $75. The net flow is from me to you, $25.
What does this have to do with cold and warm objects? It points out a very important distinction, that of the difference between individual flows of energy and the net flow of energy, and it relates to the definition of heat.
Looking at Figure 1, instead of exchanging dollars, think of it as two bodies exchanging energy by means of radiation. This is what happens in the world around us all the time. Every solid object gives off its own individual flow of thermal radiation, just as in the upper half of Figure 1. We constantly radiate energy that is then being absorbed by everything around us, and in turn, we constantly absorb energy that is being radiated by the individual objects around us.
“Heat”, on the other hand, is not those individual flows of energy. Heat is the net flow of energy, as represented in the bottom half of Figure 1. Specifically, a heat flux is the net flow of energy that occurs spontaneously as a result of temperature differences.
Now, the Second Law of Thermodynamics is only about net flows. It states that the net flow of thermal energy which we call “heat” goes from hot to cold each and every time without exception. However, the Second Law says nothing about the individual flows of energy, only the net flow. Heat can’t flow from cold to hot, but radiated energy absolutely can.
When an object emits radiation, that radiation goes on until it hits something that absorbs it, whereupon it is converted to thermal energy. The individual temperatures of the emitting and absorbing objects are not significant because these are individual energy flows, and not the net energy flow called “heat”. So there is no violation of the Second Law.
Here’s the thing that keeps it all in balance. If I can see you, you can see me, so there are no one-way energy flows.
Which means that if I am absorbing radiation from you, then you are absorbing radiation from me. If you are warmer than me, then the net flow of energy will always be from you to me. But that says nothing about the individual flows of energy. Those individual flows only have to do with the temperature of the object that is radiating.
So how do we calculate this net energy flow that we call “heat”? Simple. Gains minus losses. Energy is conserved, which means we can add and subtract flows of energy in exactly the same way that we can add and subtract flows of dollars. So to figure out the net flow of energy, it’s the same as in Figure 1. It’s the larger flow minus the smaller flow.
With all of that as prologue, let me return to the question that involves thermal radiation. Can a cold object leave a warm object warmer than it would be without the cold object?
While the answer is generally no, it can do so in the special case when the cold object is hiding an even colder object from view.
For example, if a person walks between you and a small campfire, they hide the fire from you. As soon as the fire is hidden, you can feel the immediate loss of the radiated energy. At that moment, you are no longer absorbing the radiated energy of the fire. Instead, you are absorbing the radiated energy of the person between you and the fire.
And the same thing can happen with a cold object. If there is a block of wood between you and a block of ice, if you remove the wood, you’ll get colder because you will be absorbing less radiation from the ice than you were from the wood. You no longer have the wood to shield you from the ice.
Why is all of this important? Let me offer up another graphic, which shows a simple global energy budget.
Figure 2. Greatly simplified global energy budget, patterned after the Kiehl/Trenberth budget. Unlike the Kiehl/Trenberth budget, this one is balanced, with the same amount of energy entering and leaving the surface and each of the atmospheric layers. Note that the arrows show ENERGY flows and not HEAT flows.
These ideas of individual flows, net flows, and being shielded from radiation are important because people keep repeating over and over that a cold atmosphere cannot warm the earth … and they are right. The temperature and the radiation are related to each other by the Stefan-Boltzmann equation. When we apply the S-B equation to the 321 W/m2 of downwelling “back radiation” shown in the graphic above, it tells us that the effective radiating level is somewhere around freezing, much colder than the surface.
BUT a cold atmosphere can leave the earth warmer than it would be without the atmosphere because it is hiding something even colder from view, the cosmic microwave background radiation that is only a paltry 3 W/m2 …
And as a result, with the cold atmosphere shielding us from the nearly infinite heat sink of outer space, the earth ends up much warmer than it would be without the cold atmosphere.
To summarize …
• Heat cannot flow from cold to hot, but radiated energy sure can.
• A cold atmosphere radiates about 300-plus W/m2 of downwelling radiation measured at the surface. This 300-plus W/m2 of radiated energy leaves the surface warmer than it would be if we were exposed to the 3 W/m2 of outer space.
My best regards to all,
w.
My Usual Request: When you comment, please QUOTE THE EXACT WORDS THAT YOU ARE DISCUSSING, so that we can all understand the nature of your objections.
My Second Request: Please keep it civil. Speculation about the other person’s motives and cranial horsepower are greatly discouraged.
Further Reading: My post entitled “The Steel Greenhouse” looks at how the poorly-named “greenhouse effect” work, based on the principles discussed above.
Math Notes: There’s an excellent online calculator for net energy flow between two radiating bodies here. It also has the general equation used by the calculator, viz:
with the following variables:
and Q-dot (left-hand side of the equation) being the net flow.
Now, when the first object is totally enclosed by the second object, then area A2 is set to a very large number (I used a million) and the view factor F12 is set to 1. This is the condition of the earth completely surrounded by the atmosphere. For the general case, I’ve set area A1 to 1 square metre. Finally, I’ve made the usual simplifying assumption that thermal IR emissivity is 1.0 for the surface and the atmosphere. The emissivity values are greater than 0.9 in both cases, so the error is small. With those usual assumptions, the equation above simplifies as follows, courtesy of Mathematica:
But sigma T ^ 4 is simply the Stefan-Boltzmann radiation for the given temperature. That is why, in the energy budget above, we can simply add and subtract the energy flows to produce the budget and check to see if it is balanced.
The USAF used blow torches to cool down the skin of SR-71 Blackbirds returning from high altitude, high speed (high Mach number) recce missions.
why?
Actually, to reduce the rate at which the skin cooled to ambient. Cool almost any very hot metal too rapidly, and you get fracturing, from microscopic ones up to spectacular explosions. (So saying that they were cooling the metal is technically accurate; its temperature was going down while they were playing the torch over it – but they were cooling it more slowly than the naked air would.)
Link, please.
With your steel greenhouse I take the outer shell down to a 1 molecule thick layer and place it adjacent to the outer layer, no gap but as molecules do not touch there is still radiation.
The outer layer still emits to space the heat of the surface.
I do not “see” the surface as being twice as hot to send back the same amount of energy as it sends out but we know it does send back the same amount.
Hence double that amount must be coming through for it to emit that amount and send it back but the temperature we measure for that energy is purely that of the outgoing energy.
Something to consider for the shell argument?
After all there is a lot of internal radiation, can we call it down welling radiation, in a conductive blackbody.
Should it be called conduction when a lot of it is radiative, just not over a very long distance and not visible.
This is simply the “net” argument all over again.
An object at a particular uniform temperature will radiate from all surfaces.
And yet: if you have (in isolation, in total dark vacuum) an inner energy source constantly radiating energy X, warming up a passive shell surrounding it, the outside of that shell will also radiate a TOTAL of X.
Always. Without fail. The universe could not exist otherwise.
Exactly. And, since the shell will always have a larger surface area than the inner energy source object, the shell must always be at a lower temperature than the inner energy source object. Same total energy radiated from larger surface area = lower flux (and also, all else being equal, lower temperature). So, a 235 W/m2 inner sphere resulting in a 470 W/m2 outer shell? Not going to happen.
Black holes are not real?
Dolphins are not real?
Thanks Anne, but as I pointed out, there is a paradox.
A paradox that is resolved once you understand that the surface of the outer shell cannot possibly come to double the temperature of the sphere in the first place, for the reasons explained. The steel greenhouse analogy is flawed in this regard from the very outset.
(since the sphere is the only energy source, the only way it could warm to a temperature producing more than its maximum radiative output of 235 W/m2, is if the outer shell was warmer; yet the outer shell must always be at a lower temperature than the inner energy source object due to its greater surface area)
Good luck. It seems that there are a lot of people who do not understand the argument. Temperature (of a region) is proportional to the average kinetic energy of the molecules (in that region). But there are always molecules with below average kinetic energy in the same region with molecules of average and above average kinetic energy.
I’ve added a note to the head post with a link to my post on the Steel Greenhouse. It goes through these questions from a different perspective.
w.
You want to try nested steel greenhouses.
YES!! Thought experiments (I am presuming no one here has actually built a steel greenhouse but if you have lets see the pictures please) should always be tested via logical extension. Nested steel greenhouse is the next step. What results do we get? What conclusions can we draw?
“Now imagine that the planet gets completely surrounded by a thin black steel shell, located a few thousand metres above the surface, as shown in a cutaway view in the picture above”
Let’s change it. Imagine if earth had this “thin black steel shell, located a few thousand metres above the surface,”
Now there are few ways to do this and it does not matter *much* which way it’s done.
I going to pick having magically strong thin black steel shell which withstand more than 1/2 atm
of outward pressure and shell is going to be at 1/2 atm pressure- 5.5 km elevation.
So other chopping off a few mountain tops, I can keep earth, earth. Though missing half it’s atmosphere mass, though has the same surface air pressure of 1 atm.
Of course Earth isn’t warmed much from it’s molten core, and has to get it’s warmth from the steel shell.
So how warm is steel shell.
Well if it was was an ideal thermally conductive blackbody it “should” be about 5 C and radiate
1367 / 4 = 341 75.
But ideal thermally conductive blackbody is magic- and you are using steel which only magic added is it’s strong enough to withstand a huge amount of force. Such magic might gotten from nanotechnology which if build something molecule by molecule you could theory make a material stronger- though we have get into to how thin is thin and unless thin is meter thick, mere nanotechnology might not enough magi- but details we will ignore.
But steel or copper, silver or even diamond doesn’t give enough thermally conductivity- one could invoke some kind magical plasma based system, perhaps. But let’s look at steel as it’s the namesake.
So steel with a good blackbody coating is going to have lower average temperature than the uniform temperature of an ideal thermal conductivity blackbody.
Or in sunlight with sun at zenith, an ideal blackbody will be about 5 C. With steel it’s going to be much hotter than this and radiate into space much more energy as compared to the ideal.
At Earth surface and where sunlight would be at zenith, the steel sky will be quite warm, and elsewhere steel sky would be cooler.
How big is the hot spot, how big is the hottest spot of hot spot,
Roughly hot spot is same as solar peak hours- 3 hours before and after noon.
An hour is 15 degree longitude. So 45 degree both east and west of the point of zenith.
Each degree is about 111 km. 111 times 45 is 4995 km. So roughly a radius of 5000 km.
So if sun over equator 45 degree north, south, east and west. So if in UK, you don’t see the hotspot. You have wait for summer to see it- and then you will see it for couple hours a day.
Now at the equator you see it for 6 hours of 24 hour day.
How does the hot spot effect someone standing a equator at noon. More than the entire sky is
hot. How hot? Well roughly the hot as it could be is about 120 C,
It takes some time for the outer surface at 120 C to heat the inner surface- and the thinner the steel the less time. Let’s assume it thin enough to do this fairly quickly.
Next, there is 1/2 atm of pressure on the inner side- how cold it would be would effect how much heat is transferred to the air. The air not going to heat well because it’s like hot air against the ceiling of room in a house.
For rough idea with current earth, if air temperature was 20 C, it’s 6.5 C cooler per 1000 meter.
So 5.5 times 6.5 C is 37.75.
So air temperature was 20 C at surface it’s 17.5 C at shell. Now can’t have air at say 100 C and a foot away have air at -17.5. Or 100 C will make gradient of heat, maybe 20 C per meter.
So like the steel it takes some time to form this gradient.
So skip some these details what the effect if entire sky is 120 where closest is 5.5 km from you?
Well, it would seem if you were only 3 km from shell the radiant heat would more than if 5.5 km from it, though air is 3 time 6.5 C cooler than 20 C if surface air was 20 C= .5 C
So could like being in winter and stand next to a fire.
Or if at surface or at 3 km elevation there would direction the heat is coming from, and if at 3 km more heat is coming directly over head but unlike being at surface more heat coming all side save the below you.
Anyway rough fairly warm- but unless the air is warm, not particularly uncomfortable- or living constant darkness would be more of problem.
And if air was 20 C, the radiant heat would warm surface and warmed surface would warm the air.
And it seems the highest air temperature would not be at surface but at the ceiling.
Well I cut it short. And to remain brief, I think if lived in such steel greenhouse, you want to live on the ceiling to remain warmer.
Steel is a thermal conductor. Air is a thermal insulator.
Steel is a thermal conductor. Air is a thermal insulator.
Yeah, but steel is also insulator.
Or steel has conductivity of 54 W/m K
Copper is 386 W/m K
Air is 0.024 W/m K
Water is 0.58 W/m K
diamond is 1000 W/m K
I guess to quantify, look at formula:
https://www.engineeringtoolbox.com/conductive-heat-transfer-d_428.html
And just using calculator there
Per square meter which is 1 meter thick
temperature difference of 120 C to 0 C
Copper can conduct:46320 watts
Steel: 6480 watts
Air: 2.88 watts
Water: 69.6 watts
Diamond: 120000
But air which 1 mm thick: 2880 watts
1 cm: 288 watts:
So steel is 120 C it warm air near, and develop heat gradient,
Likewise the steel has gradient.
So if steel is 1 cm thick [very thin]
And top is 120 and bottom is 119 C
It gives 5400 watts
2 cm is :2700 watts
What about 120 to 119.5 how much watts go thru 1 cm:
That’s also 2700 watt.
So have 2 cm steel and sunlight of 1367 watts can heat surface to 120 C
and can heat 1/2 way down to 119.5 [or more]
And 119.5 to 119 C is again 2700 watts
Where is it 119.9 C in the 2 cm of of steel?
120 to 119.9 at 1 cm depth is 540 watts
And 1/2 cm should double, and it is: 1080
and 3.9 mm it’s 1385 watts
Somewhere around .1 K per 3.9 mm
Or 10 mm is 119.7 and 20 mm is .5 C
So 20 mm or 2 cm it’s 119.5 if surface it 120 C and radiating
close to 1367 watts, but some of that 1367 watts or
119.5 C is 392.65 K is 1348 watts. some of the 1348 watts
is lost to warming the air. And the warmer air gets, the less
heat is lost. Or if air within couple mm is close to 119.5 C
it won’t conduct heat to it- but doubt it would get to that point.
But warmer the air was say 1 meter away from surface, it seems
more likely it could happen.
Of course if there convection due gravity, hotter and lighter gases
would rise. But lesson of fire safety is if in a fire, crawl out rather than
walk out, because you can get very hot air closer to ceiling.
So if there isn’t any wind [and there could be] one can trap a small amount of
hot air near the ceiling.
This assuming, less an inch of the magical steel is strong enough.
Fig 2 is complete nonsense. The surface does not absorb twice as much energy from the atmosphere as it does from the sun.
“If I can see you, you can see me, so there are no one-way energy flows. Which means that if I am absorbing radiation from you, then you are absorbing radiation from me.” More nonsense. What you are seeing is light reflected from the other person. Switch off the light and you can’t see the other person, even if he is radiating infra-red.
“More nonsense. What you are seeing is light reflected from the other person. Switch off the light and you can’t see the other person, even if he is radiating infra-red.”
So you are saying that (say) a pyrometer in a room at ambient 20C cannot see an object at the bottom of a chest freezer at -20C?
Or that when pointed at a clear sky it cannot see/measure it’s temp?
Roy Spencer shows you can here …
http://www.drroyspencer.com/2010/08/help-back-radiation-has-invaded-my-backyard/
No I didn’t say that. I was talking about people seeing each other – ie just the visible spectrum, which is what people see as a reflection off other people, not an emission from them.
That is correct. Take a totally dark room. Shine a light on you. I can see you, you cannot see me.
Oh, for chrissakes, Phiip!
He was just expressing colloquially the FACT that the path that electromagnetic radiation takes from Point A to Point B is the same as it takes from Point B to Point A. That (technically true wavelength by wavelength), combined with the fact that absorptivity and emissivity are the same for each wavelength, validates his argument.
Willis is correct. For those who think that the temperature of a body receiving radiant energy does not affect the temperature of the hotter radiating body, here is a simple experiment.
Connect an electric resistance heating element to a constant power output power source. Attach a temperature sensor (e.g. a thermocouple) to the heating element and measure its temperature in a large open area until it comes to equilibrium. Say a 1000 Watt element heats to 500 C in this situation.
Now suspend the heating element in a small sealed ceramic box under vacuum (so the only mode of heat transfer is radiation) and monitor the temperature of the box surfaces and the heating element from the time the element is energized. You will see that the temperature of both the box and the element increase until the heat loss from the box equals the heat input to the element (i.e. 1000 watts). If the box starts at 20 C and the element at 500 C and the box inside surfaces end up at 300 C then the heating element must increase in temperature to 549.43 C in order to maintain the steady 1000 watt output. And the outside of the box must transfer 1000 watts to its surroundings although this heat flux will be by a combination of radiation, convection and conduction.
Initial condition SB Temperature term = (500+273.1)^4 – (20+273.1)^4=3.49 E11
Equilibrium SB Temperature term = (549.427+273.1)^4 – (300+273.1)^4=3.49 E11
This experiment is relatively easy to do if you have access to a laboratory vacuum furnace. The results would be only slightly different in a furnace with air at ambient pressure.
In my opinion Willis has used the wrong relationship to answer the question he has posed. He should have used Planck’s Law, not the Stefan-Bolzman’s law.
The reason is that he is considering an object A emitting at wavelength X and
B an object emitting at wavelength Y
where Y is longer (cooler) than X
Has object B caused object A to emit at a wavelength shorter (warmer) than the wavelength A would emit at if B did not exist?
Or alternatively, if object B were to cease to exist, would object A emit at wavelength longer than X?
Willis need three assumptions, no heat transfer
by conduction
by convection or
by reflection.
The Stefan-Bolzman Law cannot answer this question because it is derived from Planck’s Law by integrating across all wavelengths.
This is explained on another skeptical blog, The Science of Doom,
here https://tinyurl.com/y8d7gor8
here https://tinyurl.com/y77qnqpl
The experiment proposed by Rick C PE describes a closed system. My understanding is that Willis is not imposing that condition.
Well done. I see the usual arguments above, all stemming from the completely incorrect “greenhouse” tag on the effect of atmosphere (any atmosphere). It should properly be called the “shield” effect, allowing a more rational discussion.
For instance – it is obvious that making your shield more effective – i.e., adding a gas that makes it reflect (or absorb and emit) more heat towards the surface – will make the surface warmer than it would be otherwise.
No rational “skeptic” denies that adding more shielding CO2 does increase the surface temperature. Less energy escapes directly to the far colder surroundings with that additional CO2 in the atmosphere.
The inconvenient truth, however (for alarmists) is that the net effect is very small. Two facts come into play – the effect of additional CO2 is logarithmic. Generously, humans have added approximately 150 ppm of CO2 to the atmosphere, bringing its concentration to around 400 ppm. In order to add as much CO2 shielding effect again as we already have, we would have to burn more than twice as much fuel as we have already in our entire history since the start of the industrial revolution. That actually is not an easy thing to accomplish, even if we try very hard. Which, looking at the recent history from even the rather dodgy measurement systems, we aren’t.
So the alarmists rely on the slight increase in temperature from the CO2 shield effect to vaporize a far more effective shield gas – dihydrogen monoxide (water). This is touted as the “fatal multiplier” that will roast our world.
But the alarmists just can’t win. Mother Nature (or God, or quantum physics; whatever fountain you drink from, it’s the same flavor) has arranged things so that the “fatal multiplier” is actually a “saving divisor.” The slightly higher temperature at the surface does evaporate more water into the atmosphere. Fortunately (for humanity in general, not the alarmist’s grant prospects, or long term investors in “green” companies), this temperature also creates stronger upwelling as the hot air rises, carrying the water vapor along with it. Until it reaches the upper atmosphere – where the water vapor condenses, releasing energy where there is virtually no shield between it and the far cold reaches. This natural brake on the temperature means that, at the most, the Earth can warm an average of 14 degrees (Fahrenheit – about 7 degrees Celsius). Not that we can manage to add enough CO2 to get that high – you need 7,000 ppm for that – burning forty-five times as much fuel as we have already. Their best hope is for massive vulcanism, although that just might tip us over into an ice age from aerosols before the additional CO2 can get to work.
(The honest climate researcher, of course, knows that even the above is highly simplified – but covers the most significant drivers of climate. The “fiddly bits,” such as changes in albedo, lower in the less icy northern latitudes – somewhat offset by more water surface; outgassing of CO2 from warmer water; sinking of CO2 both geologically and biologically; et cetera, et cetera, et cetera.)
“No rational “skeptic” denies…”
No true Scotsman denies…
‘Tis a bit of a difference between denying who is the rightful King, and denying a physical phenomenon. As all too many Scots with too much Scotch in them have discovered. You might escape the usurper’s men for your intemperate words, but not the ground when you trip on your feet instead of your tongue.
No, yours was a perfect example of the no true Scotsman fallacy.
Looking at your other posts, it should not surprise that you have no better idea of the “no true Scotsman fallacy” than of physics.
Not having the patience of a Job – or even a Willis – last reply here.
Sorry you were busted.
https://en.m.wikipedia.org/wiki/No_true_Scotsman
I would point you to the last few words of that definition (which is a good one, by the way). The objective facts of physics do not change – not for alarmists – not for skeptics – not for Scotsmen.
Tony, you embarrass yourself.
More clearly-
“The No True Scotsman (NTS) fallacy is a logical fallacy that occurs when a debater defines a group such that every groupmember possess some quality. ”
Writing observer did not state or imply that all skeptics are rational, so in order to be a “true” skeptic one must have that quality. By including the qualifier “rational” before the word “skeptic”, he implies that some skeptics are NOT rational.
The group was skeptics, the exclusion is to dismiss those that disagree with the GHE as irrational. Not difficult to understand.
Even disregarding the effect of H2O vapor upwelling, it seems that there is a leap of logic in alarmists’ reasoning. Are they ignoring the logarithmic characteristic of CO2 (which I assume is due to saturation effect?), or (if not ignoring that) are they assuming (or do they possibly have evidence for) that the increase in H2O vapor due to (CO2 caused) warming would not follow the CO2 increase linearly?
The narrative is that more CO2 causes more H2O, which causes more H2O, which causes even more H2O.
If the pesky molecule would only stay put on or near the surface, they’d be right.
“The narrative is that more CO2 causes more H2O, which causes more H2O, which causes even more H2O.”
Which will cause more convection of wetter air flattening the lapse rate and causing more condensation removing more heat from the surface. It isn’t even clear that the H2O feedback is positive.
It is, in the minds of climate modelers, tty. Observations do confound them, though.
“No rational “skeptic” denies that adding more shielding CO2 does increase the surface temperature. Less energy escapes directly to the far colder surroundings with that additional CO2 in the atmosphere.”
I deny that, CO2 slows the cooling, the incoming visible radiation provides the warming. Never will you see the GHG affect warm the earth at night…never (ignoring conduction and convection). CO2 is a blanket, not a heating source. The incoming radiation warms the earth, CO2 slows its cooling.
I’m leaning your way. It slows the cooling.
Yep, unless it is from a chemical reaction where energy is changed in form, you won’t heat a body above the temperature of the radiating body.
For some reason I like the word retards rather than slows. Here’s a question: Does radiant heat act like electric current? IN other words, for there to be radiant energy transfer (heat) does there have to be a difference in potential of some form? Kind of like current (analogous to heat) doesn’t flow until there is a voltage differential. Might sound like a dumb question to some people, but yano I’m just a student of the masters.
Great point, the atmosphere is like an electrical resistor. It impedes the path to outer space.
ICISIL:
My heat transfer texts, in their chapters on radiative heat transfer, are full of “electric circuit” diagrams with resistors for what they call “network analysis” of these transfers.
Remember that a “transparent” atmosphere (no GHGs) provides no resistance to radiative energy transfer.
“Slows” is a misleading term; “retards” may a little better. I prefer “reduces”, myself.
Excuse me? You just restated exactly what I said.
Less energy escapes directly to the far colder surroundings… Emphasis added!
No, CO2 does not heat the surface (nor does any other component of the atmosphere). But what both Willis and I are trying to get through heads is that the atmosphere affects the net heat flow, reducing the net flow of heat from the surface (whether day or night) to space.
Ignoring, of course, the effect of convection, which transports surface heat (in the form of latent heat in water vapor) to an altitude where there is far less atmosphere – thus increasing the net outflow of heat from the entire system.
It’s called a Thermal Insulator. From a grade school website:
“Materials that are good conductors of thermal energy are called thermal conductors. Metals are very good thermal conductors. Materials that are poor conductors of thermal energy are called thermal insulators. Gases such as air and materials such as plastic and wood are thermal insulators.”
“One way to retain your own thermal energy on a cold day is to wear clothes that trap air. That’s because air, like other gases, is a poor conductor of thermal energy. The particles of gases are relatively far apart, so they don’t bump into each other or into other things as often as the more closely spaced particles of liquids or solids. Therefore, particles of gases have fewer opportunities to transfer thermal energy. Materials that are poor thermal conductors are called thermal insulators.”
https://www.ck12.org/book/CK-12-Physical-Science-Concepts/section/5.17/
I’m sorry Writing Observer-
You state first-
“No rational “skeptic” denies that adding more shielding CO2 does increase the surface temperature.”
Then state- “No, CO2 does not heat the surface (nor does any other component of the atmosphere).”
There is a difference between something that “heats” something else-causes it’s temperature to increase-and something that slows down the rate at which something COOLS.
“But what both Willis and I are trying to get through heads is that the atmosphere affects the net heat flow, reducing the net flow of heat from the surface (whether day or night) to space.”
Affect it YES….reduces the net flow of heat from the surface to space? Nope…merely slows the rate of flow. Why? Because the atmosphere ALSO reduces the flow of heat from the SUN to the surface. It slows down the rate at which the Sun can warm the surface in a 12 hour period, and then slows down the rate at which that warmth returns to space, but it does NOT reduce the NET flow of heat from the surface or from the SUN. It does not “cool” the Sun anymore than it “warms” the Earth. It’s a SUCKY conductor of thermal energy. It’s an insulator.
If I pour hot coffee (100 F) into a plastic thermos (that contains an air pocket inside it’s walls for insulation) that is “cold” (compared to the coffee) but is at room temperature (let’s say 70 F), and then put that thermos outside in a snow bank in “colder” 10 F weather, does the “cold” previously room temperature thermos have the ability to WARM the already HOT coffee by “hiding it” from the colder snow and outdoor temps or does it simply slow down the rate at which the coffee INSIDE THE CONTAINER cools???
If I poured the original 100 F coffee onto a saucer made of metal, and placed it in the same spot outside in 10 F, the rate at which the coffee cools would be MUCH faster. Because metal is a great conductor of thermal energy. Far better than the air pocket inside the plastic thermos was. But the fact remains that the coffee began to COOL the instant the heat source (stove, coffee pot) was removed.
The net heat transfer in both scenarios is exactly the same…one just took a lot longer than the other. The net transfer of $25 dollars to “you” from “me” in Willis’s scenario would have been the EXACTLY the same whether “me” paid “you” one penny a day for 2,500 days or handed me $1.00 for 25 days, or wrote me a freaking check for $25 and handed to me in 20 seconds!!!
Words matter. Scientific principles matter. Spontaneous heat flow will ALWAYS be from a warmer object to a colder object. Net heat transfer does not END between two objects of different temperatures until both are in equilibrium and reach the same temperature.
Yes, the whole question is does a change of 100 ppm CO2, 0.0001% of the atmosphere, alter the Net heat flow, that is the whole question. Using the thermos example, you fill the Thermos with Hot Coffee, and then let it cool over night, and they you add the same amount of energy over the next day to warm the Coffee again, and then let it cool. Over time, does that does adding the same amount of energy to a thermos throughout the day result in gradual warming because the energy leakage is less than the energy added each day. The marginal difference is what would result in the warming.
Aphan, you say: “Because the atmosphere ALSO reduces the flow of heat from the SUN to the surface.”
No! The whole point of the “greenhouse” metaphor (and it is just a metaphor) is that the atmosphere is far more transparent to solar (visible and shortwave IR) radiation than it is to terrestrial longwave IR.
Yes!! The atmosphere DOES absorb incoming heat and radiation. Learn something:
https://earthobservatory.nasa.gov/Features/EnergyBalance/page4.php
“About 29 percent of the solar energy that arrives at the top of the atmosphere is reflected back to space by clouds, atmospheric particles, or bright ground surfaces like sea ice and snow. This energy plays no role in Earth’s climate system. About 23 percent of incoming solar energy is absorbed in the atmosphere by water vapor, dust, and ozone, and 48 percent passes through the atmosphere and is absorbed by the surface. Thus, about 71 percent of the total incoming solar energy is absorbed by the system.”
“and it is just a metaphor”
No, Its a FALLACY !!
Only statement above that has meaning in the real world:
So just to recap these ‘radiation budgets’ are undoubtedly scientifically correct…they are also to all extents and purposes irrelevant or to put it another way…negligible.
http://tropic.ssec.wisc.edu/tropic.php
I think that you’re dead right Willis. AFAICS, the only reason that the Second Law of Thermodynamics isn’t written in terms of net flow is that the second law was derived in the nineteenth century from the ideal gas equation pV=nRT using mental models where all heat flow was by convection/conduction. Back flows are (usually) still there with convection and conduction. But they are difficult to observe and measure.so the second law is stated in terms of heat flows without the term “net”. They are, in fact, net flows.
When one deals with energy transfer by radiation, one has to make a few adjustments to one’s mental models including recognizing that flows are net flows, and also including the property of emissivity at the source and destination. If one doesn’t do that, one will find that perpetual motion is not only possible, but easy.
I think many of us might prefer to live in a world where perpetual motion machines can be built and do useful work. But it’s clearly not the world we actually live in.
Hi Willis! Your argument is correct for two emitters with independent energy sources (e.g. two glowing light bulb filaments, two Suns, etc.). However, the Sun does not emit any significant far-infrared radiation at 667 cm^-1 which can be absorbed by CO2 in the atmosphere. The 288 K surface of the Earth, however, emits far-infrared radiation whose Planck black body curve peaks near 667 cm^-1. CO2 is such a powerful absorber that 667 cm^-1 radiation emitted from the Earth’s surface is almost 100% absorbed within metres of the surface. By Kirchhoff’s Law that a good absorber is a good emitter, CO2 will emit almost 100% at 667 cm^-1 (parabolic dish antennas are not only good radio receivers, but are also good radio emitters/radiators). But the excited CO2 molecules formed on absorption of 667 cm^-1 photons can also lose their excitation energy during radiationless collisions with the main molecules of the air (N2, O2, Ar) which cannot and do not re-emit any significant amount of infrared (IR) because their molecules do not possess any permanent electric dipole moment. The energy does not go away, but ends up as translational and rotational kinetic energy of the departing molecules. The heat capacity at constant pressure for linear molecules like N2, O2 and CO2 is 7k/2 per molecule, where k is Boltzmann’s constant. Since N2 and O2 together outnumber CO2 by 1,000,000:400 = 2500:1, almost all of the absorbed energy ends up warming N2, O2 and Ar molecules (and is stored as enthalpy, heat content, H). This is the mechanism of the greenhouse effect, and has nothing to do with back-radiation. Back-radiation does exist, but it can be thought of as energy flow that just balances an equal energy flow for two surfaces at thermal equilibrium, with no net change in the temperature of either. Some might argue that excited CO2 molecules can also be formed on collision of ground state CO2 molecules with high-energy (fast-moving) air (N2, O2, Ar ) molecules, and that powers the back-radiation. But this energy would come from the air molecules, whose average kinetic energy (temperature) must decrease; no, at thermal equilibrium there can be no net warming of the Earth’s surface by back-radiation alone. Here is an analogy that might make this more clear: suspend two flat parallel black metal plates in a blackened vacuum chamber cooled with liquid helium to 4 K, close to the 3 K of the cosmic background microwave radiation. If one metal plate contains a 100 W heating coil, and the other is a passive radiator without a heating coil, at thermal equilibrium 50 W will be emitted from the outside surfaces of the two plates, for a total of 100 W outward toward the 4 K inner walls of the enclosing vacuum chamber. But the space between the two plates will have 50 W exchanged back-and-forth, so there is no net heating or cooling of the plates. If the two plates are moved together until touching, the total output will still be 100 W, with 50 W outward from each surface. But the back-radiation will, like the forward radiation it balanced in the previous gap, disappear with no change in temperature or output. Note that when the two plates are separate, 50 W is emitted from both sides of the plate with the heating coil, and 50 W from both sides of the passive plate, for a total of 200 W emission. How can this be powered by the 100 W heating coil? The Law of Conservation of Total Energy is not violated because the 50 W back-radiation from the passive plate to the plate with the heating coil just balances 50 W in the opposite direction, so the back-radiation photons can be considered to be photons initially emitted from the powered plate and then reflected back to the power source. This is not the same as photons reflected from a mirror, because a mirror does not emit in the forward direction, so the mirror analogy is wrong for CO2 in the atmosphere which absorbs and re-emits in the forward direction.
Suppose the surface temperature of the single powered 100 W plate is T0. If another identical black 100 W plate is brought close to and parallel to the first, this time the surface temperature will rise by a factor of the 4th root of 2 = 1.1892. This time, “back-radiation” will heat up both plates because those photons come from separate power sources. This may be seen when the two plates are brought together, forming a single plate with a 200 W internal power supply, so that 100 W will be emitted from each of the two remaining surfaces to the surroundings. This is double the 50 W from each of the two surfaces of the original 100 W plate, requiring a higher surface temperature for emission by the Stefan-Boltzmann Law.
Seriously good post (From a mobile?) however, could do with some formatting, just a bit too busy for my eyes in one big chunk as it is.
Here is an observation that demonstrates ‘delayed’ cooling. Wait for a warm, calm, sunny day and note the temperature. As the sun sets below the horizon the temperature starts to fall. While the sky remains clear overhead the temperature continues to fall. Then a cloud bank arrives overhead. The temperature stops falling! In fact it will start to rise. I have noted a temperature increase of over 5C under these conditions. Remember, it is pitch dark, the sun is on the other side of the world. So what is causing the warming? It is not the wind which I said was calm, it is in fact back radiation from the cloud base. This radiation from the cloud, impacting the surface, is reducing the cooling rate of the surface. Subsurface heat from the previous day is still rising. As the surface temperature rises the net radiation increases, the back radiation from the cloud increases.
The heat capacity of the ground is considerable. To really experience this go live in a desert region for a while. Here sunlight, day after day, really does warm up the ground yet just before dawn the local temperature will be close to freezing. The outer surface of the rock is trying to shrink. Often a piece of rock spats off with a loud crack. Find that rock and feel the new exposed surface. It will be quite warm!
Eventually there will be a lot of sand. Dig into that sand on a cold dawn and you will fell a lot of warmth. Dig a little trench, lie down, and enjoy some wonderful star gazing. Mind the scorpions!
There are clearly 2 schools of thought.
1. Willis is right and ALL e-m radiation that “hits” an object is absorbed and thermalized. Temperature changes are thus based on NET radiation (in versus out).
2. Willis is not even wrong. ONLY e-m radiation from a hotter object can be thermalized in a cooler object.
This needs to be resolved before one moves on to any discussion about CAGW, atmospheric physics, etc. You need to get the FUNDAMENTAL workings of radiation and heat transfer correct FIRST. Otherwise you will be building everything else on a false foundation.
For those so inclined I invite you to invent an experiment to test which explanation is correct, or perhaps find the results of an experiment which has already been done. The experiment must clearly differentiate between the 2 fundamentally different explanations of course.
This is a useful excercise generally as you will often find arguments in science where there are 2 competing views. They are not always resolvable via reference to literature as each camp obviously has well developed arguments for its own case. In all cases it pays to carefully study BOTH sides even if you are pretty sure yourself what you believe. I recommend it in this case.
Good luck with it.
I would be interested to hear of any experiments regarding e-m radiation being thermalized always/sometimes.
If em was thermalized all the time the hottest spot in America would be the base of the antenna of a 50000 W radio station. But it isn’t.
mkelly,
If you live just under that station, you can light a fluorescent tube by only adding an antenna at one side and a ground cable at the other side…
That may be true but it does not make it hot. We used to do that with radars in the Navy. The old APS20 would if given a chance heat soup but it would not hezt air.
The Reverend,
Many thousands of CO2 lasers prove daily that point 2 is wrong. These emit IR at between 9.600 en 10.640 micrometer. That is comparable to the peak radiation of a black body at around a human’s body temperature, not really “hot”.
According to point 2, that could warm up steel to maximum the same temperature, not further up.
In the real world CO2 lasers heat up steel and melt it at 1200ºC…
Which proves that Willis is right…
http://hockeyschtick.blogspot.com/2014/11/why-cant-radiation-from-cold-body-make.html?m=1
“The radiation from the cooler source doesn’t have the high frequency light-wave energy components which would be required to fill up the higher-energy microstates that the warmer source already has filled up, let alone the higher frequency states beyond to result in an increase in temperature. You can see it a little bit in the Planck curve plot above, that the higher-temperature object has a microstate population that is “activated” at higher frequencies, i.e. shorter/smaller wavelengths, than the lower-temperature object has. (The warmer one also has a larger population in each frequency microstate as compared to the cooler one.)
The integration over all the active microstates of a system determines its macrostate, and the macrostate is the thing you actually measure to take a temperature. If you activate higher-frequency microstates, then you shift the macrostate to a higher temperature. But you can only activate the higher-frequency microstates with the frequencies required to activate them, which are obviously their frequencies.
And so because the radiation from a cooler source lacks the higher-frequency microstates that the warmer source of radiation already has, then the cooler source can not have the effect of raising the temperature of the warmer source. And obviously the same would be true of two systems with identical temperatures – they could have no effect on each other’s microstate population, hence can not heat each other.”
All the math is there too, and a link to the steel greenhouse debunking.
Aphan:
you can only activate the higher-frequency microstates with the frequencies required to activate them, which are obviously their frequencies.
A CO2 laser only sends IR at a frequency of around 10 μm. Steel at 1000 K has its peak frequency at around 5 μm, thus the CO2 laser beam can only activate some 30% of the iron atoms in the steel, if that story was right.
That means that only individual atoms with a “temperature” – or energetic level – between 0 K and about 300 K (the corresponding frequency level of the beam) can be heated by the laser beam. Or an average increase to about 1050 K for steel as a whole. Not enough to melt it. In reality steel is melted by a laser beam of only 10 μm waves up to 1500 K.
Something wrong with the theory?
Ferdinand, What is the Power of the Laser, not the Frequency?
Ordinary Cold Water can be used to make a hole through steel, how do you think that works?
It works by having enough Pressure, is not Power the same as Pressure?
A C Osborn,
A water “beam” can dig a hole in steel due to kinetic energy. That is mechanical energy, as good as drilling or using a hammer and chisel.
Power is not kinetic energy, it is electrical energy which in a laser is transformed into electromagnetic energy, which if absorbed by an object is transformed into vibrational energy of the atoms/molecules. Vibrational energy is temperature…
“BUT it can leave the hot object warmer than it would be if the cold object weren’t there. ”
No need for a large article
The cold object provides a temperature differential potential to allow heat to move from the hot object to the cold object, no colder object, no loss of heat from the warmer object to the colder object.
Done
Mark,
Well said – at last a sign of sanity in this crazy debate. Life is so simple unless one has a compulsive desire to over-complicate it. It seems that almost everyone else here is bent on grand-standing and mindless prevarication.
Simple standard physics says…
1. The surface of an object X at a temperature Tx will assert a radiative potential Px where Px is proportional to the 4th power of Tx (S-B equation).
2. The surface of an object Y at a temperature Ty will assert a radiative potential Py where Py is proportional to the fourth power of Ty (S-B equation).
3. If the surfaces of the two bodies X and Y are facing one another, and if Tx > Ty, then the rate of transfer of radiative energy between them is simply Px – Py; and the direction of energy flow is from the warmer body X towards the cooler body Y.
In the whole of physics is there anything easier to grasp than that? It applies under all situations everywhere in the universe. In particular it applies in the case of the earth’s warmer surface facing its cooler atmosphere.
In Willis’s energy budget diagram, Px = 392W/m^2 and Py = 321W/m^2. One conceptual trick that I use is never to think of these two numbers as independent energy flows, but as calculated potential flows. After all, they clearly do not exist in isolation. On the contrary they are completely bound up together by the geometry of the two facing surfaces. If one thinks of them as independent, one can easily get into the silly position that some people have done of thinking that the 321W/m^2 of ‘back radiation’ is somehow violating the 2nd law because it is larger than the incoming Sun’s radiation of 169W/m^2 absorbed at the surface. In reality it is always only the net radiation (in this case 392 – 321 = 71W/m^2) that counts.
It is also vital to appreciate that all of the above is true irrespective of any non-radiative energy transfers that may also be occurring at the same time. In Willis’s energy budget diagram, the non-radiative energy transfers from surface to atmosphere are 22W/m^2 (sensible heat) and 76W/m^2 (latent heat). Adding these on to the net radiative energy flow of 71W/m^2 makes a total of 22+76+71 = 169W/m^2. This exactly balances the amount of Solar radiation absorbed by the surface, as must be the case for steady-state temperatures. So the non-radiative and radiative energy flows coexist happily.
Bingo!
David, a simplified discussion and a question:
1) Addition of greenhouse gasses (GHG) to the system would increase back radiation, thereby reducing the net 71W/m^2 surface emissions.
2) The surface must heat in order to restore the 71W/m^2 necessary to balance input/output.
3) The net contribution of man’s production of CO2 (net of sinks) would necessarily heat the surface in order to get H2O feedbacks, positive, negative or neutral.
4) There are massive and constantly changing energy flows within the earth’s climate system. Much of this is the result of chaotic weather systems.
Is the radiative contribution of man’s net CO2 lost in the noise?
David, thank you for simply stating what should be obvious to an educated person.
It should be noted that fluxes in Trenberth’s diagram are based on averages and are presented as in a steady state. But our water world is turbulent and chaotic, with all of Trenberth’s noted fluxes changing on all time scales, except possibly the isolation of the growth of CO2 equivalent fluxes for analytical purposes.
I’m not a researcher so I keep asking simple questions: In isolation, how much of the indicated back-radiation is the result of man’s contributions to CO2 equivalent fluxes? What is the resulting theoretical warming from such contributions, in isolation. Is this theoretical warming even discernible in our turbulent and chaotic water world?
Again, it’s had a name and definition for decades. It’s called a Thermal Insulator. Thermal insulators are the opposite of thermal conductors. Air is a lousy conductor of thermal energy.
“The temperature and the radiation are related to each other by the Stefan-Boltzmann equation” Only for black bodies and by extension for the mythical gray bodies, too. For objects where absorption / emission is frequency dependent (that is, quite a LOT of them), where the radiation is not fully thermalized (which is the case for the atmosphere), it does not apply. Ex falso, quodlibet.
Pseudo sciences crap on the boundary conditions and will apply at ease for example theorems even if the assumptions they start from when proved are plain false. With the principle of explosion, such pseudo sciences can derive anything they intend to.
All the ‘extra’ radiation held within an atmosphere as a consequence of the heat capacity of the atmosphere is utilised in potential energy form ( not heat) for the purpose of holding the atmospheric gases off the surface against the force of gravity.
Thus none of that ‘extra’ energy is available for further warming the surface above S-B.
Such further warming of the surface arises not from DWIR but from the return of potential energy to kinetic energy beneath descending convective columns of air.
The cold atmosphere does heat the warmer surface but due to the gas laws that warming is a result not of DWIR but of such conversion of PE to KE
Quite – there is a whole industry of it on both sides of the debate. A great shame.
Adrian, you say: “For objects where absorption / emission is frequency dependent (that is, quite a LOT of them), where the radiation is not fully thermalized (which is the case for the atmosphere), it [the SB equation] does not apply.”
Absolutely not true! Explanations of radiative heat transfer almost always start with blackbodies, next covering graybodies, because these simplifications make it easier for the beginning students to focus on the important concepts rather than the mathematical minutiae.
If you look at the equation Willis posted for radiative transfer between graybodies, everything inside the big left parentheses is needed for graybodies and not for blackbodies (as he shows).
For “real world” bodies, the emissivity is not constant over wavelength, as it is for idealized blackbodies (e=1.0) and graybodies (e less than 1.0). So you must evaluate each wavelength band invidually. The MODTRAN database does this to moderate resolution; the HITRAN database does this to high resolution. But the physical principle is absolutely the same in all cases.
That’s a bunch of anti-logical arguments. You either don’t know what Stefan-Boltzmann law is or you only play ignorant. Stefan-Boltzmann is about total energy, not about ‘lines’. And it’s about full thermalization, not about non-equilibrium situations. For your info, energy going ‘in’ for a line can go out through other ‘lines’. Despite radiation not being fully thermalized, quite a bit of CO2 collisions will lead to non radiative transitions. The energy going ‘in’ can go ‘out’ after a sensible time, it does not need to go out instantly and there is no ‘radiative balance’. That renders Stefan-Boltzmann law useless. The energy can go ‘in’ as radiation in a place at one temperature and can go out in a different place with quite a different temperature. Stefan-Boltzmann law requires a single thermodynamic temperature, that is, equilibrium, which is not the case for Earth’s atmosphere.
MODTRAN/HITRAN is a red herring and for your info it has enough error in results (especially where H2O is more involved) to be able alone to give the ‘garbage in’ for models to be exponentially amplified in the exponentially grown pile of shit the models output.
Adrian:
You say that the SB equation applies “only for blackbodies and by extension for the mythical gray bodies, too.” It’s only slightly more of an extension to “objects where absorption / emission is frequency dependent”.
If you’re trying to be a pedantic nitpicker, you should be correct in your nitpicking. You should be talking about absorptivity and emissivity, which are different things from absorption and emission.
Besides, the graybody approximation is a VERY good approximation for most solid and liquid substances at earth ambient temperatures.
You imply that when there is frequency dependence of a and e, it is because “radiation is not fully thermalized” (although it might be awkward phrasing on your part). Two points. First, a/e frequency dependence is fundamentally independent of full thermalization. They are separate issues.
Second, the atmosphere throughout the full troposphere fully thermalizes absorbed radiation, counter to your assertion. A molecule excited by absorbed radiation is about a million times more likely to collide with another molecule while still excited than it is to re-emit before a collision. (Of course, this is even more true of solid and liquid substances.)
Besides, your arguements are fundamentally irrelevant to the qualitative argument presented.
“the graybody approximation is a VERY good approximation for most solid and liquid substances at earth ambient temperatures” It is extremely bad and it’s again a red herring, the cliamastrological religion has delusions about the atmosphere, which has a spectrum which is very far from that of a black body.
“the atmosphere throughout the full troposphere fully thermalizes absorbed radiation” As I explained, it doesn’t do that as required for a black/gray body, so your anti-logic is anti-logic and nothing else.
“your arguements are fundamentally irrelevant to the qualitative argument presented” They are fundamentally relevant, because such ‘qualitative’ anti-arguments will treat wildly non-linear systems as being linear, non-equilibrium systems as being at equilibrium, white as black and false as truth.
Ex falso, quodlibet.
Oh, and since you appear to have absolutely no idea what thermalisation means, start from here to rise yourself above the total ignorance you exhibit: https://en.wikipedia.org/wiki/Thermalisation Maybe you’ll figure out why I stressed that it’s not equilibrium.
Transfer of kinetic energy. Temperature is a misnomer. We must talk of it in terms kinetic energy
Temperature can be cumulative in directional kinetic energy application, increasing pressure but it requires an external input of kinetic energy to achieve this, it does not happen in passive kinetic energy exchange.
Passive kinetic energy exchange is decided by the difference in kinetic energy between the two objects,
Leave two blocks next to each other,
*one cooler one hotter,
*The kinetic energy transfer is passive, the warmer block loses kinetic energy to the cooler block (difference potential)
Smash the cooler block into the warmer block
* You are inputting kinetic energy into the exchange that is greater than the kinetic energy of passive transfer * The cooler block uses the kinetic enegy from momentum to warm the warmer block.
All about passive vs non passive transfer of kinetic energy
The cooler block uses the kinetic enegy from momentum to warm the warmer block *on impact
Which increases the rate of kinetic energy. KE\Time is what changes and overrides passive transfer
Ok, this is a subject I must confess to struggling with – but I very much appreciate this:
“These ideas of individual flows, net flows, and being shielded from radiation are important because people keep repeating over and over that a cold atmosphere cannot warm the earth … and they are right. The temperature and the radiation are related to each other by the Stefan-Boltzmann equation. When we apply the S-B equation to the 321 W/m2 of downwelling “back radiation” shown in the graphic above, it tells us that the effective radiating level is somewhere around freezing, much colder than the surface.”
Terms like “backradiation” and “downwelling” sort of add to the confusion by making it sound like one should be able to feel it – when it’s just shielding us from the cold of outer space – which is what we learned in grade school.
At the same time, the atmosphere as a whole functions, not just to shield us from the cold of outer space, but as a heat dispersal system as well.
It is impossible physically for a cool object to warm a warmer object unless the cooler object can make use of additional energy from outside of the two object equation.
Otherwise kinetic energy will always be passive and always be dictated by difference potential.
Exactly!! There has to be an additional source of energy to perform the work required.
I grew up in my fathers steel fabrication business. By the time I was 12 I was welding and fabricating and sometimes blacksmithing using a forge. One thing I know is that if one heats the end of a piece of a bar of carbon steel of any solid shape or compound that is a foot or a few feet long until it is cherry red and then quenches the hot end while holding the cooler end, the end that they are holding will become hotter very quickly after they have submerged the heated end in water or oil.
I love the way 5% carbon steel can be made into a chisel (Brown) or a spring (Blue) simply based on heat during hardening and colour during tempering. Now, this is a very very old memory.
You can straighten bowed or twisted “I” beams, even very heavy ones, using only a rosebud tip on an oxy-acetylene torch and some water soaked rags. One just has to know where to apply the heat.
It’s because steel is an excellent conductor of thermal energy, liquids less so, air even less than that.
The truth is, as things stand, it is impossible for us to even remotely know what earth’s energy balance is.
Impossible for us to know where all the kinetic energy goes.
Any claims otherwise are either utterly delusional or dishonest.
Logical examination is dead in science, your solution must be logical first and foremost
Too many get lost in the mathematics and lose all logical structure in their solution
See: actual physical singularties, produced from thin air by some of our “greatest minds”. There is obviously no such thing in the physical world
Relative mass is bunk, it requires transfer of kinetic energy between objects to manifest the extra kinetic energy from momentum
So an object moving at say half the speed of light has no relative mass, the mass is exactly the same as a static object, the energy that is changed is kinetic energy and that is only manifested when there is a kinetic energy transfer.
The so called extra mass that is in relative mass, does not belong to or come from that object, it is external kinetic energy provided by another source.
I can assure you that even the relatively modest speed of an electron in a CRT (like in an old-fashioned TV) will make its movement through the electric field come out significantly wrong if you don’t take the relativistic mass increase of the electrons into account. I did an experiment on this when reading freshman physics, so I know it from personal experience.
you obviously dont understand, relative mass is mass+pseudo mass ie mass and momentum
they can be calculated as one, BUT the Electron does NOT have more mass
As I clearly explained, if there is physical interaction then this kinetic energy comes into play, and in a CRT that is the case.
If the electron does not physically interact then the case is different, maybe read first then reply 😛
If you work this problem of energy balance in terms of kinetic energy, it becomes clear that we have no way to work out this problem and are left with mathematical fantasies and severe top level averaging that does not represent the problem, or provide a solution.
We’re dealing with ghosts here
Can A Cold Object Warm a Hot Object?
No. End of discussion about radiation.
Can a cold object make a warm object warmer than it would have been if the cold object wasn’t there?
Only if it insulates the warm object. If you insulate a warm object, the rate of cooling decreases.
What has this got to do with Carbon Dioxide’s contribution the ‘Greenhouse Effect’?
Hardly anything in practice. Carbon Dioxide is not an effective insulator for two reasons:
1. It is a poor insulator by itself.
2. There is not enough of it to make an appreciable difference.
To Willis’s figures:
Figure 1:
This is NOT a good analogy!
According to Willis, doubling the number of ‘Yous’ would give $150 to the ‘Me’. Hence, by implication, ‘warming’ the ‘Me’.
But RADIATION IS NOT HEAT.
Here is a counter- question: How many cold objects does it take to warm a warmer object?
Answer – it can’t be done.
In Fig 1, the ‘net’ flow (Heat) is always from ‘Me’ to ‘You’, no matter how many ‘You’s there are.
In the climate debate, doubling the ‘You’ (CO2) will have no measurable effect on the temperature of the ‘Me’.
Why?
Because the spending power of the $75 will never overcome the spending power of the $100. ‘Me’ is richer than ‘You’. ‘Me’ can afford to give all the ‘You’s their $75 back and ‘Me’ would STILL be richer than all the ‘You’s. Equate richer for warmer and you have the true picture.
Ask yourselves this:
Why does the Sun’s radiation heat the Earth? Because the energy of the radiation from the Sum is sufficient to increase the thermal energy of the Earth’s receiving molecules.
Can atmospheric CO2 not do that? NO.
Hence Figure 2 is also wrong:
The 321 W/m2 that REACHES the surface is NOT absorbed for thermal gain. The ‘absorbed by the surface’ is irrelevant in a thermal context if the Earth is warmer than the atmosphere.
This is why you can surround a hot object by billions of cold objects and the hot object will never (EVER) get hotter.
Don’t conflate insulation with insolation.
Comparing atmospheric CO2 with a blanket, thermos, whatever, is like saying that a string vest made of 99.96% air and 0.04% cotton is an effective insulator. Even that would be wrong because the atmospheric CO2 doesn’t actually prevent heat loss, so the effectiveness of the cotton is further reduced. CO2 does not trap heat. Backradiation is not heat. Remember the ONLY source of energy in this process is the Sun.
<i.["Anthony Watts: "I get so tired of hearing this stupid argument from the anti-ghg crowd. Thanks for quantifying it."]
Mr Watts, I have a lot for respect for the work you have done in making this website. However, just because you don’t understand an argument doesn’t make that argument stupid. Wilis has not ‘quantified’ your objection.
The science is far from settled!
Kind regards to all.
Arfur
“Can a cold object make a warm object warmer than it would have been if the cold object wasn’t there?
Only if it insulates the warm object. If you insulate a warm object, the rate of cooling decreases.”
And you don’t think blocking the warmer object from the coldness of space is insulation?
BTW a real definition of “insulation” might be useful here.
[“And you don’t think blocking the warmer object from the coldness of space is insulation?”]
A blocking material would act as an insulator but CO2 does not block anything, therefore it cannot either warm the Earth’s surface nor effectively reduce the rate of cooling. CO2 emits radiation as effectively as it absorbs. CO2 may delay the LWIR loss to space but the time delay is tiny compared to the length of time taken to increase the amount of CO2.
As to the definition, I am (obviously) referring thermal insulation, which I would define as a material capable of reducing the rate of cooling of a warm object.
Arfur: I agree mostly but I think that when an insulator is in equilibrium between the heat gain from the warm object and the heat loss to the cold surroundings it does not any more slow the cooling of the warm object. I guess that is the situation in the atmosphere.
I sort of agree but the rate of change (build up) of the insulating material is important in this context. It is not as if we have suddenly surrounded the Earth with a thermos-type shell (or a steel shell for that matter). Our situation is a very slow increase in a trace gas which has virtually no insulating properties and which exists in sparse form (each CO2 molecule is surrounded by approx 2500 non-radiative molecules).
Also, this transfer should be considered in a realistic time-frame. It takes a comparatively long time to make the measured increase in atmospheric CO2. The Mauna Loa dataset shows an increase of 90 parts per million in about 70 years. Thats just over ONE part per million each year! Even if CO2 was an effective insulator (and it’s not), how much effect could it have at that rate of increase? That is a whole different debate! 🙂
Willis, good post, and thank you for attempting to clear this up.
I suspect the effort is in vain, however. Sad as that may be.
The people that refuse to let this penetrate their dense outer bony layers of cranium have heard it before…it bounces off like rain from a duck’s back.
Now, if you could extend you laser-like focus to the subject of dowsing…
Oops, sorry…I did not see your second request on my first reading.
My bad.
Another interesting experiment. A 3mm mild steel plate about half a square metre. Lay this on a slab of polystyrene. Place these outside, a couple of feet above ground. I had a digital thermometer screwed to the plate. The plate is exposed to the sky but insulated from the ground below. Note local air temperature and temperature of plate. Needs a reasonably calm clear morning. As the sky turns blue overhead BEFORE sunrise, note the temperature of the metal plate. It will start to rise sooner than the local air temperature.
That blue sky overhead is warming stuff on the ground. Be interesting to know how blue sky warming effects the oceans, especially as that ‘blue’ colour will have good penetration into water.
It may be blue but it is not very bright.
Get a light meter, point it at the blue re-dawn sky.
Then do the same during full daylight an hour later, and again at noon.
BTW, did you do this experiment Richard?
What was the change in temp?
I did the experiment more than once. The mild steel plate weighed 11.5 kg and warmed up about 2 degrees more than the surrounding air. After that the temperatures rose together. I assume the plate was now losing heat to the air. The point being that ‘blue’ light is diffused sunlight and has energy to do some work. If the sun was allowed to shine on the plate then the temperature rocketed.
It doesn’t really matter a lot if a photon goes direct or bounces off a molecule of air before being absorbed, so the result is not very surprising. Also the air isn’t really that much warmed by the sun, but rather by re-radiated and conducted heat from the ground.
Another interesting phenomenon. In still air the ground will normally be coldest slightly after sunrise. This is because it has very high emissivity in the IR spectrum, but reflects visible light fairly well, so it takes a while before the sunlight catches up with the IR emission.
Glass has very high IR emissivity, which is the reason that you may have to scrape ice off it even when the air (and ground) temperature in the morning is slightly above freezing.
You may have to put the light meter at the bottom of a long tube.
Richard: The thermometer on the slab of polystyrene but without the steel plate. How would the temperature behave?
Esa-Matti: Have previously photographed polystyrene at night and notice it is highly reflective of the IR emitted by the camera, thus assume this is part of the insulation property therefore the polystyrene is unlikely to warm.
Richard: I was asking because I think that the slab of polystyrene insulates also the thermometer from the cooler ground. Therefore, my guess is that you would have noticed that thermometer without the steel plate would also show higher temperature than the local air temperature.