Shock News : Cooler Objects Can Raise The Temperature Of Warmer Objects

It is called an insulator.

You are cold – you put a jacket on – you get warmer. Everyone knows this.

You get hypothermia, your core temperature drops to 97 degrees. You put on a space blanket, your core temperature increases to 98.6 degrees.

There is a principle that heat always flows from warmer objects to cooler objects, which is sometimes abused by skeptics in a mindless attempt to debunk a theory which is inconvenient for them to believe. A cooler object most certainly can increase the temperature of a warmer heat source.

Scientists need to understand the difference between temperature and heat.

About Tony Heller

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91 Responses to Shock News : Cooler Objects Can Raise The Temperature Of Warmer Objects

  1. Latitude says:

    yep……….

  2. Bernd Palmer says:

    I have put a jacket over a tea kettle filled with cold water. Why can’t I get my tea-water to boil?
    An insulator doesn’t heat a cold(er) object, it just reduces the loss of heat by reducing convection and radiation.

    • Until you understand the difference between heat and temperature, you are going to continue to be clueless.
      Apparently you didn’t understand a single word I wrote.

      • Bernd Palmer says:

        Heat is energy. Temperature is a measure of the energy.
        If temperature is measured in Kelvin degrees, then the value of temperature is directly proportional to the average kinetic energy of the molecules of a substance.
        Prove me wrong!
        But I admit, I still don’t understand what you want to demonstrate with your experiment.

        • Gail Combs says:

          Bernd Palmer says:

          “Temperature is a measure of the energy.”

          Not quite. If you heat a kettle of water on a stove on high, that is with the same energy per unit time, the kettle will warm to 100C (at sea level). At that point the TEMPERATURE will remain constant and the energy from the stove will go into the phase change, changing the water from liquid to vapor. Once all the water is evaporated the kettle will again increase in temperature until it is cherry red. (The reason I set an egg timer.)

          This is why a 32C (90F) temperature reading in the desert at 10% humidity is NOT EQUIVALENT energy wise to a 32C (90F) reading at 98% humidity here in North Carolina.

          As some skeptics will be heard to mutter, air temperature is a piss poor measure of energy making all the screaming and yelling over a couple tenths of a degree change quite comical if it wasn’t driving policy.

        • Bernd Palmer says:

          Gall Comps: The clearest definition of heat vs. temperature can be found here
          http://physics.about.com/od/glossary/g/heat.htm
          “Heat always refers to the transfer of energy between systems (or bodies), not to energy contained within the systems (or bodies). “

    • Bernd, your argument appears to me to be one about whether “causing” needs to involve a physical transfer of properties or whether it refers to a series of events.

      So, e.g. you would argue “pulling a plug out of a bath doesn’t cause the water to flow out … its the gravity that causes it”.

      The normal way of talking about such things is that we assume the context and then if something occurs as a result of doing another thing then it “causes”.

      In the context of heat, the insulator does not itself originate the heat that warms up the hotter body. But starting from a condition with no insulator and adding the insulator results in the hotter body getting hotter. It therefore causes the hotter body to heat up.

      However, obviously it only works if you define the context and don’t separate the “cause” from the context. Because … to use an euphemism … holes only exist where there’s soil.

      (Although I reserve the right to give my wife a Xmas gift of “more space” … an empty box … or at least I threaten it every so often)

      • A C Osborn says:

        Yes it is a trick of words. The insulator does not heat up the human body, chemical burning of energy does. If the person is dying of malnutrition and hypothermia an insulator will not make the body hotter if there is no energy to burn, the person will still die without food.

        • In the context it is correct. Out of context – such as an object without an internal heat source then it is incorrect.

          But it is true of any situation: you use the level of detail that is necessary. If you know the context, then it is not necessary to use a very pedantic description even if the very pedantic description is technically more precise in the detail.

          However, you do have a point if an inference is made from a simple model – out of context – where the new context makes it wrong. In that situation, then we would need to carefully go through what is happening in a detailed way to ensure the conclusions are scientifically based.

      • Bernd Palmer says:

        In a sequence of events there is typically a cause and there is typically an effect. The cause is an event, the effect is a physical consequence of this event, so Causes and effects are typically related to changes, events, or processes.
        In you example with the bath tub, the gravity is there before you pull the plug and you are not changing the gravity by pulling the plug.
        But here we play with philosophical questions, with words; and that’s also how I see the initial blog post. There is nothing to learn.

        “A cooler object most certainly can increase the temperature of a warmer heat source” is most certainly wrong. Show us the physical equation.

      • Michael 2 says:

        “But starting from a condition with no insulator and adding the insulator results in the hotter body getting hotter.”

        ONLY in the circumstance that something is adding heat to the object you have just insulated.

        If I remove a pot from the stove and wrap it in a perfect insulator, the pot will not get warmer. It also won’t get colder.

        Insulators cannot heat but can improve the effectiveness of a heat source acting on something colder.

        A cold object can SEEM to heat a warm object in the following scenario:

        1. The object is being heated (as from sun or fire).
        2. The object is exposed to extreme cold, that is to say, nothing reflects any of its radiant heat back to itself. It will stabilize at a temperature where head being added is balanced with heat lost to the environment.
        3. You place shiny cold metal in the vicinity, perhaps surrounding the object. It has a longwave infrared emissivity of only 0.1, meaning the radiant heat from the warm object is reflected back to it. The “coldness” of the metal is irrelevant, what the pot sees is its own radiance and acts like it has been surrounded by an object of nearly its own temperature. This will dramatically reduce heat loss but does not add heat to the pot. Only the fire adds heat to the pot. But since you have reduced heat loss, a new equilibrium will be achieved at a much higher temperature when heat loss is once again the same as heat gain.

        Climate equivalence: A clear sky at night is very, very cold. Recently I purchased an infrared thermometer. It is fun to use. Pointing it at a cloud at night shows about 25 degrees F. Pointing it at the stars shows about -60 F.

        Consequently, clouds reduce nighttime heat loss, dramatically so in low humidity areas such as a desert. This is not the same as “warming” the earth; only a heat source of higher temperature than the Earth surface at that moment and place can do so.

        • No more stupidity, please.

        • Michael 2 says:

          Stevengoddard says “No more stupidity, please.”

          That would be a fine thing and then bloggers would have nothing to write about. But I wonder why you have attached your comment to me? A few more words would help.

          Scottish Skeptic wrote “But starting from a condition with no insulator and adding the insulator results in the hotter body getting hotter.”

          To which I explained (rather than just saying this is stupid) ONLY in the circumstance that something is adding heat to the object you have just insulated.

          Self-heating in other words. If you insulate a heat source then the source will probably get hotter (all other things being equal it will definitely get hotter). A living person is self-heating, so adding insulation on a cold day will increase skin temperature. A dead person is not self-heating, so adding insulation to a dead person accomplishes nothing.

    • If you put a tea kettle on and turn the burner on, and put your jacket on the kettle, you will burn your house down. Then your water will boil. However, you will still be an idiot.

      • Michael 2 says:

        There was no mention of turning on a burner. The idea is to challenge the idea that insulation, by itself, makes a thing warmer. Wrap a thermometer in fiberglass insulation and see if it gets warmer. It won’t. Insulating a pot on a stove is actually a very good idea and some special pots exist for that purpose, usually for hikers and campers.

        • My tolerance level for incredibly stupid comments like this is gone.

        • Michael 2 says:

          stevengoddard in response to Michael 2: “My tolerance level for incredibly stupid comments like this is gone.”

          That’s twice in a row you seem to have mistaken me for someone else. Once is a mistake, twice is shame on you. It is becoming clear that the value you bring to this debate is not what I had hoped for. A proper response is for you to argue what you believe is my error. That you have not suggests you cannot.

          Well, as I have said many times on other blogs, while it seems warmists can be lumped into a consensus of obligatory beliefs, skeptics come in all flavors – “everything else” and there’s no guarantee that any two skeptics will agree with each other, much less with a warmist.

          I had hoped that the inflexible nature of science would *compel* some agreement among skeptics but that appears to be an optimistic view.

          What an unexpectedly interesting development!

        • Michael 2 says:

          It is likely you met my father at Kanab. He was and continues to be a staunch environmentalist and helped stop the Kaiparowits power plant.
          http://earthjustice.org/features/kaiparowits-power-plant

          I wish you would not tar all of your opponents with the same brush. Anyone that thinks he has nothing to learn from others has a disease called “hubris” and that closes the door on persuasion and communication. Your mileage may vary.

  3. BBould says:

    According to Hansen then, if you throw a blanket over a dead body it will warm up.

  4. A C Osborn says:

    Sorry, you are wrong.
    http://hockeyschtick.blogspot.co.uk/2014/11/scientist-proved-214-years-ago-that.html

    In your example the core body temperature does not get hotter, only your skin.

    • Apparently you don’t understand what hypothermia is. Look it up.

    • A C Osborn says:

      You are not heating the body with the blanket, you are blocking heat loss, the body is warming itself by burning energy, the blanket has only slowed down the heat loss and bears no relationship to your original argument about DWIR and you know it.

      • Curt says:

        You are not heating the [earth] with the [greenhouse gases], you are blocking heat loss, the [earth is warmed by the sun], the [greenhouse gases have] only slowed down the heat loss and bears [every] relationship to your original argument about DWIR and you know it.

    • Curt says:

      The Pictet experiment had no separate source of power. It gets very tiresome to keep explaining the fundamental difference in situations.

      No separate source of power: difference in temperature of the two bodies, by controlling the rate of heat transfer between them, simply affects the rate at which the temperatures of the bodies converge as heat is transferred from hotter to colder body.

      Separate source of power to hotter body: difference in temperature of the two bodies, by controlling the rate of heat transfer between them, does affect the steady state temperature of the hotter body. A cooler body that is still warmer than an alternate colder body reduces the heat transfer from the hotter body to the colder body. With the transfer here reduced below the separate source of power, the body’s internal energy and therefore temperature increase until the transfer to the cooler body once again matches the power input from the separate source.

  5. Brad says:

    THAT, was the best damn description yet. Thanks Tony.

  6. Bad Andrew says:

    Analogies don’t prove anything scientifically. The just get you to think a certain way.

    Because jackets warm people doesn’t say a damn thing about how the atmosphere works.

    Andrew

  7. Paul Clark says:

    It is called an insulator.

    The distinction is: an insulator for EMR can’t make anything warmer than it was originally, even if that thing’s a heat source.

    If you take a light bulb in a vacuum painted black and put another few layers of black paint on it, the light bulb doesn’t get any warmer, despite thicker “greenhouse blanket”.

    A jacket only warms you because it’s stopping heat loss by convection to the air. None of a jacket’s warming effect comes from EMR blocking.

    ..a mindless attempt..

    It’s far easier to believe in the greenhouse consensus than buck the trend.

    ..which is inconvenient for them to believe..

    In a sense, greenhouse non-belief simplifies the AGW debate since CO2 climate sensitivity is zero. But intellectually it’s hard; the 2nd law of thermodynamics and radiative physics are evidently hard topics, not easy topics, for folks to fathom.

    As for space blankets, they cool by reducing emissivity. CO2 increases earth’s emissivity, albeit at higher, cooler layers of the atmosphere.

    • The religious zealotry of some skeptics is every bit as bad as that of alarmists.

      • A C Osborn says:

        Some people can’t rationalise being proved wrong, I asked you on another of your threads, if you believe in DWIR warming the erath’s surface you must believe it does WORK.
        So try and prove it, everybody knows DWIR COOLS an object not warm it.

        • “Scientists need to understand the difference between temperature and heat.”

        • A C Osborn says:

          No, you need to understand science better.
          Prove DWIR produces HEAT in an object and I will convert to your brand of science.

        • Scientists need to understand the difference between temperature and heat.

        • A C Osborn says:

          Repeating it doesn’t make it true.
          You are just ignoring a question you can’t possibly answer.
          Therefore you are admitting to losing the argument.

        • I have already answered it many times. You din’t understand the difference between temperature and heat.

        • A C Osborn says:

          Sorry Heat can do work, it can be used. You accuse me of not understanding, but I understand too well.

        • Insulators don’t warm. They slow the cooling. Why is that hard to understand?

        • A C Osborn, I think you’ve bugged Steven enough over what appears to me to be a pretty minor point. However, I’m not sure what was originally said, so perhaps if you came to my blog at Scottish Sceptic we could discuss it there.

        • Michael 2 says:

          “Prove DWIR produces HEAT in an object and I will convert to your brand of science.”

          Trivial. You can see a demonstration here:
          http://wattsupwiththat.com/climate-fail-files/gore-and-bill-nye-fail-at-doing-a-simple-co2-experiment/

          Anthony uses what looks like a Fluke VT04 imaging infrared thermometer. A heat source provides longwave infrared which is absorbed by the glass of a nearby cookie jar. This absorption is controlled solely by emissivity of the glass jar, not by its temperature. Thus, if the rate of incoming infrared is sufficient to raise the jar 1 degree, then it will be raised 1 degree whether from 20 to 21 or from 40 to 41.

          This will be true even in the case that the source of infrared is COLDER than the object to be warmed. However, it will also be true that more energy will be radiating from the glass jar to the colder object. But if you removed the colder object (leaving NO source of infrared, namely 0 degrees Kelvin) in its place, the warm object will see no back-radiation and consequently cool much faster.

          On average, heat only goes from hot to cold objects. However, a photon arriving from a cold object being absorbed by the warmer object will produce a tiny area of increased energy — temperature — heat — from cold to hot. The cold object just got one photon colder, and the hot object just got one photon warmer.

          It went the wrong way because photons DO NOT CARE. They are emitted at random from hot or cold. Just a lot more often from hot. So, on average, hot flows to cold, but at quantum levels, there is neither hot nor cold — only quantum energy levels. In fact, the Planck curve allows for occasional high-energy (high temperature) photons to emit from almost any temperature blackbody radiator.

          This can be shown experimentally although I haven’t tried it. Maybe when it gets down to -20 where I live I’ll try it. Take two jars at 50 c water. One of them leave exposed to radiate to the -20 environment. It would help to prevent convection. The other jar will be exposed to the same environment but nearby place one or more jars of cold water. The cold water will radiate into the hot water, imparting just a bit of energy (heat), but of course the hot water is losing heat to the environment. But the *addition* of heat from the cold water jars radiating to the hot water jar will produce a reduced rate of cooling in the hot water jar. The net energy flow is still hot to cold, but cold is imparting enough energy to slow down the rate of heat loss.

          Of course, the cold water jars will see the heat from the hot water jar, they will also see ambient, and eventually everything will be -20.

    • Paul Clark says:

      ^meant to say: space blankets “warm”, not “cool”.. by reduced emissivity

  8. Robertv says:

    That’s why the natural place for people to live is the tropics.

    http://youtu.be/LMA5qHAzD8I

  9. Tim says:

    But Tony can that jacket increase the body temperature above 98.6?

  10. Edmonton Al says:

    My response to this statement…….
    “Scientists need to understand the difference between temperature and heat.”
    Response:
    WHY NOT BACKRADIATION? THE AMAZING NATURE OF LIGHT
    Written by Joseph E Postma on 21 Nov 2014
    The Basics
    It seems that a major source of confusion stems around this equation for radiative transfer of heat energy:
    Q = ?T24 – ?T14
    The term ‘Q’ is not the incoming solar energy nor does it represent a source of energy at all. From that incorrect interpretation of the equation arises all sorts of further misinterpretations and bad physics. It’s where the whole incorrect idea of backradiation heating arises and all of the various arguments about cold helping to make something warmer hotter still. I address that misinterpretation of the equation many times on this blog, but here I do it up front:
    http://climateofsophistry.com/2013/12/08/revisiting-the-steel-greenhouse/
    ‘Q’ is the heat flow between the Sun and Earth and so is not the solar energy. The solar energy flux would be a term on the right hand side, ?T24 say, but factored for distance. How this is done is demonstrated in the link above. ‘Q’ is actually zero if we consider the Earth to be in energy equilibrium with the solar input, which it should be within a small margin.
    It is also discussed here:
    http://climateofsophistry.com/2013/05/27/the-fraud-of-the-aghe-part-12-how-to-lie-with-math/
    So to repeat, ‘Q’ can not be the solar heat input, when T1 and T2 are supposed to be the temperatures of the atmosphere and surface. That’s not what that equation is about at all.

    I’ll let Rosco explain, and I was going to blockquote his comment from a previous thread but with my own editing now finished, I’ll just acknowledge that this next section comes largely from him:
    The Physics
    ‘Q’ is not an actual radiant emission corresponding to any temperature and hence is not subject to any valid algebraic manipulation as if it supposed to be a conserved quantity from some source. It is not a conserved quantity and it does not represent a source.
    ‘Q’ is actually Q(net) – the difference between two emission powers from 2 objects.
    This is easily observed on a Planck curve diagram.
    ?T14 is the area between the Planck curve and the x axis for T1.
    ?T24 is the area between the Planck curve and the x axis for T2. If T2 > T1 then the area of the Planck curve for T2 completely includes all of ?T14.
    But ‘Q’ = Q(net) = ?T24 – ?T14 is not a radiative flux at all. It is the area between the T2Planck curve and the T1 Planck curve!

    Only the T1 and T2 terms have any relationship to the Stefan-Boltzmann equation because they are explicitly derived from Planck’s equation as the integral from 0 to infinity of Planck’s equation in either df (frequency) or dlambda (wavelength) terms.
    If T1 = T2 then Q(net) = 0 this simply says that two objects at the same temperature have zero net energy exchange and therefore no thermal effect on each other.
    If T1 is less than T2 then object 2 is heating object 1 and raising its temperature. This demonstrates the ridiculous nature of all “greenhouse effect” “physics” – they simply play algebraic tricks without any recognition of what the terms actually mean.
    F1 = ?T14 is a valid flux for T1; F2 = ?T24 is a valid flux for T2.
    Q = ?T24 – ?T14 is just a number, just the difference of the fluxes – nothing more! It is not a source in itself. It is not conserved. You can’t hold it constant and say that an increase in the cooler T1 will cause an increase in the warmer T2, i.e., that cold can heat hot.
    To manipulate this expression by algebra and claiming that ‘Q’ has to remain fixed because it is the energy from the Sun is entirely nonsensical by the terms and logic of the equation itself.
    The Question
    So this is the question that confuses just about everybody, and it is also where the error of backradiation heating and the misinterpretation of the heat flow equation originates:
    What happens to the energy from the cooler portion if it travels to the hotter side but does nothing? How can it do nothing? Does it even travel to the hotter side at all? How can it not? How could it know not to? How could the photons from the cooler side either A) not cause any heating when they get to the hotter side, or B) not travel to the other side at all? These are all related questions.
    To be sure, the equation we’ve been discussing for ‘Q’ absolutely, most definitely, 100% says that radiative heat energy only goes from hot to cold and thus that temperature can only be increased by something hot warming up something cooler. It specifically does not result in something like the supposed “steel greenhouse effect”, as debunked previously. And this equation is the correct equation from radiative transfer theory and thermodynamics, when used and interpreted properly. It says what happens, and so it appears that the problem is that it doesn’t say why that happens.
    The Answer
    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.
    For example, if we add the two Planck thermal radiation curves from the above plot together, then the existing microstate populations are increased, but it doesn’t activate any new higher-frequency microstates than the warmer curve already had activated, and thus the temperature doesn’t get increased:

    Now you can’t see it because the lines all overlap each other at high energy (i.e. short/small wavelength), but the cooler curve has zero contribution, zero activated microstates, at the high energy, high frequency, short wavelength end of the plot (left side) where the warmer curve does have populated microstates. Thus, adding the cooler radiation with the warmer radiation doesn’t result in a macrostate that has microstates activated at higher frequencies as compared to the original warmer radiation (T2), and thus, the result doesn’t have a higher temperature than the original warmer source of radiation .
    Also note, and this you can see, that the peak of the population of microstates for the sum of the warm and cool source radiation is at longer wavelength and thus lower energy than the original warm source radiation. Adding cool to hot doesn’t result in something hotter…but adding hot to cool certainly does result in something hotter.
    But still the question can be asked: Why doesn’t the addition of energy from the cooler source to the warmer source result in an increase in temperature for the warmer source? If you add any energy to anything, doesn’t that have to result in an increase in internal energy, and thus temperature?
    The answer to that, is “No”, because that is not how thermodynamics works. It’s not what the radiative heat transfer equation says or how it works.
    It is not that simple.
    More Answer
    We actually shouldn’t have added those fluxes together at all because that is not how heat radiatively transfers. If we add those fluxes, then is that now the flux that the hotter object emits, or is it the cooler objects? Or is that the radiation field between the two objects? None of those are correct. It was wrong to add those fluxes together. It is easy to make that mistake, but on the other hand, we’ve always had the equation for radiative heat transfer at hand and so we should simply refer to it and obey it. The energy field flowing as heat between the hot and cool objects is only ‘Q’, only the difference between the hot and cool fluxes.

    You can see a little more clearly now that ‘Q’, the heat flow, merges to the hotter curve at high frequency/small wavelength, which again is all about how only the hotter source has active microstates at high frequency, and the cooler curve doesn’t. If you add the ‘Q’ curve, the heat flow, to the T1 object’s curve, then you get the T2 curve, which would indicate that object 1 has come to the temperature of object 2 in thermal equilibrium, and then the heat flow ‘Q’ would be zero, meaning that no more temperature changes can occur. Object 1 & 2 become a unified thermal system on the side facing each other, and object 1 then simply emits the energy supplied by object 2 out from the side facing way from object 2.
    When you have say, two planes of material, touching each other, then the heat flow would be purely diffusive, meaning purely by conduction. The addition of another layer physically touching the existing warmer layer doesn’t make the warmer layer warmer still – the new layer simply gets heated to the temperature of the original layer, and then it becomes the new surface of the original warmer plane. Nothing about this changes when there is a gap between the layers so that the heat transfer between them become radiative. It’s not like you would remove the added layer from touching the original one to create a vacuum gap in between, and suddenly this would cause the warmer layer to become warmer still, when this is not what happened when the layers were touching. The modes of heat transfer obey the same limitations and the same general rules of heat flow.
    The situation for the Sun and Earth is a little different. Scaled for local intensity at the distance of the Earth from the Sun, the solar spectrum has much less intensity at terrestrial frequencies and thus the Earth can locally shed heat energy in the direction of the Sun. You can see this in idealized Planck curves given the solar and terrestrial effective temperatures:

    However, as we calculated in the steel greenhouse debunking thread, the radius of heat potential for the Earth only extends to 56.8 million kilometers before the terrestrial heat becomes indistinguishable from the cosmic background thermal radiation. Therefore, the Earth can not heat the Sun with terrestrial backradiation (because the Sun is 150 million kilometers distant), and that will always be a general result in any scenario.
    Of course, as you go back towards the Sun, the solar flux spectrum increases in intensity and so well before reaching the Sun, the flux from the Earth will become dominated by the outward flux from the Sun at those wavelengths. The flux intensities from the Earth vs. the Sun, at the wavelength of the peak of the terrestrial spectrum, become equal at 27.1 million kilometers distance from the Earth towards the Sun, which is well inside the heat envelope of Earth.

    • Michael 2 says:

      “object 1 then simply emits the energy supplied by object 2 out from the side facing way from object 2.”

      Objects have no idea where the energy came from. An object at a given temperature will radiate in all directions, including right back to the source of its heat, returning some of that energy but even then it isn’t exactly a “return”, the photons have no idea of their history, where they came from or where they ought to go.

      “Heat” is the effect of energy on an object, so I understand that heat only flows from hot to cold as it is always “net”. Energy as photons flows from everything above zero degrees Kelvin completely ignoring the temperature of the object that subsequently captures the photon.

      So if you give me ten photons and I give you two, Energy has been exchanged but heat (8 photons worth) flowed from you to me.

  11. gymnosperm says:

    Jeez, I so wish I figure out how to make you a water vapor and CO2 jacket to wear so you could see how that worked out for you.

  12. philjourdan says:

    Reminds me of the blonde joke, A blonde, just finding out the purpose of a thermos, puts a popsicle and a cup of hot soup in one for her lunch.

  13. Richard Smith says:

    Slayers have the strange belief that the greenhouse effect is contrary to the laws of thermodynamics. This is because they cannot seem to grasp the simple idea that insulation of an article will raise its temperature even though there is no additional energy being supplied to it.. The temperature rises until equilibrium is restored. Heat still flows from warmer to cooler.

    • Paul Clark says:

      Insulation against EMR doesn’t work the same as insulation from convention/conduction. You’ll find that every example of an what’s called an “insulator” involves blocking of physical heat flow.

      In the case of EMR you can have a low emissivity coating, and this does affect temp (e.g. vacuum flask), and this also works in a vacuum. But this low emissivity effect is not the same as a greenhouse effect, the latter relying on the internal flux of EMR to boost temperature.

      In a vacuum, which earth is in, only a reduction in emissivity can increase temperature, without increasing earth’s energy content.

      Since greenhouse gas increases earth’s emissivity, the greenhouse effect implies a violation of the 1st law; for how does greenhouse gas create the extra energy (for all the “worse storms” and so on)?

      Equilibrium of what? Electromagnetic radiation (“backradiation”) from a cooler object (sky) can not warm a warmer object (ground). This is embodied in the 2nd law.

      • Michael 2 says:

        Paul Clark says “In a vacuum, which earth is in, only a reduction in emissivity can increase temperature, without increasing earth’s energy content.”

        Emissivity is also an absorption index. Materials with low emissivity also have low absorption. Changing a blackbody from emissivity of 1 to 0.5 means it will emit less radiation, but it also absorbs the exact same amount less, and thus will remain at the same temperature (provided all heat is obtained by absorbing radiation).

        “Since greenhouse gas increases earth’s emissivity, the greenhouse effect implies a violation of the 1st law; for how does greenhouse gas create the extra energy (for all the “worse storms” and so on)?”

        It violates first laws only in the simple model you have imagined. Where carbon dioxide has its increased emissivity is at the TOA, Top Of Atmosphere, where being 70 degrees below zero the increased emissivity is working on dramatically reduced temperature. Since emissivity is linear whereas emission is the fourth power of temperature, the reduction in temperature far outweighs the advantage of increased emissivity at Top Of Atmosphere.

        Your simple model also ignores the atmospheric window from 8 to 15 micron wavelength whose emissivity is that of the surface of the earth or of water (0.98 or so). This window is unaffected by water vapor or carbon dioxide and is sort of a “sieve” for some energy direct from surface to space.

        So adding carbon dioxide does two things simultaneously — as you say, it increase emissivity at the top of atmosphere, but it also retards or delays heat reaching TOA in the first place, keeping it closer to the surface.

        The actual, net effect of these two phenomenon depends on a great many factors and is why observed sensitivity is neither the zero you believe exists nor the 4 to 7 that the most striden warmists believe exists.

      • Michael 2 says:

        Paul Clark says: “Presumably this means you’ve read Postma’s post and found something wrong with it? What in particular?”

        Start with these statements by Postma:

        “To be sure, the equation we’ve been discussing for ‘Q’ absolutely, most definitely, 100% says that radiative heat energy only goes from hot to cold and thus that temperature can only be increased by something hot warming up something cooler.”

        and

        “Scaled for local intensity at the distance of the Earth from the Sun, the solar spectrum has much less intensity at terrestrial frequencies and thus the Earth can locally shed heat energy in the direction of the Sun.”

        The problem is “radiative heat energy”. Heat isn’t energy, and energy isn’t heat, although you can easily go from one to the other and back again (in fact, there seems to be no way to prevent it).

        Energy (radiation, photons) are emitted in every direction possible from a warm object. Postma tries to explain that the Earth can radiate to the sun because of its distance; but that is irrelevant. He explains that the spectrum from the sun is different than that of the Earth, which is true but still irrelevant.

        Photons from a cold object DO reach the warmer object. However, at the exact same time, many more photons are leaving the warmer object and impinging upon the colder object. The net effect of this is heat flowing from warm to cold — but at a reduced rate as compared to an object so cold it radiates nothing.

        This can be seen with hand-held infrared thermometers. But I’ll be discussing more of that at further down the threads.

    • Michael 2 says:

      ” This is because they cannot seem to grasp the simple idea that insulation of an article will raise its temperature even though there is no additional energy being supplied to it..”

      Tell you what. Make a video. Start with a thermometer not insulated until it stabilizes at room temperature. Wrap a thermometer in an insulator and watch it get hotter (or not) until it explodes. Show me the video.

  14. Paul says:

    Steve,
    Scientist proved 214 years ago that a cold body cannot make a hot body hotter, proved opposite
    A paper published in the American Journal of Physics describes the simple experiment of Pictet in 1800, demonstrating that a cold body cannot make a hot body hotter, in fact quite the opposite. It wasn’t until 51 years later in 1851 that Lord Kelvin & Clausius introduced the 2nd law of thermodynamics to explain the theory behind this phenomenon, requiring a one-way transfer of heat only from hot to cold that is necessary to maximize entropy production.

    http://hockeyschtick.blogspot.fr/2014/11/scientist-proved-214-years-ago-that.html

    • daveburton says:

      Your language is imprecise, Paul: “Hotter” than what?

      Your error is in confusing “hotter than it was” with “hotter than it otherwise would be.” You’ve erroneously applied to the later a principle which is only true for the former.

      A cold reflective or insulative body can certainly make a hot body hotter than it would otherwise be. For example, take a pair of identical hot objects, or heat sources, in otherwise identical surroundings. Surround one of them with shiny aluminum foil, at a distance great enough to ensure that the foil stays comfortably cool. Track the temperatures of the two objects. The object surrounded by foil will be warmer than the other, no matter how cold the foil is kept.

      For another example: If you’re sleeping outdoors in a sleeping bag, on a frigid night, and you become too cold, you may wrap or cover your sleeping bag with a blanket. That will help. You will be warmer than you otherwise would be, despite the fact that the blanket stays way below your body temperature.

      • Bernd Palmer says:

        “A cold reflective or insulative body can certainly make a hot body hotter than it would otherwise be.” It’s not “make it hotter” but “keep it hotter”. You see the difference?

        • Michael 2 says:

          I have no doubt he sees the difference but isn’t arguing it. Rather, the sleeping bag retains heat that is already in your body, keeping it warmer than it would be without the insulation.

          The application to climate is that keeping more of the daily heat from sunshine, during the day but particularly at night, produces a warmer surface. This is easily perceived by a cloudy night being much warmer than a clear night, particularly in the desert where water vapor does not stabilize the temperature.

  15. gator69 says:

    Then let’s call it the “Jacket Effect”. 😉

    • Michael 2 says:

      Jacket effect conveys only half the picture, the insulating half, and it works well enough for that. But insulators keep heat OUT as well as IN, which is why some desert dwellers wear heavy clothing even in, or especially in, extremely hot dry climates.

      The greenhouse magic is that it lets heat IN but not OUT. More precisely, it lets more in than out, since glass does stop some of the heat from coming in (and special glass exists that keep a LOT of heat from coming in) and prevent heat from escaping. What actually happens is the glass absorbs outgoing infra-red, warming itself in the process and shoots half of the energy back in and the other half back out. So it isn’t strictly a one-way mirror kind of thing, and neither is carbon dioxide.

  16. rah says:

    Ok I grew up in a steel fabrication business. I know that if one heats the end of a metal rod red hot on one end and then puts that end in the water the other end that your holding will get get warmer.

    • daveburton says:

      If you cool the red-hot end, the other end (that you’re holding) will not get as hot as if you hadn’t cooled the red-hot end. It might still get too hot to hold, though.

      A better example:
      1. heat up a pair of metal rods (not enough to char cotton).
      2. wrap one rod in a cotton blanket.
      3. set both rods on the cold concrete floor: one bare rod, and one rod wrapped in a blanket.
      4. Wait 10 or 20 minutes.
      5. Check the temperatures of the two rods.

      You will find that the unwrapped rod cooled much faster than the other one. The cotton blanket has caused the much hotter rod to remain much hotter than it otherwise would be, even though the cotton blanket (most of it, anyhow) is not as hot as the rod.

      • Gail Combs says:

        Dave says:
        …You will find that the unwrapped rod cooled much faster than the other one. The cotton blanket has caused the much hotter rod to remain much hotter than it otherwise would be, even though the cotton blanket (most of it, anyhow) is not as hot as the rod.
        ……
        However that cotton blanket did not make the temperature of the hotter rod hotter. Since it is passive it only slowed the cooling of the rod.

        I do not think anyone disagrees that the atmosphere and IR active gases keep the earth from cooling down as much as would happen if the Atmospheric Window through which the earth radiates directly was a lot bigger.

        Dave go look more closely at Dr. Happer’s slide #16.

        Since you have his contact info. perhaps Dr Happer can clear up the question about whether a blanket causes a hot object to get hotter, and if that is a good description of the Earth + sun + IR active gases in the atmosphere. Or if it the NET EFFECT is just a retarding of the earth’s heat loss to space.

      • rah says:

        “If you cool the red-hot end, the other end (that you’re holding) will not get as hot as if you hadn’t cooled the red-hot end. It might still get too hot to hold, though.”

        I will disagree. That is not the way I remember it. For some reason putting the hot end in water seems to speed the conduction to the other end of rod. Perhaps because I’m talking about carbon steel and not just any metal and quenching it changes it’s physical properties significantly actually making the quenched portion denser and slightly contracted.

        If for example you have a carbon steel I or H beam that is bowed. One can a series of heat triangle shaped areas on the flanges starting at the apex of the bow with the point of the heated triangle area towards the web of the beam and quench them and straighten that beam. One just needs a rosebud tip on the torch then once heated have a five gallon pail of wet rags ready to slap on the heated areas. This is done with the beam sitting with it’s flange flat on the table and the web vertical so has nothing to do with using gravity to help. It takes some practice to get it right but it works great on the lighter beams if you know what your doing. Just don’t do it to the heavier primary load bearing structural members of a building. How would that technique work if the metal was not actually contracting in the quenched areas?

      • Bernd Palmer says:

        remain much hotter than it otherwise ” Further up, you claimed it would make it hotter. see the difference?

  17. bwdave says:

    The big problem with the atmosphere as a blanket model is that radiative cooling of the surface is only a small portion of total surface cooling; the bulk being convection and conduction. The GHE becomes outright fraud when it is claimed that without it, the surface would be
    `33°C cooler..

    Whether 0.06% of atmospheric mass that is CO2 has a warming or cooling effect on the mostly wet surface of the Earth is still debated; with reasonable arguments for either. But really, how much heat can we be talking about?

    • Curt says:

      bwdave: If the atmosphere were completely transparent to radiation, there would be several significant implications.

      First, the only way the earth could reject as much energy as it receives from the sun is for the surface to radiate straight to space.

      Second, since radiative emission equals radiative absorption, an atmosphere that does not absorb radiation cannot emit it either, so it has no way of transferring heat to space. It can only exchange heat with the surface, and this zeros out on average. So there would be not net conductive/convective transfer from the surface to the atmosphere — there would be as much in the other direction on average.

      If the surface, with near unity emissivity, is radiating to the effective near absolute zero of space as much power as it receives from the sun, its temperature would be far below what we see in our world. If it were uniform over the surface of the earth, it would be at about 255K (-18C) as is often stated. But the less uniform the temperature is over time (day vs night) and area (tropics vs poles) the lower the average temperature, so in many ways the 33C warming is an understatement.

      (But most of that is from H2O, which is about 1% of the atmosphere on average.)

      • Baa Humbug says:

        If the surface, with near unity emissivity, is radiating to the effective near absolute zero of space as much power as it receives from the sun, its temperature would be far below what we see in our world. If it were uniform over the surface of the earth, it would be at about 255K (-18C) as is often stated.

        Do you have any evidence that Earths surface is “near” unity emissivity (?)?
        if ? is not 1, but 0.99, 0.95, 0.9 or 0.8, there will be an error for the surface radiation of 3.9, 19.5, 39 or 78 W/m2 respectively.
        As you can see, the smallest difference in ? makes a large difference in flux. This is not a trivial matter. Surface emissivity MUST be determined accurately if we are to believe radiation budgets as presented by K&T et al.

        • Michael 2 says:

          Swallowing camels while straining at gnats. It hardly matters what is the emissivity of the Earth; make it 0.98 (sea water), it will absorb more and emit more. Make it 0.8 (dirt). It will absorb less and emit less. Either way the temperature of the surface will be about the same, it would be exactly the same except for the internal heat source.

          Emissivity becomes important when you actually start computing heat flows but this conversation hasn’t gotten that far.

  18. bwdave says:

    Just a thought about the jacket that is put on. Suppose the jacket has been soaked in water,and frozen solid..

  19. Mack says:

    Meanwhile , back in the real world, I’m wondering where this blanket might be, and what it is made of. I’ve heard some people say that the blanket keeps the Earth’s surface warm….so does the blanket keep your tootsies warm? Some basic gas physics for you Steve…Gases do not add energy but disperse it. All gases dissipate heat. If you’ve got that hair-dryer packed in your luggage…., don’t use it under a blanket held close over your head .

  20. higley7 says:

    The downward radiation form the radiative gases may indeed slow down the heat loss from the surface, such that the surface will have a higher temperature at a particular time of day than otherwise, but that is different from warming the surface—it is not a source of energy, just a reflector in effect. Also, the wavelengths available to the CO2 are woefully sparse, such that CO2 probably has undetectable effect. It is indeed water vapor that might have an effect. However, the water cycle and the conduction and convection of the huge atmospheric heat engine swamps the small effect water vapor might have as an insulator. It is important to not forget that the atmosphere is quite dynamic and this “insulation” is fleeting and the atmosphere cools it rises, such that the surface is warming the gases a lot more than the gases are radiating energy back downward to be reflected back upward.

  21. bwdave says:

    Water vapor is about the worst insulator there is.

    • usJim says:

      Depends on the concentration …

      • Michael 2 says:

        Water vapor draws heat by conduction and convection and absorbs longwave infrared. A cold day in Virginia is a LOT colder on the skin than the same temperature in Nevada. 40 degrees F. in Nevada is just about shirt-sleeve temperature, but in Virginia I wear a parka at 40 degrees, or at least a pea-coat.

  22. Rick Fischer says:

    I can hardly stop laughing at your success in trolling the less educated. You use terms like “heat source” or “emit” without specifying whether the object is or is not generating heat energy. The resulting confusion is hilarious.

    An warm object can be a source of heat energy to its cooler surroundings by virtue of simply being warmer than its surroundings, but only to the point where cooling brings the temperature difference is zero. Or it can continue to be a source of heat energy if it generates more heat energy to replace what is being “emitted”.

    If I put a steak into a freezer, it is a source of heat to its surroundings, but only until it is well frozen. If I wrap the steak in a blanket, it emits heat more slowly and freezes later, but the temperature of the steak does not rise because of the blanket.

    If I put on a coat and walk into the freezer, I will be a continuing “emitter” of heat, a continuing “heat source”, because I generate heat energy to replace what I lose. My skin temperature depends whether my rate of emission of heat energy is greater or less than my generation of heat energy.

    On second thought, Steve, perhaps you are using the Socratic method, presenting you students with incomplete information and letting them discover the truth by thrashing through their errors?

  23. Stealth says:

    Being a layman I can’t use all the technical terms but I think I grasped the concept from the start. Would it be right to say that “temperature is measured with a thermometer” while “heat is measured as an amount of energy”?

    A given amount of “energy” can cause a certain “thing” to become, say 10 deg C in temperature while another “thing” may only become 5 deg C using the same amount of “energy”? (Assuming 100% insulation.)

    Temperature is then obviously completely different than heat, but under proper known circumstances the 2 can correlate. Absolute zero is where there is no energy (I think) but adding “x” energy to a certain something will raise it’s temperature under “perfect insulation” while a different something will not have the same temperature with the same energy input. Thus heat and temperature are obviously different things…

    This is hard to verbalize… Maybe this?

    If you have a freezer with 100% perfect insulation, once it and the contents reached it’s temperature setting it would never run again, thus not using anymore energy. If the insulation was less than 100% it would need more energy (heat) at times to maintain the set temperature… or conversely the same with an oven. So the less insulation “efficiency” the more energy (heat) is necessary to maintain the temperature setting.

    Did I get close even?

    • Michael 2 says:

      Stealth, I believe what you are describing is “specific heat” although I now see that it is called “heat energy” (not just heat any more which comes as a bit of a surprise to me).

      That’s the amount of energy it takes to add one degree C to something or to remove that 1 degree C. Wikipedia calls it “heat capacity” but I remember it being called specific heat but that was long ago.

      http://en.wikipedia.org/wiki/Heat_capacity

      I see in this article an explanation for much confusion I see here — once upon a time “heat” was a substance, and an object had a certain amount of this substance which wikipedia is now calling “heat energy” not just “heat”. Heat, by itself, appears now only during movement from hot to cold. That should produce plenty of confusion.

      So anyway heat energy can be measured variously, used to be calories or Kilocalories (food calories are actually kilocalories, the amount of energy needed to raise 1 gram of water 1 degree C; so a kilocalorie raises a kilogram 1 degree).

      Anyway, you cannot really tell by looking at an object how much “heat energy” it has. It won’t give up any of it unless it gives it to something colder. It isn’t something you can measure directly.

      Anyway, if you have an amount of energy, and you have a substance with a known “specific heat capacity”, you can calculate its increase of temperature if you succeed in imparting all of that heat energy to the object.

      The relevance to climate science is that water vapor has a rather large (and non-linear) heat capacity. So the same amount of solar energy in a dry desert will raise temperatures rapidly but there’s not much “heat energy” — actually, about the same as Virginia where temperatures rise slowly — but it is a lot of work to cool that warm, moist air. Most of the heat energy is hiding in the water vapor and comes out when you condense it.

      That’s a bit rambly but will give you something to study.

  24. Michael 2 says:

    I think I made a mistake somewhere above. Maybe it won’t get posted. I had claimed that radiation from a cold object could actually warm something already warm, that appears to be a mistake. The cold object does radiate to the warm object, and the warm object does capture those photons and convert to phonons (vibrational energy), BUT at the exact same time the warm object is radiating much faster thus losing heat and reducing temperature. The best that can happen is nearby cool objects slow down the net energy loss of the warmest object, as compared to having absolute zero in the vicinity thus nothing returned to the warm object.

    Hardware store science:

    I’ve learned a lot from a hardware store infrared thermometer. I have one with adjustable emissivity and it can read down to -60 F. What I have been studying is how it can do that. Does it not “see” its own heat? Indeed it does, but not for long. The sensor is radiating and receiving at the same time, and because of its tiny size, reaches equilibrium in milliseconds.

    The sensor is a “micro thermopile” and you can google it for examples. It generates electricity when one side is warmer than the other. Usually you aim it at something hot, the infrared heats the front and the heatsink on the back stays relatively unchanged (and is measured by a thermistor anyway). The voltage reveals temperature of the source. But what happens when you point it at COLD? It works just as well but the polarity is reversed — the front is now the cold side and ambient is now the hot side. That means that the front of the micro thermopile actually cools below ambient, in the same way that the windows of your automobile cool at night below ambient and condense dew or frost.

    It works amazingly well. I open my freezer, aim at the ice (which has near unity emissivity), and read -24 F almost instantly. Whether the sensor actually becomes -24 F is not certain since many factors are involved, but it will be nearly so.

    What is happening is that the hot/cold side of the thermopile are reversed. The front of the thermopile “sees” only the ice, and radiates its own infrared to the ice. The ice of course is radiating its own infrared at the sensor, the thermopile. Because of the tiny size of the sensor, and bathed in argon to reduce convection effects, it gives up its energy almost instantly, a time constant of 15 to 30 milliseconds depending on the model. It takes about 5 of these time constants to reach equilibrium.

    So, amazingly, it appears that the thermopile will actually become as cold, or hot, as what you aim it toward. It does this through radiance. A cold object IS radiating to the thermopile even though the thermopile is much warmer. The thermopile is radiating to the cold object at exactly the same time but, because of its higher temperature, at a much higher rate. In milliseconds both achieve equilibrium (with hardly any effect on the thing being measured) and you can directly read the voltage on the thermopile which is now the difference between the cold exposed surface and the ambient heatsink.

    Another purpose exists in bathing the sensor in argon; since it tracks “cold” it will also condense moisture so you have to keep cold air away from the sensor.

    You can experience this in a grocery store. You can “feel” the cold radiating from a freezer long before convection brings cold air to your face. But what you are feeling is a gap in the normal bath of ambient infrared that is everywhere around you and can be measured by that little infrared thermometer.

    So when you face something cold, your skin radiates to the cold object, but it does not radiate much back, and your skin is NOT in thermal equilibrium, and will cool. But in a normal household environment, your skin radiates energy but also receives the same amount from everything around you.

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