Chris Mooney reports on the latest attack on the hockey stick. Joe Barton, chair of the Committee on Energy and Commerce has sent out a set of letters, supposedly “requesting information regarding global warming studies”. However, if you look at the letters, you will find that the only study he is interested is Mann, Bradley and Hughes from way back in 1998 (the “hockey stick” study); and the questions are loaded ones of the form: “Can you explain why you made all the errors detailed in Mcintyre and McKitrick’s Energy and Environment paper?”

It is probably just a coincidence that Joe Barton has received $574,000 in campaign contributions from the oil and gas industry, more than any other congressman.

Update: Reaction from:

  • Atrios: “The appropriate response to this is ‘Bite me, Congessman’.”

  • teece: “This is the kind of tactic you would have expected in Soviet Russia.”

  • Kevin Drum: “Joe Barton is harassing scientists who have the temerity to publish results he finds inconvenient”

  • Josh Rosena: “This is an anti-climate science Congressman trying to get material for a smear against Mann.”

  • john m. lynch: “The interference continues.”

  • Paul from Wizbang: “I’m guessing the creators of the global warming hockey stick are –shall we say– pucked.”

  • Steve Verdon: “there seems to be a pattern with regards to climate scientists and their willingness to share data”

  • Mark Trodden: “Dear Congressman Barton, … I am extremely concerned by the tone and implications of these letters and consider them a thinly-veiled attempt to intimidate honest scientists into avoiding work that might lead to an opinion different from the current administration on topics that are politically sensitive.”

  • de Selby: “I expect industry whore congressmen to create false controversies. When they abuse their power at the expense of individual citizens, I call it McCarthyism”

  • David Appell: “This is unprecedented, as far as I know, and has the air of a scientific witch-hunt.”

  • PZ Meyers: “Joe Barton is an arrogant pissant”

  • James Annan: “I suspect that a witch-hunt like this could have serious repercussions for scientific research in the USA”

Comments

  1. #1 Tim Lambert
    July 25, 2005

    Ah, your silly free money offer. You don’t actually know the answers to the questions you asked, do you?

  2. #2 David
    July 26, 2005

    So that to weight by area, the input to PCA has to be weighted by the square root of area. I retract my sugestion that von Storch might have been mistaken ? he found an error in MBH98, though he does not seem to think that it was important.

    Those calling MBH’s failure to wait by latitude an error are kidding themselves. Both area weighted and non-area weighted PCA are used freely in climate because which is prefered depends on your goal. If you simply wish to perform a data compression (reduce the DOF) of a dataset such as was the primary goal of MBH98 this is more precisely achieved through non-weighted PCA rather than weighted PCA.
    This trivial issue is nothing compared to M&M’s radian versus degrees blunder…[McIntyre wants to make it clear that the degrees/radians mixup was McKitrick's error and had nothing to do with McIntyre. TL]

    BTW can someone explain why the sceptics are still so focused on a nearly 10 year old study, when 8 subsequent studies have broadly confirmed Mann et al, and all contradicted M&M’s fanciful recontruction.

    David

  3. #3 Dano
    July 26, 2005

    BTW can someone explain why the sceptics are still so focused on a nearly 10 year old study, when 8 subsequent studies have broadly confirmed Mann et al, and all contradicted M&M’s fanciful recontruction.

    Do you mean besides mindlessly parrotting predigested talking points, David?

    No.

    D

  4. #4 Eli Rabett
    July 26, 2005

    Mark, btw, http://farside.ph.utexas.edu/teaching/sm1/lectures/node69.html http://gaia.ecs.csus.edu/~reardonf/MET140/UandHforIdealGas.htm In the limit of small changes in temperature or ideal gas behavior Cv is a constant.

  5. #5 Robert P.
    July 27, 2005

    Eli, IIRC Mark argued that in general the internal energy of a gas is a function of its chemical composition as well of its temperature, so that Lambert’s formula was incomplete. That’s correct as far as it goes, although I do not think it goes very far – since the major constitutents of the lower atmosphere are well mixed except for water vapor, the only consequence is that one might want to take relative humidity into account when averaging temperatures. (Put another way, the effective heat capacity depends on relative humidity.) My guess it that the effect of this would be pretty small compared to other uncontrolled factors. I suppose sea-surface temperatures would get a larger weight while desert stations would get a smaller one when one computed the “average global temperature”, but the trend of the curve with time ought to be pretty much the same.

  6. #6 Mark Bahner
    July 29, 2005

    Tim Lambert writes (regarding my free money offer):

    “Ah, your silly free money offer. You don’t actually know the answers to the questions you asked, do you?”

    Heh, heh, heh! Good one, Tim! Ever the arrogant twit, eh? But seriously:

    1) Yes, I think I do know the answers to the questions I asked you.

    2) But I’m even more certain that you DON’T know the answers.

    3) And I’m even more certain, your arrogant twittering aside, that I’ve forgotten more about thermodynamics than you’ve ever learned (especially as related to the questions at hand).

    4) Finally, I’m quite certain, your laughably hypocritical assertion aside, that you will NEVER admit your error.

    As a result of those 4 items, I’m prepared to make you an Even More Astounding Free Money Offer:

    1) I will raise my offer to $30 if you will even ANSWER my questions, plus this additional question: Which has more internal energy: a cube of atmosphere 10^12 kilograms in mass, in the northeast wall of a Category 5 hurricane, or cube of air of equal mass at the same temperature on a clear calm day over Phoenix, Arizona?

    Note that you do NOT have to answer ANY of the questions correctly to get this $30you just have to attempt to answer them.

    3) I will send you $10 if you can offer me solid evidence that you’ve ever taken university-level course in Thermodynamics. (Solid evidence might include the approximate dates you took the course, the name of the university, and the professor’s name, or the textbook name, or something like that.) And I will send you $40 if you can send me similar evidence that you’ve ever taken a university-level course in Heating, Ventilation, and Air Conditioning. (THAT’S $40 that’s completely safe…or else you slept through the entire course!)

    3) I will send you $30 if you ever write: “I wrote (regarding the relationship between internal energy in the atmosphere and temperature) that: ‘I guess we’ll just have to ditch the entire field of thermodynamics then. In fact, Temperature T and internal energy U are related by the formula U=Tmc where m is the mass and c the specific heat.’ I realize now that I was wrong. Furthermore, I realize I also was wrong when I told Mark Bahner that, ‘The equation I gave is actually a very (close) approximation. Do you also complain that Newtonain physics is the wrong way to describe the atmosphere because it doesn’t account for relativistic effects?’ So I was (at least) twice wrong.’”

    So there you have it. I’ll give you: 1) $30 if you’ll even attempt to answer my questions (you don’t have to get the answers right), 2) $10 if you provide evidence of university-level coursework in Thermodynamics, and $40 more if you provide evidence of university-level coursework in Heating, Ventilation and Air Conditioning, and 3) $30 if you admit that you were (at least) twice wrong in this matter.

    Finally, I have a bit of free advice: when you’re stepping out of your particular areas of expertise, you might do well to have a bit more humility, and be a bit less condescending.

  7. #7 Eli Rabett
    July 29, 2005

    Robert, technically, Cp includes the effects of composition, so Tim’s formula was correct, it is just that the value of Cp is a function of composition. However, as you point out this is a distinction without a difference.

    We can, of course, dial in the numbers for dry and wet air. For dry air, at sea level the specific heat is 29.10 J/mole-K (essentially 3.5 R. The vibrational contribution is well under a tenth of a percent.) The specific heat of water vapor is 33.16 J/mole-K (essentially 4R).

    The maximum saturated vapor pressure of water vapor is less than 4%. Assuming 4% the heat capacity of the saturated air would be 29.26 J/mole-K. That is a difference of 0.5%. Splitting hairs.

  8. #8 Mark Bahner
    July 30, 2005

    Tim,

    Feel free to get all the help you want from your friends (e.g., Eli Rabett) on these questions:

    1) Which has more internal energy, a kilogram of air at 20 degrees Celsius and atmospheric pressure that is a) dry (i.e., 0% relative humidity[RH]), b) at 30% RH, or c) at 80% RH?

    2) Which has more internal energy, a kilogram of air at 30 degrees Celsius and at atmospheric pressure that is a) dry (i.e., 0% relative humidity, or RH), b) at 30% RH, or c) at 80% RH?

    3) How much internal energy would need to be removed to bring a kilogram of air at 30 degrees Celsius and atmospheric pressure down to 20 degrees Celsius if the air is: a) dry, b) at 30% RH, and c) at 80% RH?

    4) Which has more internal energy: a cube of atmosphere 10^12 kilograms in mass, in the northeast wall of a Category 5 hurricane, or cube of air of equal mass at the same temperature on a clear calm day over Phoenix, Arizona?

  9. #9 Ed Snack
    July 30, 2005

    quote “Ed, from the way you vanished from this thread, it is pretty obvious that you didn’t know what you were talking about on the weighting issue.” Gee thanks Tim. Just so obvious you knew all about it by your first reply. As for it being important or not, why don’t you work it through and find out ? Same goes for david, and it is not as if this was the only error in MBH is it, there is a whole corrigendum full of specific errors. And that doesn’t cover the problem of using the Bristlecone Pine records as a temperature proxy, or the concealing (or non-release) of unfavourable statistical tests.

    There’s plenty of meat for you to get your teeth into Tim, why don’t you ?

    And, just a quick note, Eli, are you still sure that Mann had released all his code ? Even our Tim no longer thinks that, and he is unwilling to correct his commenters although he does ask that others do so.

  10. #10 Tim Lambert
    July 31, 2005

    Ed, the fact McIntyre hasn’t bothered to work out the effect on the reconstruction is evidence that he doesn’t think it likely that it makes much difference.

    I specifically refered you to my comment where I corrected Eli on releasing all the code. Your next comment you claimed that it did not correct him. As far as I can tell, you have knowingly made a false statement. Please explain your conduct.

  11. #11 Mark Bahner
    July 31, 2005

    Tim Lambert writes, “As far as I can tell, you have knowingly made a false statement. Please explain your conduct.”

    Tim, why all this insistence that other people correct (what you think are) THEIR false statements, when you completely refuse to correct your OWN false statements? (Especially since you insist that you DO correct your false statements!)

    Let’s review one more time (from my first fabulous free money offer to you):

    1) Lambert quotes Paul Georgia of Tech Central Station:

    “Moreover, temperature and energy aren’t the same thing. The internal energy of a system can change without changing the temperature and the temperature can change while the internal energy of the system remains the same. In fact, this occurs all the time in the climate because the two variables are fundamentally different classes of thermodynamic variables and there is no physical law that requires that they move together.”

    2) To which Lambert responds:

    “Wow. I guess we’ll just have to ditch the entire field of thermodynamics then. In fact, Temperature T and internal energy U are related by the formula

    U=Tmc

    where m is the mass and c the specific heat. It is true that it is possible for internal energy to change without affecting the temperature if there is a phase change, but the atmosphere stays way above the temperature of liquid nitrogen, so this makes almost no difference to temperatures.”

    3) After I pointed out your response was wrong (your “nitrogen” comment was particularly hilarious!), you responded to me:

    4) “I mentioned nitrogen because that is what the atmosphere mostly is. Water vapour is less than 0.5% of the atmosphere. And the equation is true for water vapour as well if there is no phase change. The equation I gave is actually a very (close) approximation. Do you also complain that Newtonain physics is the wrong way to describe the atmosphere because it doesn’t account for relativistic effects?”

    5) However, you have refused to answer any of my followup question, even when I’ve offered you money to do so, and have completely withdrawn the requirement that your answers even be CORRECT in order to get the money!

    Do you STILL insist that the temperature and energy of the atmosphere are directly related by your magical equation, to a degree of accuracy that’s as close as neglecting relativistic effects?

    Or have you realized that Paul Georgia was basically right, and you are wrong? If so, why haven’t you ADMITTED that you are wrong?

  12. #12 Tim Lambert
    July 31, 2005

    Mark, I never wrote that the accuracy was the same as ignoring relativistic effects. That’s your misinterpretation.

    In order to encourage you to answer your “free money” questions, I make my own “free posting” offer: Answer your questions and gain the privilege of being able to post comments here.

    Until then, you may not post here.

  13. #13 John McCall
    August 1, 2005

    re: 203 “This trivial issue is nothing compared to M&M’s radian versus degrees blunder”

    You mean like your blunder of attributing a Michaels and McKitrick error (now corrected) in one paper, to McIntyre and McKitrick in another.

    Care to name the 8 subsequent studies that “have broadly confirmed Mann et al”, I presume they’re independent of MBH and RealClimate —- what you define as “broadly confirmed Mann et al” — and what you define as “all contradicted M&M’s fanciful reconstruction” — and can you show, McIntyre & McKitrick’s or is it Michaels & McKitrick’s reconstruction that you are referring to?

    Have you actually read these papers in entirety, or are you like Mr. Connelley, proud of that he hasn’t read them?

  14. #14 Mark Bahner
    August 2, 2005

    Tim Lambert writes, “In order to encourage you to answer your “free money” questions, I make my own “free posting” offer: Answer your questions and gain the privilege of being able to post comments here. Until then, you may not post here.”

    Heh, heh, heh! You’re a hoot, Tim!

    I offered you *****$30***** just to answer my questions! That offer is good even if you don’t get a single answer correct! Are you willing to make me the same offer? If not, why not? Is it because you know that I do know the answers?

    What in the world are you afraid of? In responding to Paul Georgia, you made a big deallike you knew thermodynamics cold (so to speak)! What’s the problem? Is the problem that you really are completely clueless regarding thermodynamics, such that all you know is what you read on Wikipedia?

    If so, why don’t you ask one of your friends who does know, like William Connolley?

    Or is it that you DO know the answers, and know that they’d utterly destroy your argument that your equation closely models the relationship between energy in the atmospere and temperature?

    Oy, vey!

    But I’m a very generous man, (Far more generous than you, obviously.) Not only will I keep the offer of $30 in place for you to answer the questions?whether your answers are RIGHT OR WRONG!?I will even be so generous as to answer question #4 for you.

    1. Question: “Which has more internal energy: a cube of atmosphere 10^12 kilograms in mass, in the northeast wall of a Category 5 hurricane, or cube of air of equal mass at the same temperature on a clear calm day over Phoenix, Arizona?”

    Answer: The cube of atmosphere 10^12 kilograms in mass, in the northeast wall of a Category 5 hurricane…obviously! (How hard can that be to answer, even for a clueless amateur?)

  15. #15 Tim Lambert
    August 3, 2005

    Mark, here’s a definition of internal energy (my emphasis):

    The total kinetic and potential energy associated with the motions and relative positions of the molecules of an object, excluding the kinetic or potential energy of the object as a whole. An increase in internal energy results in a rise in temperature or a change in phase.

  16. #16 Eli Rabett
    August 3, 2005

    Well, to the precision given in the “problem” statements and excluding all condensed phases in the various chunks of air, the answer is that the internal energy of the masses of air compared is the same in most of Markies’ fables, including the “hurricane and Phoenix”. That one has a precision of one decimal unit, ie, 1 E 12 kg so the internal energy is 1 E 12 kJ in both cases.

    OTOH because we are given no guidance as to the presence of condensed water, supersaturation, etc. several of the problem statements are unclear and you could play with them to give different answers. Kind of like the “physics student and the barometer” http://oldeee.see.ed.ac.uk/~slm/barometer.html

    The reason, of course, is that the saturated vapor pressure of water at non-Venusian temperatures is less than 4% and the difference in heat capacity between water vapor and oxygen or nitrogen is about 1/8 the heat capacity of dry air. Multiply those two together and you get an upper limit for the heat capacity difference between dry air and hot air at 100% relative humidity of about 0.5% in favor of the wet air, but that is pretty small.

    Another thing to think about is that the same volume of wet air weighs less than dry air, because H2O has a molecular weight of 18, N2 28 and O2 32. Comparison of wet and dry air on a mass basis rather than a molar** one, slightly shifts the heat capacity ratios towards the wet air.

    The big energy transfer process that water vapor brings to the atmosphere is, again of course, condensation. If one mole of water vapor condenses to liquid water the net decrease in the internal energy is -44 kJ/mol, or -2442 kJ/kg. By comparison, cooling a kg of water vapor 10 C changes the internal energy of the water vapor by -10 kJ/kg.

    If someone wants to throw a tortoise, a hare, perhaps an ogre or two into the hurricane we have the makings of a fractured fairy tale here.

    **For those who have forgotten their general chemistry a mole is Na=6 x 10^23 molecules. If you divide the weight of a chemical in grams by Na you get the weight in atomic units (now called Dalton) of a single molecule. Na, Avogadro’s number, allows us to “count out” the number of molecules by weighing them. It forms a bridge between the lab, where we weigh things, and the nanometer molecular basis where chemicals react. This costs a lot less then trying to count molecules one by one.

  17. #17 ÐanØ
    August 3, 2005

    With all due respect, and I hope this isn’t taken the wrong way Tim, but why does anyone engage Mark Bahner except to see what kind of rant he’ll produce?

    Best,

    an

  18. #18 Robert P.
    August 3, 2005

    Eli, I get the same result using a slightly different methodology: the difference between the molar heat capacities of dry air and saturated air at 298 K comes out to about one half of one percent.

    In general terms (Eli knows this but I thought a summary for lurkers might be helpful): the internal energy of a general thermodynamic system can be expressed as a function of its temperature, volume, and chemical composition:

    E = E(V, T, N1, N2, N3, ….)

    where N1 is the number of molecules of type 1, etc. (The answers.com article only considers the case of a single-component system.) So in general, it is quite possible to change the energy without changing the temperature, by varying one of the other parameters.

    However, we’re not interested in “general thermodynamic systems” here, we are interested in air. At atmopheric pressure, air behaves like an ideal gas at our level of precision, and a fundamental property of an ideal gas is that its internal energy does not depend upon its volume. The dependence on total number of particles is trivial, because E is extensive: two moles have twice the energy as one mole. That leaves chemical composition: the relative number of molecules of each type. Now different molecules have different heat capacities, so the internal energy of an ideal gas mixture does depend on its chemical composition. However, it once again comes down to almost nothing. All of the major components of the earth’s troposphere are well-mixed – the chemical composition is the same everywhere – except for one, water vapor. Water has a slightly different heat capacity than oxygen or nitrogen (because it is a triatomic rather than a diatomic) but when you put in the numbers, as Eli and I did, you end up with a difference of 0.5 percent between dry air and air saturated with water.

  19. #19 Mark Bahner
    August 3, 2005

    Tim Lambert writes, “Mark, here’s a definition of internal energy (my emphasis):”

    …blah, blah, blah. (You simply quote Wikipedia.)

    Wonderful, Tim! I see you can copy/paste from Wikipedia! Why don’t you use your (Wikipedian) knowledge to ANSWER MY QUESTIONS? (My emphasis.)

    MY answer to question #4 was “yes.” Is YOUR answer to #4, “No, they have the same internal energy”?

    If so, I have an IMPORTANT NOTE TO CLUELESS AMATEURS (my emphasis): I don’t know what planet you live on, but here on the planet earth, “a cube of atmosphere 10^12 kilograms in mass, in the northeast (eye)wall of a Category 5 hurricane” has quite a bit of WATER in it, in both VAPOR and LIQUID form! Again, emphasis is addedjust in case that changes your answer!

    Once again, MY answer to #4 was “yes.” Is YOUR answer to #4, “No, they have the same internal energy”?

    And if your answer to #4 is “no,” why don’t you go ahead and answer the first three questions:

    1) Which has more internal energy, a kilogram of air at 20 degrees Celsius and atmospheric pressure that is a) dry (i.e., 0% relative humidity[RH]), b) at 30% RH, or c) at 80% RH?

    2) Which has more internal energy, a kilogram of air at 30 degrees Celsius and at atmospheric pressure that is a) dry (i.e., 0% relative humidity, or RH), b) at 30% RH, or c) at 80% RH?

    3) How much internal energy would need to be removed to bring a kilogram of air at 30 degrees Celsius and atmospheric pressure down to 20 degrees Celsius if the air is: a) dry, b) at 30% RH, and c) at 80% RH?

  20. #20 Mark Bahner
    August 3, 2005

    Eli Rabett writes, “OTOH because we are given no guidance as to the presence of condensed water, supersaturation, etc. several of the problem statements are unclear”

    Heh, heh, heh! Good one, Eli! You must live on the same faraway planet as Tim Lambert! (See my previous comments to Tim.)

    Here on EARTH, you have been given “guidance” when someone writes, “a cube of atmosphere 10^12 kilograms in mass, in the northeast (eye)wall of a Category 5 hurricane” versus, a “cube of atmosphere of equal mass at the same temperature on a clear calm day over Phoenix, Arizona.”

    If you can’t figure out how those two situations differ, you need to visit our planet! (Or at least watch The Weather Channel.)

    Heh, heh, heh!

    You can go back to your calculations now.

    P.S. Does my telling you that clouds are made of water, and that hurricanes have lots of clouds, change any of your calculations at all?

  21. #21 Mark Bahner
    August 3, 2005

    Robert P. writes, “However, it once again comes down to almost nothing. All of the major components of the earth’s troposphere are well-mixed – the chemical composition is the same everywhere – except for one, water vapor. Water has a slightly different heat capacity than oxygen or nitrogen (because it is a triatomic rather than a diatomic) but when you put in the numbers, as Eli and I did, you end up with a difference of 0.5 percent between dry air and air saturated with water.”

    Please see all my comments to Tim Lambert and Eli Rabett, in case that changes your mind, Robert P. But in the event it doesn’t, do I correctly understand YOUR answers to be as follows?

    1) “A kilogram of air at 20 degrees Celsius and atmospheric pressure that is a) dry (i.e., 0% relative humidity[RH]), b) at 30% RH, or c) at 80% RH all have the same internal energy.”

    2) “A kilogram of air at 30 degrees Celsius and at atmospheric pressure that is a) dry (i.e., 0% relative humidity, or RH), b) at 30% RH, or c) at 80% RH all have the same internal energy.”

    3) “The amount of internal energy that would need to be removed to bring a kilogram of air at 30 degrees Celsius and atmospheric pressure down to 20 degrees Celsius if the air is: a) dry, b) at 30% RH, and c) at 80% RHis ?????????” (Presumably the same value?)

    4) “A cube of atmosphere 10^12 kilograms in mass, in the northeast (eye)wall of a Category 5 hurricane, and a cube of atmosphere of equal mass at the same temperature on a clear calm day over Phoenix, Arizona both have the same internal energy.”

    If I understand YOUR answers to be as shown above, please confer with Tim Lambert, and get him to either agree or disagree that those are HIS answers. (It’s important, because I offered him $30 if he would answerREGARDLESS of whether or not the answers were correct. But first I want to make sure he really HAS answered. So far, all he does is quote Wikipedia.)

    P.S. By the way, I’d like to get your/Tim’s number (or numbers) for item #3, where there are question marks.

  22. #22 Mark Bahner
    August 3, 2005

    Dano writes, “…but why does anyone engage Mark Bahner except to see what kind of rant he’ll produce?”

    Oh, brother.

    I can tell you why YOU should “engage” me, Dano. You might learn something.

    Like remember how you used to go into spasms of ululation every time I mentioned my probabilistic predictions for methane atmospheric concentrations, CO2 emissions and atmospheric concentrations, and resulting temperature increases? You would bray that predicting such things was like predicting the results of the Powerball?

    Remember how I very patiently and generously explained and showed you that the Powerball are random, and that atmospheric methane concentrations, CO2 emissions and atmospheric concentrations, and resulting temperature increases are….um….

    NOT?

    Did you learn anything from that, or are you still just as clueless?

    P.S. Speaking of probabilistic projections, maybe y’all could explain to Kevin Vranes how it might be “relevant” as to whether there is a 99% chance of the earth warming less than 1.4 degrees Celsius, or a 99% chance of the earth warming more than 5.8 degrees Celsius?

  23. #23 Tim Lambert
    August 3, 2005

    Mark, the definition of internal energy I quoted comes from The American Heritage Dictionary of the English Language, Fourth Edition Copyright 2004, 2000 by Houghton Mifflin Company.

    It does not come from Wikipedia.

    The only one of your questions that is even relevant is question 3 and we both know the reason why you don’t want to post the answer — the differences are small enough that they are usually ignored.

  24. #24 Eli Rabett
    August 3, 2005

    Ah yes, old Mark again. As we pointed out the answer to your first and second question is that to to the precision of the stated problems (two significant figures) they all have the same internal energy. The difference is in the fourth significant figure.

    The answer to the first two parts of the third question is the same. The answer to the third part depends if condensation occurs or the air is supersaturated.

    As to the fourth you are contradicting yourself as atmosphere refers to the gas phase inspite of your posturing. I suggest you consult a dictionary before blathering.

    http://www.m-w.com

    Main Entry: at.mo.sphere
    Pronunciation: ‘at-m&-”sfir
    Function: noun

    Etymology: New Latin atmosphaera, from Greek atmos vapor + Latin sphaera sphere

    1 a : the gaseous envelope of a celestial body (as a planet) b : the whole mass of air surrounding the earth

    2 : the air of a locality

    3 : a surrounding influence or environment

    4 : a unit of pressure equal to the pressure of the air at sea level or approximately 14.7 pounds per square inch (101,325 pascals)

    5 a : the overall aesthetic effect of a work of art b : a dominant aesthetic or emotional effect or appeal.

    Me, I’m a humble sock puppet, but Robert, he knows for sure what he is talking about. Mark, you don’t

  25. #25 Mark Bahner
    August 4, 2005

    Tim Lambert writes, “The only one of your questions that is even relevant is question 3 and we both know the reason why you don’t want to post the answer ? the differences are small enough that they are usually ignored.”

    Please, Tim, do NOT try to characterize what is relevant and what isn’t. As I pointed out, according to Kevin Vranes, it isn’t “relevant” whether there’s a 99%+ chance that the earth will warm by less than 1.4 degrees Celsius in the 21st century, or a 99%+ chance that the earth will warm by more than 5.8 degrees Celsius. So I do NOT agree with leftists’ characterization of what is or isn’t “relevant.”

    Do I correctly understand your answers to the FIRST TWO questions (let’s at least get SOME questions answered) to be:

    1) “A kilogram of air at 20 degrees Celsius and atmospheric pressure that is a) dry (i.e., 0% relative humidity[RH]), b) at 30% RH, or c) at 80% RH all have the same internal energy.”

    2) “A kilogram of air at 30 degrees Celsius and at atmospheric pressure that is a) dry (i.e., 0% relative humidity, or RH), b) at 30% RH, or c) at 80% RH all have the same internal energy.”

    ????

    And what about the fourth question, do I correctly understand your answer to be:

    4) “A cube of atmosphere 10^12 kilograms in mass, in the northeast (eye)wall of a Category 5 hurricane, and a cube of atmosphere of equal mass at the same temperature on a clear calm day over Phoenix, Arizona both have the same internal energy.”

    ?????

    Finally, on the third question, do I correctly understand your answer to be:

    3) “The amount of internal energy that would need to be removed to bring a kilogram of air at 30 degrees Celsius and atmospheric pressure down to 20 degrees Celsius if the air is: a) dry, b) at 30% RH, and c) at 80% RH…are all within…less than 3 percent of one another?”

    ????

    If you say that I correctly understand all your answers, I’ll send you the $30. (Though I gotta tell you, I wouldn’t hire you as a consultant, given your extreme reluctance to actually provide definitive answers!)

  26. #26 Scott Church
    August 4, 2005

    Tim,

    Based on the history of this thread, I’d say Dano has a point. So far, I’ve been unable to discern any coherent train of thought in Mr. Bahner’s comments that is relevant to the original subject, and he’s been abusive and downright juvenile. Short of using profanity and making death threats I’m not sure what else he could do to qualify for Troll status. I’d welcome any thoughts he has regarding global temperature trends and the relevance of Barton’s inquiries to the Hockey Stick controversy, provided they’re offered with civility and professionalism. But unless this is forthcoming, this thread is going to get sucked into a pit of troll quicksand before anything constructive can be accomplished. I’d say it’s time to implement the Troll Filter. What do you think?

    All the best.

  27. #27 ÐanØ
    August 4, 2005

    Mr Church sir,

    a much more likely explanation is that Mark’s caps lock is merely stuck, rather than there being a swirling descent into trollishisms.

    I’ll bet you $30 it’s so.

    Best,

    an

  28. #28 Mark Bahner
    August 4, 2005

    Scott Church writes, “So far, I’ve been unable to discern any coherent train of thought in Mr. Bahner’s comments that is relevant to the original subject,…”

    I came to this thread because I saw Tim’s false statement, “When I make a mistake, I correct it.”

    But you’re right, this exchange is getting tiring. Tim Lambert has completely refused to answer my simple questions. I’ve even gone so far as to provide what I think his answers are, based on his cut/pasting of definitions from a dictionary.

    Since it’s been more than 24 hours, and you haven’t even confirmed or denied the answers I thought you were making, Tim, I count you as being unresponsive. Therefore, you do NOT get the $30.

    NOTE: It’s very interesting to me that Robert P. ALSO didn’t confirm or deny what I thought were HIS answers. That’s a little surpising, because I thought Robert P. was actually going to respond.

    Since y’all have not responded, I have no reason to give all my answers, but I will anyway. You’ll notice that I’m changing the wording, from “internal energy” to “enthalpy.” That’s the term that Paul Georgia should have used:

    1) At 20 degrees Celsius, a kilogram of air at 80 percent relative humidity has more enthalpy than a kilogram of air at 30 percent or zero percent relative humidity.

    2) At 30 degrees Celsius, a kilogram of air at 80 percent relative humidity has more enthalpy than a kilogram of air at 30 percent or zero percent relative humidity.

    4) A cube of atmosphere 10^12 kilograms in mass, in the northeast (eye)wall of a Category 5 hurricane, has way, way, way, WAY more enthalpy than a equal mass at the same temperature on a clear calm day over Phoenix, Arizona.

    3) Cooling 1 kilogram from 30 degrees Celsius and 80 percent relative humidity to 20 degrees Celsius and 80 percent relative humidity requires…

    *****3.5 TIMES AS MUCH ENERGY*****

    …to be removed as cooling 1 kilogram at 30 degrees Celsius to 20 degrees Celsius, if both are dry (at 0 percent relative humidity).

    I assume you all agree with all 4 of my statements?

    (Hint: You should, because I actually know what I’m talking about. As compared, for example, YOU, Tim Lambert.)

  29. #29 Eli Rabett
    August 5, 2005

    You seem to be confusing internal energy and enthalpy. Internal energy is NOT enthalpy. When you heat a gas at constant pressure, some of the heat energy goes to doing work in expanding the volume and some to internal energy. Your problems were stated in terms of internal energy.

    Second, your answer to problem 3 assumes that some of the water vapor will condense when cooled. Almost all of the energy difference comes from the condensation, not the cooling of the vapor.

    And yes, we did answer your poorly stated problems. And now we have read your incorrect answers. Thanks for the giggles.

  30. #30 Mark Bahner
    August 5, 2005

    “And yes, we did answer your poorly stated problems. And now we have read your incorrect answers. Thanks for the giggles.”

    Please identify which one of my statements was incorrect. You can’t. They were all absolutely correct.

  31. #31 Mark Bahner
    August 5, 2005

    Eli Rabett writes, “And yes, we did answer your poorly stated problems.”

    Yeah, right, Eli.

    OK, if you feel cheated, I’ll give you $10 if you can correctly label these assertions as “true ” or “false”:

    “Surface air temperature alone is inadequate to monitor trends of surface heating and cooling. The SI units for temperature are degrees Celsius, and the SI units for heat are Joules. The surface temperature can go up while the enthalpy goes down or remains the same. The surface temperature can go down while the enthalpy goes up or remains the same. The surface temperature can remain the same while the enthalpy goes down or up.”

    And I’ll give you another $20 if you correctly answer by what percentage the enthalpy of a given mass of air goes up or down if:

    1) It goes from 20 degrees Celsius and 50 percent relative humidity to 30 degrees Celsius and:

    a) 10 percent relative humidity, or
    b) 30 percent relative humidity.

    2) It goes from 20 degrees Celsius and 30 percent relative humidity to 15 degrees Celsius and:

    a) 80 percent relative humidity, or
    b) 60 percent relative humidity.

    3) It stays at 20 degrees Celsius, but the relative humidity changes from 40 percent to:

    a) 20 percent, or
    b) 70 percent.

    I’ll be working/playing this weekend, and I won’t have time to bother with this. So you can take your time, and get your answers right.

  32. #32 Ian Gould
    August 6, 2005

    Actually the SI unit for temperature is the Kelvin, not the degree Celsius.

    This is, of course, a minor nit-picking point but no more so than most of your own.

  33. #33 Tim Lambert
    August 7, 2005

    Mark, thanks for answering some questions, even though it took you over a year and you changed the questions.

    You claim that

    “At 20 degrees Celsius, a kilogram of air at 80 percent relative humidity has more enthalpy than a kilogram of air at 30 percent or zero percent relative humidity.”

    I am not convinced that your answer is correct. Please state what you think the actual value of the enthalpy is for theose three cases.

  34. #34 Robert P.
    August 8, 2005

    Actually, Tim (and Mark and Eli), Mark’s first two questions do not have a unique answer. In thermodynamics, energies and enthalpies are not absolute quantities – their numerical values are defined with respect to a standard state. For pure substances there is a single convention for the standard state – the stable state of the substance at 1 bar pressure – so one can compare enthalpies of pure substances without ambiguity. For mixtures, however, more than one standard state is in common use.

    For a solution of two components that have more or less similar properties, such as oxygen/nitrogen or ethanol/water, the usual choice of standard state is the pure components. That is, the enthalpy (and energy, and free energy) of the solution is measured with respect to the enthalpies of the pure solvents in their stable states at 1 bar pressure. The “heat of solution” then comes out to be the difference between the enthalpy of the solution and a weighted average of the enthalpies of the components.

    If the two components have very different properties, however, and if one component is normally present in low concentrations, a different reference is commonly used. The standard state for the solvent is again taken to be the pure solvent in its stable state at 1 bar, but the standard state of the solute is defined in terms of an extrapolation of the solute’s properties to infinite dilution. This is sometimes called the “Henry’s Law standard state”, because it amounts to using the properties of the solute in the regime where Henry’s Law is accurate (high dilution.) This is what’s generally done for solutions of solids in liquids, e.g. sugar in water (ionic solutions involve further complications.)

    So if we ask, “what is the enthalpy of humid air”, we need to specify what standard state we are using for the water. If it’s the pure substance, then the standard state is liquid water, and the enthalpy of humid air differs significantly from the enthalpy of dry air, the difference coming almost entirely from the heat of vaporization of the water. If we use the Henry’s Law standard state for the water, the standard state is water vapor and the molar enthalpy of humid air differs from that of dry air by about 0.5 percent, the difference coming from the slight difference in heat capacities between water and O2 or N2.

    The two choices can be thought of as referring to two different experiments. In the first, you take a small amount of liquid water and let it evaporate into a large volume of air, large enough so that it evaporates completely. The air temperature will decrease in order to make up for the heat of vaporization. If you then heat it back up to its original temperature, the amount of heat supplied is the difference between the enthalpy of humid air and dry air. In the second experiment, you take a corresponding amount of water vapor and let it mix into a volume of air. Substituting water molecules for air molecules changes the enthalpy only very slightly.

    Mark’s first two questions are a bit of a distraction, therefore. His third, however, is very much to the point. Let me restate the point, as I see it, by considering a variant of Tim’s original argument way back when. Suppose we thermally link two volumes of air, one dry and one humid, at different temperatures. What will the final temperature be ? If condensation does not take place, then the final temperature is a simple weighted average of the initial temperatures. The weighting is primarily given by the masses, with a very small correction arising from the heat capacities. If condensation occurs, however, you need to take into account the heat of vaporization as well when you calculate the final temperature. The final temperature is still a linear function of the original temperatures, but there will in addition be a constant term.

  35. #35 TCO
    August 8, 2005

    There is a lot of posturing here and trying to put people on the spot by quizzing them vice making assertions. If either of you is so convinced of your thermo that you can make an argument, please go ahead and do so. I’m a little worried that neither of you really feels confident on the nitty gritty (which can be tricky). That said, I think JohnA came accross as the silliest. (But I don’t have the thermo down nat’s ass either…would appreciate a bit of exposition/teaching vice all the posturing…)

  36. #36 Scott Church
    August 9, 2005

    It seems to me that this entire 5 page long thermodynamic debate misses the point. The original subject of this thread was Joe Barton’s inquiry into the activities and funding Mann et al, their Hockey Stick work, and whether or not his pursuit of them constitutes a witch hunt. The subject of the research being challenged is historical temperatures and whether or not the Hockey Stick represents them accurately.

    Internal energy, enthalpy, moist vs. dry air and all aside, temperature is a “state” function. In other words, 50 deg. F is 50 deg. F regardless of the path taken to get to that point (the same is not true of entropy or enthalpy change for instance). Regardless of whether we’re discussing dry air, moist air, peanut butter, personal lubricant, or whatever, 50 deg. F is 50 deg. F. It is entirely meaningful to speak of average temperatures on a zonal, hemispherical, or global basis without bringing up the thermodynamics of moist vs. dry air. The real difference is that at 50 deg. F water and moist air will both hold more total energy then dry air. Thus, they will be a better medium for transferring latent heat from one location to another. Alternately, it will take more energy to produce a given temperature change in them (which is of course, why coastal climates tend to be more moderate than inner continental ones). But either way, 50 deg. F is still 50 deg. F. Tim has already dealt at length with the failure of McKittrick’s claim that there is no basis for the concept of average temperature. That discussion does introduce thermodynamic concepts into the debate, but McKittrick’s claims would fail whether these were mentioned or not.

    Furthermore, the whole argument is rendered moot simply by considering the entire earth/ocean system. Globally averaged temperature increases will raise the total energy content of the complete atmosphere/ocean system regardless of how energy is distributed regionally within it. It’s straightforward to show that increases in the total global energy content of the atmosphere will result in an increase of the global average tropopause height. Evidence suggests that this is in fact occurring (Santer et al., 2003, Science 301). Likewise, the total global energy content of the world’s oceans is also increasing (Hansen et al., 2005, Science 308; Levitus et al., 2000, Science 287; Hansen et al., 2002, J. Geophys. Res. 107).

    Unless Mr. Bahner and other Barton supporters are prepared to argue that a meter is not really a meter and there’s no meaningful definition of altitude, these facts remain no matter how many capital-letter laden challenges and bet spectacles they indulge in.

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