 2011-02-21 14:24:40 John McLean Miscalculation Alan aj.flint@bigpond... 121.219.99.104 Thought you might be interested in another McLean rant that seems to have its basic physics that he claims to know so much about all wrong.http://www.quadrant.org.au/blogs/doomed-planet/2011/01/alarmists-ignore-physicsHere he is claiming 1m of seawater has the same heat capacity of 3300m of atmosphere.A few calculations seem to prove that the heat capacity he claims is out by a factor of 855. But it is a while since I did tertiary physics so I wouldnя't mind a cross check.Values for specific heat capacity CpAir: 1.0035 J/(g*K)fresh Water: 4.1813 J/(g*K)The density of air atВ  sea level and at 15Г‚В°C according to ISA (International Standard Atmosphere), air has a density of approximately 1.22521 kg/m3. At 3300m it is about 72% of sea level, basically linear reduction with height. So in the first 3300m density is an average 86% of SL or 1053.5 kg/m3.1m of Seawater; density is 1025kg/m3, specific heat capacity 3993 J /kg/K @20 degrees.В В В В В В В В В В В В В В В В В В В В В В В В В В В В В  В waterВ В В В В  airSLВ  density/m3В В В В В В  1025В В В В В В В В  1.22521densityВ factorВ В В В В В В В В В В В В В  1В В В В В В В В В В  .86av/density kg/m3В В В В 1025В В В В В В В В В 1.0535В В  heightВ В В В В В В В В В В В В В В В В В В В В В В В В  В 1В В В В В В В В  3300 actual kgВ  "x"В В В В В В В В В В  1025В В В  В В В В  3477В В В  kj/kg/KВ В В В В В В В В В В В В В В В В В  3993В В В В В В В В 1003.5 kj/KВ for \"x"В В В В В В В 1028588В В В В В  3563464В  В Ratio q air/H20В В В В В  .2886В В В В В В В В В В В  В Ratio qВ H20/airВ В В В В В В 3.464В В В В В В В В В В В В В В So 3300m ofВ airВ has a bit more than a quarter of the heat capacity of 1m of water.So McLeansВ claim that 3300 m of air has the same heat capacity ofВ 1 m of waterВ is wrongВ by a factor of 3300/3.464 or 952.5It seems heВ can't even claim a decimal point error with any more conviction than the rest of his bodgy unqualified theories!Alan Flint 2011-02-21 14:37:28 Correction, heat capacity of water overstated by 3.4 times. Alan aj.flint@bigpond... 121.219.99.104 Correction, the heatВ capacity of water spould read of course as overstated by 3.4 times as McLean stated it was equivalent.Can anyone confirm?Alan 2011-02-21 19:11:12 MarkR Mark Richardson m.t.richardson2@gmail... 134.225.187.80 I got a bit confused here: actual kgВ  "x"В В В В В В В В В В  1025В В В  В В В В  3477В В В  kj/kg/KВ В В В В В В В В В В В В В В В В В  3993В В В В В В В В 1003.5В   Shouldn't you multiply mass by specific heat capacity to get total heat capacity?1025 x 3993 =В  4.1E63477 x 1003.5 =В  3.5E6 2011-02-21 19:42:50 Errors nealjking nealjking@gmail... 84.151.62.32 Alan,At least one of your numbers is wrong:1,028,588 = 1025 x 1003.5 ; whereas you should be comparing: 1025 x 3993 = 4,092,825 withВ В В В В В В В В В В В В В В В В В В В В В В В В В В В В В В В В В В В В В В В В В В  1003.5 x 3447 = 3,489,169.5 These are not a perfect match, but it is not an unreasonable comparison: McLean doesn't seem way off. But actually, there is a more sophisticated way to do this comparison:The question is, What altitude of atmosphere has equivalent heat capacity (Cp) to 1 meter of sea water?The heat capacity of 1 meter of sea water / unit area (HC/A)_sw = 1 (m) *В  (density) * (specific heat water)В  В  В  В В В В В В В В В  = 1 * 1025 * 3993 В В В В В В В В В В В В В В В  = 4,092,825For the atmosphere from ground level up to altitude h,(HC/A)_h = Integral (density) dz * (specific heat air)В  В  В  В  В  В  В  =В  1.0035 * Integral (density) dzThese are equal when:Integral (density) dz = (HC/A)_h / 1003.5 = (HC/A)_sw / 1003.5 =В  4,092,825/1003.5 = 4,078.55We can use the equation of hydrostatic equilibrium to figure this out:dP/dz = - (density) * gВ  , so:Integral (density) dz = - Integral (dP/dz) dz * (1/g) = (P(0) - P(h))/gTherefore:P(h) = P(0) - g*(4,078.55) =В  101,325 - 9.8*4,078.55 = 61,335 (Pa)According to the wiki on atmospheric density (http://en.wikipedia.org/wiki/Density_of_air#Altitude )P(h) = P(0) * (1 - L*h/To)^(Mg/(RL)with:L = 6.5e-3 (K/m)M = 2.896e-2 (kg/mol)R = 8.31 (J/(mol*K))To = 288.15 (K)So:P(h) = P(0) * (1 - h*(2.56e-5))^(5.254)(1 - h*(2.56e-5)) = (P(h)/P(0))^(1/5.254) = (61335/101325)^(1/5.254) = 0.90888h = (1 - 0.90888)/(2.56e-5) = 3559 (m)В  OK, so I get that the lower 3560 meters of atmosphere has the equivalent heat capacity to 1 meter of sea water. McLean said 3300 meters. I can't speak for the rest of his argument, but he's well within reason for THIS estimate. 2011-02-21 19:44:48 Correction WATER?AIR heat figures transposed Alan aj.flint@bigpond... 121.219.99.104 Thanks. You are correct. I knew there was something wrong somewhere. I accidentally recycled the wrong heat capacity into this post whilst a bit tired and trying to reformat the post here from Excel.That's why peer review is so good. So the true ratio of heat capacity is approximately 4.1/3.6 seawater to 33300 cubic metres of air or 14% greater than the stated unity heat capacity of 3300 cubics vis a vis with water. That would be enough to account for his failure to observe atmospheric warming perhaps? 2011-02-21 20:02:39 nealjking nealjking@gmail... 84.151.62.32 My impression of his argument was that he derived a height of 3300 meters from requiring a heat-capacity equivalence to a meter of ocean water, not a ratio of heat capacities from a height of 3300 meters. He is roughly correct that it takes a whole lot of atmosphere to match the heat capacitance of a bit of seawater.However, his broader point is still wrong:В - Oceans do not have to get all their heating from the atmosphere: Sunlight is absorbed by the ocean, and the increased evaporation and warming increase in step, not one before (or "blocking") the other.- Even if one considers only heat transfer from the atmosphere to the ocean, the point is that the atmosphere is not just heated and then tossed against the ocean: The atmosphere continues to be warmed by the enhanced greenhouse effect, and as long as it is in contact with the ocean, its higher temperature will still mean that it transfers heat to the ocean. Yes, the air in immediate contact will cool; but convection will whirl away the cooled air and bring back the warmer air: as long as the atmosphere is warmer than the ocean surface, it will continue to provide heat to it. The direction of heat transfer is determined by temperature difference, not by heat-capacity difference.