2011-02-21 14:24:40John McLean Miscalculation


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.


Here 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 wouldnˇ't mind a cross check.

Values for specific heat capacity Cp

Air: 1.0035 J/(g*K)

fresh Water: 4.1813 J/(g*K)

The density of air at¬† sea level and at 15√ā¬įC 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.

                               water      air

SL  density/m3       1025         1.22521

density factor               1           .86

av/density kg/m3    1025         1.0535  

height                           1         3300

actual kg  "x"           1025         3477   

kj/kg/K                   3993        1003.5

kj/K for \"x"       1028588      3563464 

 Ratio q air/H20      .2886           

 Ratio q H20/air       3.464              

So 3300m of air has a bit more than a quarter of the heat capacity of 1m of water.

So McLeans claim that 3300 m of air has the same heat capacity of 1 m of water is wrong by a factor of 3300/3.464 or 952.5

It seems he 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 14:37:28Correction, heat capacity of water overstated by 3.4 times.



Correction, the heat capacity of water spould read of course as overstated by 3.4 times as McLean stated it was equivalent.

Can anyone confirm?


2011-02-21 19:11:12
Mark Richardson

I got a bit confused here:


actual kg  "x"           1025         3477   

kj/kg/K                   3993        1003.5 


Shouldn't you multiply mass by specific heat capacity to get total heat capacity?

1025 x 3993 =  4.1E6

3477 x 1003.5 =  3.5E6



2011-02-21 19:42:50Errors



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 with

                                            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) *  (density) * (specific heat water)

                = 1 * 1025 * 3993

                = 4,092,825

For the atmosphere from ground level up to altitude h,

(HC/A)_h = Integral (density) dz * (specific heat air)

              =  1.0035 * Integral (density) dz

These are equal when:

Integral (density) dz = (HC/A)_h / 1003.5 = (HC/A)_sw / 1003.5 =  4,092,825/1003.5 = 4,078.55

We can use the equation of hydrostatic equilibrium to figure this out:

dP/dz = - (density) * g  , so:

Integral (density) dz = - Integral (dP/dz) dz * (1/g) = (P(0) - P(h))/g


P(h) = P(0) - g*(4,078.55) =  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)


L = 6.5e-3 (K/m)

M = 2.896e-2 (kg/mol)

R = 8.31 (J/(mol*K))

To = 288.15 (K)


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) = 0.90888

h = (1 - 0.90888)/(2.56e-5) = 3559 (m) 

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 19:44:48Correction WATER?AIR heat figures transposed

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


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: 

- 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.