INDEX TO THIS TOPIC:

HEAT STORED BY GREENHOUSE GASES

Created on ©27 April 2007 by Biology Cabinet. Last Revision by the scientific council and the author of this paper on 05 August 2011.

The authors are grateful to TS for his kind assistance with the text (he really worked hard). The authors are grateful to Climate Audit engineers and scientists for their praiseworthy help on the
improvement of this article. However, any errors in this paper are mine alone.

Updated on August 27, 2011.

To quote this article copy and paste the next TWO lines. Please, fill in the spaces of day, month and year:

Nahle, Nasif. Heat Stored by Greenhouse Gases. Biology Cabinet. 27 April 2007. Obtained on  _____(month)  _____(day), _____(year); from http://biocab.org/Heat_Stored.html

1. INTRODUCTION

When investigating the propagation of energy, we must take into account the science of thermodynamics, which allows us to predict the trajectory of the process, and the processes of heat transfer to know the modes by which energy is propagated from one system to other systems. Heat is not the same as temperature because heat is energy in transit. Heat is energy being transferred from a warmer system to a cooler system due to a gradient of temperature, whereas temperature is the measurement of the average of the kinetic energy of the particles of a substance. The average of the molecular kinetic energy depends on the translational motion of the particles of a system.

The energy absorbed or stored by a substance could cause an increase in the kinetic energy of the particles of that substance. This kinetic energy or motion causes the particles to emit energy, which is transferred to other subsystems of the emitter or towards other systems with a lower energy density.

To understand heat transfer we have to keep in mind that heat is not a substance, but entropy that flows from one system with high entropy towards other systems with lower entropy.[1]

GENERAL FORMULAS AND LAWS:

1st Law of Thermodynamics: Energy can be changed from one form to another, but it cannot be created or destroyed.

The mathematical expression of the 1st. law is as follows:

ΔU = ΔQΔW

Where ΔU is the increase of the internal energy of a thermodynamic system, ΔQ is the amount of heat applied to the thermodynamic system, and ΔW is the change of work done by that thermodynamic system.

The formula means that the change in the internal energy of a system is equal to the heat transferred to that system minus the work done by that system in its environment.

2nd Law of Thermodynamics: In all transformations from one form of energy into another form of energy, a quantity of energy is always dispersed towards other systems as heat..

The mathematical expression of the 2nd law is as follows:

ΔS/Δt ≥ 0

Where ΔS is the increase of the entropy, and Δt is time.

The formula denotes that the change in the entropy in a thermodynamic system is always higher or equal to zero, and that time is the fundamental dimension in which the system is doing work.

The formula permits us to deduce other conceptualizations of the 2nd law which mean the same thing, for example:

1. No system can transform energy into useful forms of energy with an efficiency of 100 percent.

2. Energy cannot spontaneously rearrange from low density states to high density states.

3. Heat is never spontaneously transferred from cold systems to hot systems.

4. The entropy of any thermodynamic system is constantly increasing over time.

GENERAL FORMULAS TO CALCULATE THE VARIATION OF TEMPERATURE:

Convection:

Δq = h (A) (T1-T2)

Where Δq is heat variation, A is the surface area in square meters, h is the convective heat transfer coefficient of a given substance, and T1-T2 (in Kelvin) is the difference of temperature between the warmer systems and the cooler system.

Change of Temperature:

ΔT = q /m (Cp)

Where ΔT is the tropospheric temperature variability, q is the amount of heat absorbed by a given substance, m is the current volumetric mass of that substance, and Cp is the specific heat of that substance at P = 1 atm and T = 300 K.

Conversion from ppmv to mg/m^3:

W mg/m^3 = ppmv (12.187) (MW) / 273.15 + °C

Where W is the density of the substance expressed in milligrams per cubic meter, ppmv is the concentration of the substance expressed in parts per million by volume, 12.187 is a constant of proportionality, MW is the molar weight of the substance, and 273.15 + °C is the temperature expressed in Kelvin.

If temperature changes with time, as it does, the atmosphere's temperature:

q stored = m (Cp) (ΔT/Δt) (Pitts, Donald and Sissom, Leighton. Heat Transfer. 1998. Pg. 1, and Potter & Somerton. 1993. Page 58. (For a quasiequilibrium process with Cp rel. constant)).

Where mass (m)  is the product of volume and density (Pitts & Sissom. 1998. Page 1), Cp is the Specific heat of the substance, ΔT is the change of temperature (final temperature - initial temperature), and Δt is the change of time. The result is the relation between the energy absorbed by the substance and the energy removed from the substance. (Pitts & Sissom. 1998. Page 1) (Wilson. 1994)

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2. ALGORITHM FOR CO2:

KNOWN DATA:

The weight of 1000 L of air is 1.23 Kg, or 1.23 N (Manrique. 2002. Oxford. Page 290).

The density of the dry air at T = 0 °C and P = 101.325 kPa is 1.292 kg/m^3.

The density of the mixed air at T =  25 °C and P = 101.325 kPa is 1.18 g/L, or 1.18 Kg/m^3. (Manrique. 2002).

At ambient T = 25 °C and P = 1 atm, dry air has a density of 1.168 kg/m^3 (Pitts & Sissom. 1998. Page 344)

At present, 1 cubic meter of air contains 0.000690 Kg of CO2 (690 mg).

Δ [CO2] in the last 200 years = 101 ppmv = 0.000164 Kg / m^3

Notice that for denoting concentration we enclose the symbol of the substance between square brackets [...]

Example from nature: Data taken from the meteorological station in Monterrey, Mexico: On June 22, 2007 at 18:05 UT, at the coordinates 25º 48´ North latitude and 100º 19' West longitude, and an altitude of 513 meters ASL, the air temperature at 1.5 m above ground level was 299.65 K (26.5 °C), whereas the ground temperature was 300.15 K (27 °C).  Which is the load of heat transferred from the ground to a mass of CO2 when 1 cubic meter of air contains 0.00069 kg.

CO2 Thermal Conductivity Coefficient of CO2 (k) = 0.016572 W/m*K (Manrique. 2002.Oxford)

A = 1 m^2

T soil = 300.15 K

T air at altitude 1.5 m = 299.65 K

ΔT = 0.5 K

Δq = -k (ΔT/d) or Δq = -k [(T1-T2)/d]

Δq = -0.016572 W/m*K (0.5 K / 1.5 m) = - 0.005524 W (power transferred by conduction from the surface to the atmospheric carbon dioxide at 1.5 m above ground).

0.005524 W = 0.005524 J/s (http://www.techexpo.com/techdata/conversn.html)

CONVERSION OF 0.005524 J/S TO CHANGE OF TEMPERATURE:

The heat transfer occurred through one second, thus the amount of energy transferred is:

q = 0.005524 J/s (1 s) = 0.005524 J

We use the following formula to know the change of temperature of a given mass of any substance, carbon dioxide for this instance:

ΔT = E / m (Cp)

Known values:

E = 0.005524 J
m of carbon dioxide (cd) = 0.00069 Kg
Cp of CO2 = 871 J/Kg*°C (Pitts & Sissom. 1994. Shaum's) (Engels.1998) (Manrique. 2002.Oxford)

Introducing magnitudes:

ΔTCO2 = 0.005524 J / 0.00069 Kg [871 J/Kg*°C]  = 0.00919 °C (rounding up the number, ΔTCO2 = 0.01 °C).

Heat Stored by 381 ppmv (0.00069 Kg) of CO2:

q = m (Cp) (ΔT) (Potter & Somerton. 1993. Page 58) (For a no equilibrium process with Cp rel. constant)

Introducing magnitudes:

q = 0.00069 Kg (871 J/Kg*K) (0.01 K)  = 0.0060099 J; rounding the cipher, 0.00601 J.

0.00601 J, which will cause a change of temperature of carbon dioxide equal to:

ΔT = q /m (Cp)

ΔT = 0.00601 J /0.00069 Kg [871 J/Kg*K] = 0.00601 J / 0.60099 J*K = 0.01 K; or 0.01 °C; therefore, the calculation is correct.

Let’s apply the formula related to the heat transfer by radiation from warmer sources of entropy (heat):

Q = e σ A (Te ^4 – Ts ^4)

Where Q is the radiated power, e is the emissivity of the emitter system, A is the radiating area, Te is the temperature of the emitter, and Ts is the temperature of the surroundings.

The known values are:

e of soil at 299.65 K and 1 atm = 0.7 in average (e is a unitless coefficient)
σ = 5.6697 x 10^-8 W/m^2*K^4
A = 1 m^2
Tr or T of soil = 300.15 K [(300.15 K) ^4 = 8116212154.05 K^4]
Tc or T of air = 299.65 K [(299.65 K) ^4 = 8062266098.565 K^4]

Introducing magnitudes:

q = 0.7 (5.6697 x 10^-8 W/m^2*K^4) (1 m^2) (8116212154.05 K^4 – 8062266098.565 K^4) =

0.7 (5.6697 x 10^-8 W/m^2*K^4) (1 m^2) (53946055.485 K^4) = 2.141 W

2.141 W = 2.141 J/s

2.141 J/s is the rate of heat transferred from the soil to the atmospheric CO2.

2.141 (J/s) (1 s) =  2.141 J

From this amount of energy, the carbon dioxide absorbs only 0.002 * 2.141 J = 0.0043 J

Equivalence in change of temperature (ΔT) of carbon dioxide:

ΔT = q/m (Cp)

Introducing magnitudes:

ΔT = 0.0043 (J)/0.00069 Kg (871 J/Kg*°C) = 0.006 (J)/0.60099 (J/°C) = 0.007 °C

The thermal energy (the heat being stored and removed from any system by any means of heat transfer) contained in 0.00069 Kg/m^3 of CO2 is equivalent to 0.007 °C. Then an increase of 381 ppmv of atmospheric CO2 causes an increase of the mass of carbon dioxide temperature. However, the heat stored by CO2 is not equivalent to 0.007 °C. The reason being CO2 is a poor absorber-emitter of energy and so it cannot store the absorbed energy for long periods of time. Theoretically, we obtain a change of temperature of the atmospheric CO2 of 0.01 °C, which is not the change of temperature of the whole atmosphere, but only of the specified mass of carbon dioxide I will describe another mathematical procedure in more detail below.

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3. EXAMPLES FROM NATURE:

Earth receives 697.04 W/m^2 of infrared radiation from 1367 W/m^2 of the incoming energy (light, ultraviolet, radio, etc.) from the Sun. (Maoz. 2007. Page 36). 14% of incoming heat to Earth is absorbed by air.

Data from the meteorological station in Monterrey, Mexico located at 25º 48´ North latitude and 100º 19' West longitude and an altitude of 513 meters ASL: On 31 March 2007 at 18:15 UT the soil absorbed approximately 453 W/m^2 of IR radiation causing a ground temperature of 318.15 K (45°C). The air temperature was 300.15 K (27 °C), what was the tropospheric ΔT due to the absorptivity-emissivity of air?

For the answer, first we need to know the load of heat transferred from the soil to the mixed air. Principally, we need to obtain the Grashof Number and the Heat Transfer Coefficient for those particular conditions:

Grashof Number

When we are calculating the load of heat transferred from the surface to the air we need to know the flux of the air toward the warm surface and toward the upper levels. The rate of flux is known as the Grashof number (Gr), and it describes a ratio involving buoyancy and viscosity: buoyancy/viscosity. As a fluid adjacent to a warmed surface starts to increase in temperature, the density of that fluid decreases. The buoyancy causes the less dense fluid to lift up, so the adjacent colder fluid is conveyed into contact with the warmer surface.

Gr L = g β (Ts – T ∞) D^3 / v^2

Where,

g is the gravitational constant (9.8 m/s^2)
β is the volumetric expansion coefficient (1/T)
T1-T2 is the difference of temperature between two adjacent systems expressed in Kelvin (18 K).
D^3 is the distance between two systems to the third power (1 m)
v^2 is the kinetic viscosity (2.076 x 10^-5 m^2 / s) to the second power.

Introducing magnitudes:

Gr L = (9.8 m/s^2) (3.332 x 10^-3 K^-1) (18 K) (1 m)^3 / (2.076 X 10^-5)^2 m^4 /s^2 =
= 0.58776 m^4/s^2 / 4.309776^-10 m^4 /s^2 = 1.364 x 10^9

Heat Transfer Coefficient

The Heat Transfer Coefficient (Ћ) is the rate of heat transferred from a warmer system to a colder system. It relates to the Grashof number, the Prandtl number and the thermal conductivity of the fluid. The Prandtl number is dimensionless and refers to the ratio of momentum diffusivity (v, or dynamic viscosity) and the thermal diffusivity (a). The heat transfer coefficient is determined by the next formula:

Ћ =  (k/D^3 )(C) [(Gr)(Pr) ]^a

Where,

k is the thermal conductivity (for air, k = 0.03003 W/m*K)
D or L is the distance between the two systems
C is a factor of correction for irregular surfaces facing up (soil)
Gr is Grashof Number (obtained in the previous calculus Gr = 1.36 x 10^5)
Pr is Prandtl Number (0.697 for air)
a is a constant of proportionality for laminar natural systems (1/3 for surfaces facing up).

Introducing magnitudes:

0.03003 W/m*K
Ћ =  ---------------------------------- (0.14) [(1.36 x 10^9) (0.697)]^1/3 = 4.134 W/m^2*K
1 m^3

The heat transfer from soil to mixed air is:

q = Ћ A (Ts – T∞)

Where,

q is the heat absorbed by the colder system
Ћ is the convective heat transfer coefficient (obtained in the previous formula = 4.13 W/m^2*K)
A is the implied Area (1 square meter)
Ts - T∞ is the difference of temperature between the heated system and the colder system.

Introducing magnitudes:

q = 4.13 W/m^2*K (1 m^2) (18 K) = 74.4 W (rate of heat transfer or heat flow rate)

74.4 W = 74.4 J/s (http://www.iprocessmart.com/techsmart/conversions.htm)

E = 74.4 (J/s) (1 s) = 74.4 J

If m of mixed air = 1.18 Kg/m^3 and the Cp of mixed air at 300.15 K = 1005.7 J/kg K, therefore:

ΔT = q/m (Cp) = 74.4 J / (1.18 kg) (1005.7 J/kg °C) = 0.063 °C

Considering the mass of air, the change of temperature of air caused by 74.4 W of power emitted from the surface causes a ΔT air of 0.063 °C. This means that the temperature of air increases from 300.15 K up to 300.21 K; or from 27 °C up to 27.063 °C (27.1 °C by rounding up the number).

0.063 °C was ΔT caused by heat transfer by convection from the ground to the air.

Let's see what happened in 1998, the warmest year up to present.

Known Data on April 1998 in Monterrey, N. L., Mexico:

The change of tropospheric temperature in 1998 averaged 0.52 °C throughout the year (UAH).

Tair = 317.15 K

Tsurface = 345.15 K

Density of dry air (d) = 1.168 kg/m^3

Dry Air volumetric expansion Coefficient (β) = 3.16 x 10^-3 K^-1

Kinetic Viscosity (v) = 1.741 x 10^-5 m^2/s

Dry air thermal conductivity (k) = 0.02753 W/m K

Correlation factor (C) = 0.14

Constant of proportionality (a) = 1/3

Grashof Number:

Gr L = g β (Ts – T∞) D^3 / v^2

Gr L = 9.8 m/s^2 (3.16 x 10^-3 K^-1) (345.15 K - 317.15 K) (1 m)^3 / (1.741 x 10^-5 m^2/s)^2

Gr L = 3.0968 x 10^-2 m/s^2 K^-1 (28 K) (1 m^3) / (3.031081 x 10^-10 m^4/s^2)

Gr L = 2.861 x 10^9

Conductive Heat Transfer Coefficient:

k
Ћ = ------------- (C) [(Gr) (Pr)]^a
D^3

Ћ = [0.02753 W/m K / 1 m^3] (0.14) [(2.861 x 10^9) (0.7043)]^1/3

Ћ = 0.02753 W/m^2 K (176.82) = 4.87 W/m^2 K

The load of heat transferred by convection from the surface to the air was:

q = 4.87 W/m^2*K (1 m^2) (28 K) = 136.36 W

Considering an air density of 1.22 kg/m^3, the change of temperature of the air caused by 136.4 W, transferred exclusively by convection, in 1998, was 0.11 °C, i.e. from 317.15 K up to 317.26 K.

Let us examine another case.

April 6, 2007, AT 19:01 UT, data taken from the meteorological station in Monterrey, MX, located at 25º 48´ North latitude and 100º 19' West longitude, and an altitude of 513 m ASL:

Ts = 316.95 K
T∞ = 305.45 K
Density of dry air (d) = 1.168 kg/m^3
Dry Air volumetric expansion Coefficient (β) = 3.16 x 10^-3 K ^-1
Kinetic Viscosity (v) = 1.741 x 10^-5 m^2/s
Dry air thermal conductivity (k) = 0.02753 W/m*K
Corrective factor (C) = 0.14
Constant of proportionality (a) = 1/3

Grashof Number:

Gr L = g β (TsT∞) D^3 / v^2

Gr L = 9.8 m/s^2 (3.16 x 10^-3 K^-1) (316.95 K - 305.45 K) (1 m)^3 / (1.741 x 10^-5 m^2/s)^2

Gr L = 3.0968  x 10^-2 m/s^2 K^-1 (11.5 K) (1 m^3) / (3.031081 x 10^-10 m^4/s^2)

Gr L = 1.175 x 10^9

Conductive Heat Transfer Coefficient:

k
Ћ =  ------------- (C) [(Gr) (Pr)]^a
D^3

Ћ =  [0.02753 W/m∙K / 1 m^3] (0.14) [(1.175 x 10^9) (0.7043)]^1/3

Ћ = 0.02753 W/m∙K (1.46 x 10^8) = 3.62 W/m^2*K

The heat transfer from soil to mixed air by convection is:

q = Ћ A (TsT∞)

q = 3.62 W/m^2∙K (1 m)^2 (11.5 K) = 41.61 W

41.61 W = 41.61 J/s (http://www.iprocessmart.com/techsmart/conversions.htm)

E = 41.61 J/s (1 s) = 41.61 J

On the other hand, the temperature of the air caused by41.61 W of power transferred exclusively by convection from the surface to the atmosphere was 0.034 °C.

Here we  properly conclude that 41.61 W of power cause a change of air temperature of 0.034 °C.

The heat transferred from the surface to the air by radiation was 55.04 W; however, from this load of energy, the air only absorbed 16.51 W by radiation.

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4. CHANGE OF TROPOSPHERIC TEMPERATURE BY SOLAR IRRADIANCE (This theme is better developed at Solar Irradiance is Increasing. It was corrected in 30 June 2008 due to minor grammar errors)

The total incoming solar irradiance to the terrestrial surface is 697.04 W/m^2. From this amount of infrared radiation, the surface absorbs about 348.52 W/m^2. The atmosphere absorbs 317 W/m^2. Considering the mass of air and its thermal capacity, the Earth’s temperature should vary by 30 °C. The fluctuation of the solar irradiance in the last 300 years has been 1.25 W/m^2.  1.25 W/m^2 causes a change of the Earth's temperature of 0.56 °C, which is the maximum averaged change in tropospheric temperature achieved during the 1990s (the average of change of temperature in 1998 is 0.51 °C). (Hurrell & Trenberth. 1999)

Planet Earth would not be warming if the Sun's energy output (Solar Irradiance) was not increasing. Favorably, our Sun is emitting more radiation now than it was 200 years ago, and so we should have no fear of a natural cycle that has occurred many times over in the lifetime of our Solar System.

Heat always moves from places of higher density of heat to places of lower density of heat, thus states the Second Law of Thermodynamics (Van Ness. 1969. Page 54). In daylight (P. S. obviously under, Sunlight), air is always colder than soil (P. S. obviously, the surface of soil); consequently, heat is transferred from the soil to the air, not vice versa. By the same physical law, the heat emitted by the Sun -a source of heat- is transferred to the Earth, which is a colder system.

The capacity of carbon dioxide to absorb-emit heat is much more limited than that of oceans and soil; thus, carbon dioxide cannot have been the cause of the warming of the Earth in 1998.

A fact well known to all scientists is that the absorptivity-emissivity thermal property of carbon dioxide diminishes as its density increases and as the temperature increases. This happens because the infrared radiation absorption margin is very narrow (wavelengths from 12-18 micrometers) and so the opacity of carbon dioxide to infrared radiation increases with altitude. As the column of CO2 gains height, its opacity to infrared radiation increases.

The dispersion of transferred heat increases when the density of carbon dioxide increases because there are more microstates toward which energy can diffuse. As a result, the momentum of carbon dioxide molecules decays each time heat is transformed into molecular kinetic energy, and emitted heat disperses in greater amounts towards deep space through the upper layers of the atmosphere. This process -determined by the second law of thermodynamics -could explain the observed paradoxical phenomenon of the coldness of the higher tropospheric layers in contrast with the tropospheric layer above the Earths surface, which is always warmer than the upper layers.

When the concentration of atmospheric carbon dioxide increases, the strong absorption lines become saturated. Thereafter its absorptivity increases logarithmically not linearly or exponentially; consequently, carbon dioxide convective heat transfer capacity decreases considerably.

Power emitted from the surface during nighttime = 308.2 W.

Am I measuring backradiation or the temperature of an atmosphere absorbing 30% of emitted radiation by the surface which causes a temperature of the bulk atmosphere, during nighttime, of - 27.5 °CLet us see if the issue of backradiation warming up the surface is true:

Load of heat transferred from the surface to the atmosphere during nighttime = 308.2 W. (It decreases as night advances and it is followed by the atmosphere, not the opposite).

Load of heat emitted by the surface which is absorbed by the atmosphere during nighttime = 92.45 W.

Load of heat emitted by the atmosphere during nighttime = 61.94 W.

ΔT of atmosphere by 92.45 W, considering density and Cp of air = 0.075 K.

Allegedly, one half of radiation from the atmosphere is emitted downwards = 30.97 W.

Argued power absorbed by the surface, if and only if there were available microstates in surface = 21.7 W.

ΔT of surface by 21.7 W, considering density and Cp of surface = 0.0000122 K. J

Statistical Thermodynamics debunks very easily the argument of absorption of cool quantum/waves (lower frequency and longer wavelength than the absorbed quantum/waves) by warm mass particles. We cannot take electrons from highest energy microstates ad arbitrium to higher energy microstates out from the mass particle without breaking up the latter or without the occurrence of quantum tunneling, which has never been observed, detected or produced in the atmosphere-surface system.

T of the atmosphere if the surface were not radiating = - 72.2 °C (Almost Mars’ atmospheric temperature at 1.5 m above ground).

Is it not clear that the temperature of the atmosphere depends mostly on the surface temperature and not the opposite?

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5. ALGORITHM FOR METHANE (CH4)

Molar Mass of Methane (CH4) = 16.0425 g/mol

Current mass of CH4 in air = 1.740 ppmv:

W in mg/m^3 = [ppmv] (12.187) (MW) / (273.15 + °C) = 1.740 (12.87) (16.0425 g/mol) / 300 °C = 1.198 g/m^3 = 0.0012 Kg/m^3

Specific Heat of Methane

T (K)kJ/Kg*K
275    2.191
300   2.226
325   2.293
350   2.365
375   2.442

Concentration of CH4 = 1.74 ppmv

Converting concentration to Density = [ppmv] (12.187) (MW) / (273.15 + °C) = 1.740 ppmv (12.87) (16.0425 g/mol) / 300 °C = 1.198 g/m^3 = 0.0012 Kg/m^3

Concentration of CH4 = 1.74 ppmv

Mass of CH4 = 0.0012 Kg/m^3 (1 m^3) = 0.0012 kg

Cp CH4= 2 226 J/kg*K

Δq = 0.000535 W/m^2 = 0.00013 cal-th

ΔT = Δq / m (Cp)

ΔT = 0.00013 cal-th /0.0012 Kg (533.3 cal/Kg*°C) = 0.00013 cal / 0.64 cal*°C = 0.0002 °C

Consequently, Methane is not an important forcing gas at its current concentration in the atmosphere.

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6. THE CASE ON 14 APRIL 1998 (RADIATIVE "FORCING") (Notice the standard for CO2 of 350 ppmv was determined empirically by Friederike Wagner et al. The standard of 280 ppmv is not real).

Known data:

δ CO2 in 1998 [(ppmv) ∞] converted to density = 0.00049 Kg/m^3
δ CO2 standard [(ppmv) s] converted to density = 0.00045 Kg/m^3 (280 ppmv is a flawed standard. The lower standard determined by Friederike Wagner et al is 350 ppmv during the late Holocene).
T of air = 318.15 K

Notice that for denoting concentration we enclose the symbol of the substance or units between square brackets [...]

Formula to be applied exactly as it is applied by the IPCC:

Δ T = [α] ln [(CO2) ∞ / (CO2) s] / 4 (σ) T^3.

Introducing magnitudes:

ΔT = (5.35 W/m^2) ln ([367 ppmv] ∞/[280 ppmv] s)/4 (5.6697 x 10^-8 W/m^2*K^4) (318.15 K) ^3

ΔT = 5.35 W/m^2 (0.126)/4 [5.6697 x 10^-8 W/m^2*K^4] (318.15 K) ^3 = 1.45 (W/m^2)/7.3 (W/m^2) K =
= 0.19 K; rounding the cipher, 0.2 K, or 0.2 °C.

However, the change of temperature on April 1998 was 0.786 °C (UAH), so carbon dioxide did not cause the anomaly.

Let's now apply the REAL STANDARDS for the Northern Hemisphere on April 14, 1998:

Standard temperature for the Northern Hemisphere = 288.15 K (Potter & Somerton. 1993)
Global standard density of atmospheric CO2 = 350 ppmv (The US Department of Labor Occupational Safety & Health Administration (OSHA) and the American Society of Heating, Refrigerating and Air-Conditioning Engineers (ASHRAE) determined empirically the standard average for outdoor CO2 from 350 to 600 ppmv. The lower standard (350 ppmv) determined by Friederike Wagner et al is the one we are considering here).
Density of CO2 (in some places only) in 1998 = 367 ppmv.

Applying Arrhenius' formula we obtain (In red, Schwartz, Stephen E. 2007 adjustments):

ΔT = α (ln [CO2] ∞ / [CO2] s) / 4 (σ) T^3.

ΔT = (5.29 W/m^2) ln (367 ppmv/350 ppmv) / 4 (5.6697 x 10^-8 W/m^2*K) (288.15 K)^3 =

= 0.28 W/m^2/5.43 W/m^2 K = 0.046 K; rounding the cipher, 0.05 K.

Which is compatible with the value of the heat stored obtained by applying the algorithm used in some paragraphs above:

On April 14, 1998, the temperature changed by 0.08 K in one second. What is the heat stored by 0.00056 Kg (367 ppmv) of atmospheric CO2?

Heat Stored by 367 ppmv (0.00056 Kg) of CO2:

q = m (Cp) (ΔT/Δt) (Potter & Somerton. 1993. Page 58) (For a nonequilibrium process with Cp rel. constant)

Introducing magnitudes:

q = 0.00056 Kg (871 J/Kg*K) (0.08 K/s)  = 0.48776 J/K (0.08 K/s) = 0.039 J/s

E = p (t) = 0.039 J/s (1 s) = 0.039 J, or 0.0093 cal, which will cause a change of temperature of:

ΔT = q /m (Cp)

ΔT = 0.0093 cal/0.00069 Kg (208.17 cal/Kg*K) = 0.0093 cal/0.144 cal*K = 0.065 K; or 0.065 °C; then the calculation using real standards is correct.

This comparison between the algorithm to obtain the stored heat and the algorithm to obtain the radiative "forcing" demonstrates two important features:

1. The algorithm q = m (Cp) (ΔT/Δt) denotes the three modes of heat transfer: radiation, convection and conduction.

2. When we introduce real standards and apply the proper algorithms, the temperature increase caused by CO2 is no more than 0.1 K.

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7. CO2 SCIENCE: THE CASE ON JUNE 22, 2007 (RADIATIVE "FORCING"):

If we want to know what ΔT is caused by the q absorbed by the increase of CO2 since 1985, we need to apply the next formula:

q stored = m (Cp) (ΔT/Δt)

Where the mass m  is the product of volume and density (Pitts & Sissom. 2000), Cp is the Specific heat of the substance, ΔT is the change of temperature (final temperature - initial temperature), and Δt is the change of time. The result is the relation between the heat absorbed by the substance and the heat removed from the substance (Pitts & Sissom. 2000) (Wilson. 1994).

Introducing magnitudes:

q stored = 0.000155 Kg (871 J/Kg*K) (0.5 K/s) = 0.0675025 J/s

E = 0.0675025 J/s (1 s) = 0.0675025 J = 0.016 cal

To know the change of temperature caused by 0.0003334 cal, we need to use the next formula:

ΔT = Δq / m (Cp)

Introducing magnitudes:

ΔT = 0.016 cal / 0.000155 Kg (208.1 cal/Kg*K) = 0.00027 cal / 0.0322555 cal*K = 0.00385 K; rounding the cipher, 0.004 K, or 0.004 °C.

Since each kilogram of CO2 received 0.016 cal, the temperature of each Kg of CO2, and therefore the entire volume of CO2, increased by only 0.004 °C.

A common error among some authors is to calculate the anomaly taking into account the whole mass of atmospheric CO2, when for any calculation we must take into account only the increase of the mass of atmospheric CO2. The error consists of taking the bulk mass of CO2 as if it were entirely the product of human activity, when in reality the increase in human CO2 contribution is only 34.29 ppmv out of a total of 381 ppmv (IPCC). This practice is misleading because the anomaly is caused not by the total mass of CO2, but by an excess of CO2 from an arbitrarily fixed "standard" density. There is however no such thing as a "standard" density of atmospheric CO2.

What is the heat load transferred to 101 ppmv (0.000157 Kg/m^3) of atmospheric CO2 (101 ppmv is the total increase in the mass of carbon dioxide since 1985), when the change of temperature on 2 August 2007 was only 1.23 K through one minute (ΔT/Δt)?:

q stored = m (Cp) (ΔT/Δt)

q stored = 0.000157 Kg (871 J/Kg * K) (1.23 K/60 s) = 0.0028 J/s

E = q (t) = 0.0028 J/s (1 s) = 0.0028 J = 0.00067 cal-th

To know the change of temperature caused by 0.00067 cal:

ΔT = Δq / m (Cp)

ΔT = 0.00067 cal/0.000157 Kg (208.1 cal/Kg*K) = 0.00067 cal/0.0326717 cal*K = 0.02 K; or a change of temperature of 0.02 °C.

Since each kilogram of CO2 received 0.00067 cal, the temperature of each Kg of CO2, and therefore the entire volume of CO2, increased by only 0.02 °C. Notice that the change of temperature on 2 August 2007 was twice the change of temperature on 6 April 2007. From this we deduce that there is a link between Kelvin and energy (J) established by the Stefan-Boltzmann's Law.

Does this mean that air temperature would increase by 0.02 °C per second until it reached scorching temperatures? No, it does not, as almost all of the absorbed heat is emitted in the very next second. Thus the temperature anomaly caused by CO2 cannot go up if the heat source does not increase the amount of energy transferred to CO2.

Now let us study the extreme case on July 8, 2007:

The real radiative equilibrium temperature of Earth is 300.15 K which is caused by the oceans, not the "greenhouse" effect. The change of temperature caused by the heat transferred from the ground to the total mass of atmospheric CO2 by radiation (radiative "forcing") was:

Formula to be applied:

ΔT = [α] ln [(CO2) ∞ / (CO2) s] / 4 (σ) T^3.

Where ΔT is the change of temperature, α is the assumed coefficient of heat transfer of CO2 by radiation (5.35 W/m^2), CO2 is the current density of carbon dioxide expressed in ppmv, CO2 s is the assumed "standard" density of carbon dioxide expressed in ppmv, σ is the Stefan-Boltzmann constant (5.6697 x 10^-8 W/m^2*K^4) and T^3 is the temperature to the third power expressed in Kelvin.

Introducing magnitudes:

ΔT = 5.29 W/m^2 [ln (381 ppmv ∞/280 ppmv s)] /4 (5.6697 x 10^-8 W/m^2*K^4) (300.15 K) ^3. (In red, Schwartz, Stephen E. 2007 adjustments)

ΔT = 5.29 W/m^2 (0.308) / (4 [5.6697 x 10^-8 W/m^2*K^4] (300.15 K) ^3 = 1.63 W/m^2/ 6.13 W/m^2*K
= 0.27 K ((In red, Schwartz, Stephen E. 2007 adjustments).

0.27 K/s is only 1.24% of the temperature difference between the ground and the air, which was 21.8 K.  We can see that carbon dioxide is not able to cause the temperature anomalies that have been observed on Earth.

From the most recent observations of the tropospheric temperatures and their relationship with the density of carbon dioxide, it is possible that the “radiative forcing coefficient” of carbon dioxide is not 5.35 W/m^2 nor 5.29 W/m^2, but an amount of between 1.78 W/m^2 and 2.68 W/m^2. This would be congruent with the observations of nature made by many scientists up to date. (Monckton. 2007)

For example, in the first scenario the ΔT caused by 381 ppmv of carbon dioxide is 0.1 K:

ΔT = (1.78 W/m^2) ln ([381 ppmv] ∞/[280 ppmv] s) /4 (5.6697 x 10^-8 W/m^2*K^4) (300.15 K) ^3

ΔT = (1.78 W/m^2) 0.308 / 6.13 = 0.55 (W/m^2)/6.13 W/m^2*K = 0.089 K (0.1 K, rounding the cipher).

In the second scenario for α = 2.68 W/m^2, the change of temperature caused by 381 ppmv of CO2 is 0.1 K:

ΔT = (2.68 W/m^2) ln (381 ppmv /280 ppmv) /4 (5.6697 x 10^-8 W/m^2*K^4) (300.15 K) ^3

ΔT = 2.68 W/m^2 (0.308) / 6.13 W/m^2*K = 0.83 (W/m^2)/ 6.13 W/m^2*K = 0.13 K

The latter is congruent with observations (Decadal trend = 0.12 K from UAH data), however the values have been forced to obtain a preconceived result, the procedure therefore is flawed. The real value for alpha is 0.423 W/m^2 (Read Total Emittance of CO2), so the real change of temperature caused by CO2 is 0.01 °C.

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8. THE REAL “FORCING” OF CARBON DIOXIDE

We have seen that Arrhenius’ Formula to know the change of temperature by carbon dioxide is not functional, given the high uncertainties in the values of heat flux (α), the “standard” concentration of carbon dioxide and the “standard” temperature (T^3).

Formula to be applied:

q = e (σ) (A) [(Ts) ^4 – (Ta) ^4]

Where q is the heat transferred by radiation from one system to another, e is the emissivity of the surface that absorbs energy, σ is the Stefan-Boltzmann constant, A is the area of interchange of energy, Ta is the temperature of the absorbent surface and Ts is the temperature of the emitter.

Known variables and constants:

Data taken from the meteorological station in Monterrey, Mexico: On June 22, 2007 at 18.05 UT, at the coordinates 25º 48´ North latitude and 100º 19' West longitude, and an altitude of 513 meters ASL, the air temperature at 1.5 m above ground level was 299.65 K (26.5 °C), whereas the ground temperature was 300.15 K (27 °C).  Which is the load of heat transferred from the ground to the mass of CO2 when 1 cubic meter of air contains 0.00069 kg.

e (at 300.15 K and a partial pressure of 0.00034 atm-m) = 0.001 (it has no units because it refers to an index).
σ = 5.6697 x 10^-8 W/m^2*K^4
A = 1 m^2
Ta = 299.65 K [(299.65 K) ^4 = 8062266098.565 K^4]
Ts = 300.15 K [(300.15 K) ^4 = 8116212154.05 K^4]
Ts^4 – Ta^4 = 53946055.485 K^4

Introducing magnitudes:

q = 0.001 (5.6697 x 10^-8 W/m^2*K^4) (1 m^2) (53946055.485 K^4) = 0.0031 W

If the transference of energy occurred each second, then the equivalent energy is:

q = 0.0031 W*s

0.0031 W*s = 0.0031 J

What is the change of temperature caused by the heat transfer of 0.0031 W*s?

Formula to be applied:

ΔT = q/m (Cp)

Where q is the heat transferred from a warm system to a colder system (for this case, soil is the warm system and air is the cold system), m is the mass of the interferer system (carbon dioxide) and Cp is the Specific Heat of the interferer system (carbon dioxide) at 300.15 K and constant pressure of 1 atm.

Known variables and constants:

q = 0.0031 J
m = 0.00062 Kg
Cp = 842 J/Kg*K

Introducing quantities:

ΔT = 0.0031 J /0.00062 Kg (842 J/Kg*K) = 0.006 K

Six thousandths of one degree is a negligible change of temperature.

We can apply the formula to extreme cases, for example, the case on April 6, 2007, when the temperature of the soil was 316.95 K and the temperature of the air was 305.45 K:

Ts^4 – Ta^4 = 1386835138.99 K^4

Introducing magnitudes to the formula:

q = 0.001 (5.6697 x 10^-8 W/m^2*K^4) (1 m^2) (1386835138.99 K^4) = 0.0786 W

In terms of energy, 0.0786 W*s = 0.0786 J

Now let us apply the formula to convert heat to change of temperature:

ΔT = 0.0786 J /0.00062 Kg (842 J/Kg*K) = 0.0786 J / 0.522 J*K = 0.15 K

The change of temperature caused by 0.0786 Joules of energy absorbed by 0.00062 Kg/m^3 of CO2 in the atmosphere on April 6, 2007 was 0.15 °C through one second.

Considering that the difference between the temperature of the soil and the temperature of the air was 11.5 °C, the amount of 0.15 °C is negligible (just 1.3% of the total).

We would be mistaken if we were to think that the change of temperature was caused by CO2 when, in reality, it was the Sun that heated up the soil. Carbon dioxide only interfered with the energy emitted by the soil and absorbed a small amount of that radiation (0.0786 Joules), but carbon dioxide did not cause any warming. Please never forget two important points: the first is that carbon dioxide is not a source of heat, and the second is that the main source of warming for the Earth is the Sun.

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9. WATER VAPOR:

Known data on June 22, 2007:

The concentration of atmospheric Steam = 35387 ppmv (3.15% of atmospheric water vapor) = 0.026 Kg/m^3
∂ x v = 0.026 Kg/m^3 (1 m^3) = 0.026 Kg (Pitts and Sissom. 1994).
MW of H2O vapor = 18.0151 u

q stored = m (Cp) (ΔT/Δt) = 0.026 Kg (2059.5 J/Kg*K)  (0.5 K/60 s) = 0.45 J/s

It is evident that water vapor is a much better absorber-emitter of heat than carbon dioxide. Under the same conditions, water vapor transfers 160 times more heat than carbon dioxide.

Nasif S. Nahle
February 05, 2007

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