1. According to the U.S. DOE, each year humans are. of CO through the combustion of fossil fuels (oil,

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1 1. Accordin to the U.S. DOE, each year humans are producin 8 billion metric tons (1 metric ton = 00 lbs) of CO throuh the combustion of fossil fuels (oil, coal and natural as).

2 . Since the late 1950's spectroscopists at the Mauna Loa observatory have been collectin data on the concentration of CO in the atmosphere. The plot of these data, as ppm vs. time, is known as the Keelin Curve.

3 The Keelin Curve

4 Keelin Curve 1999 to Present 385 ppm 368 ppm / yr 1.7 ppm / yr 1999 yr 009

5 Thus CO in the atmosphere is currently increasin by 1.7 ppm/yr. To compare this with the 8 billion metric tons per year produced by humans we need to determine how many moles this is and then multiply by the molecular weiht of CO.

6 The atmospheric pressure on the earth is the force produced by the mass of air in the atmosphere divided by the surface area of the earth. Therefore to determine the mass of the atmosphere we need only multiply the surface area of the earth times atmospheric pressure: surface area atmospheric pressure = m air km 10 m / km N/m /(9.8 N/k)= k

7 The mass of air divided by the molecular weiht of air (.088 k/mole) equals the total number of moles of air: 18 m air / M air k/0.09 k/mole 0 = moles Moles of air times the enterin fraction of CO equals moles of CO enterin from all sources: 6 14 nair = moles/yr

8 Finally,moles of CO times the molecular weiht of CO equals the total (includin all sources and sinks) mass of CO enterin the atmosphere per year: 14 n CO / yr MCO moles/yr k/ 13 = k/yr = 13.6 billion metric tons per year mole

9 Thus there is a net addition to the atmosphere of 13.6 billion metric tons of CO per year as a result of all sources and sinks - compare this with the 8 billion metric tons produced by humans burnin fossil fuels.

10 4. What does the CO in the atmosphere do? Amon other affects it alters the earth's averae temperature throuh interaction with the infrared radiation emitted by the earth. First consider the nature of this radiation. Equilibrium radiation is characterized by a Planck distribution and this distribution is enerally assumed for the infrared emitted by the earth. The next slide shows a raph of a Planck distribution i.e the flux (enery per unit area per unit time) per unit wavelenth versus wavelenth for the current averae temperature of the earth (88K).

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12 To a ood approximation the distribution of enery in the radiation from the sun also follows a Planck curve - one characterized by a temperature of about 5780 K at the surface of the sun. This radiation spreads out as the distance from the sun increases so the flux decreases. When this radiation strikes the earth on the averae 0.3 (called the albedo) is reflected. The combined result is that the averae net flux from the sun at the earth's surface is 40 watts m.

13 5. What is the fate of the enery absorbed by the earth? If the earth at some time and place is emittin more or less enery than it absorbs the earth's temperature chanes until the radiative enery emitted to the cosmos equals the incomin enery from the sun. This is called a steady-state and the temperature is the steady-state emission temperature. These steady-state temperatures, when averaed over the whole earth and a year, would yield the earth's averae steady-state emission temperature which is assumed (and the assumption is confirmed by measurement and averain) to be the same as the steady-state temperature calculated for the known incomin flux of solar radiation.

14 Thus knowin the incomin flux we can calculate the earth's averae steady-state temperature, barrin some other source of enery at the earth. The Stefan Boltzmann Law relates the enery flux (F) of equilibrium radiation to T: F = T 4 Usin this law we can calculate the T at which the flux leavin the earth equals the nonreflected inward flux from the sun: T 40 watts m 8 4 o yieldin T = 55 K which is close to 0 F

15 Thus, were it not for some other effect (i.e. the reenhouse as effect), when the earth reached a steady state with the radiation fom the sun, i.e. when: enery in = enery out, the averae temperature of the earth's surface would be o about 0 F. In reality the averae temperature is 88 K o (59 F). This is because reenhouse ases redirect some of the outoin infrared radiation back to the earth's surface.

16 At the top of the atmosphere (to space): F (TOA) F atm(toa) 40 Watts/m At the earth's surface: F sun(0) 40 Watts/m from sun (after reflection) F atm(0) = 140 Watts/m from atmosphere F (0) = 380 Watts/m from earth Remember the blue arrow Note that: =

17 140 w m of the outoin infrared is redirected downward in the atmosphere by reenhouse ases (polyatomic ases with vibrational eneries in the infrared reion). H O is one such reenhouse as. Another is CO. CH and O are also. This 4 3 redirected radiant enery is represented by the blue arrow.

18 6. What is the role of CO in the redirection of the 140 w m back to the earth's surface?

19 As everyone knows the climatoloists have concluded that increases in the the blue arrow (reenhouse effect) effect are larely responsible for the currently observed warmin of the earth, and that anthropoenic CO makes a sinificant contribution to this effect. My oal here is to consider the basic physical chemistry underlyin their calculations in order to determine whether or not basic chemcial physics leads to the same conclusion. I am certain that this has been done many times and that the results are well known to climatoloists. I am also certain that these results are not widely known outside of this community.

20 The states of of CO important in this discussion are the vibrational (separated by eneries in the infrared reion) and the rotational (separated by eneries considerably lower than those separatin the vibrational states) states of CO.

21 CO is a linear molecule: O=C=O and amon its motions is the "asymmetric stretchin motion" that absorbs infrared radiation with wavelenths in the vicinity of meters (15 m or 15 microns). In what follows the reion from 9.3 m to 19 m will be denoted by. The reason it is necessary to desinate a reion is because each vibrational state (the transitions between adjacent vibrational states are the only ones important here) has associated with it rotational states such that transitions of interest are from one of the rotational states associated with the lowest vibrational state to one of the many that are associated with the adjacent vibrational state.

22 The followin slide shows a calculated and observed absorption spectrum for CO in the interval from Stull, Wyatt, and Plaas, Applied Optics, Vol. 3 No., p50 (1954), whose data will be used in a later calculation of the contribution of CO to the reenhouse effect. Note the areement between the observed and calculated spectra. Also note that there is more than one peak. Most of the peaks arise because of contributions from rotational states.

23 Theoretical and Calculated Absorbtion of CO in the Infrared 16.6 m 14.3 m 1.5 m

24 I will assume that the rotational-vibration transitions of CO that are important in the atmosphere occur between 9.3 and 19 m. This assumption is necessary because I am limited by the data available to me, but it is clear that the calculated effect would be increased, althouh not sinificantly, if the reion were increased. The enery emitted in this reion is found by takin the ratio of the total area under the Planck curve for 88 K to the area under the curve within the reion. The result is that 53.1% of the radiation emitted by the earth is in the reion of the chosen CO rot-vib absorption lines.

25 The 53.1 % is obtained from: u 3 u u / (e 1)du / u / (e 1)du where u hc / kt. This technique will be used aain in a later slide.

26 A SIMPLE EXAMPLE OF A PLANCK DISTRIBUTION The enery separatin the round and first assymmetric stretchin states of CO is at a iven temperature T 0 [CO ] / [CO ] e. : * 960/T * 960/T * Joules. For these two states at equilibrium If [CO ] / [CO ] e then CO CO photon. If [CO ] / [CO ] e then CO photon CO * 960/T * Thus the infrared radiation in the interval emitted by the earth interacts with (is absorbed and reemitted by) CO as it passes throuh the atmosphere.

27 The absorption of the infrared, because of the excitation of CO from vib-rot states into vib-rot states, results in a decrease in the radiant flux that that reaches the top of the atmosphere and oriinates from the round. This decrease is, in the simplest case of a sinle linear path of a flux from a sinle source of nearly monochromatic radiation, iven by: df kpfd, where df is the chane in flux, P is the pressure of the absorber, d is the path lenth interval and k is called the absorption coefficient. When interated this ives Beer's Law: ln(f/f o) kp

28 Beer's Law for linear absorbtion of radiation can be written: e u where is the linear transmittance and u kp is the linear absorbancc

29 However the case of radiation in the atmosphere has several complications: 1. the source is a band of eneries and is in no sense monochromatic,. all directions (not just a sinle straiht line) are of interest, and 3. the radiation that is reemitted is of interest.

30 Radiation emitted by the earth passes throuh the atmosphere interactin with the reenhouse ases up to the top of the atmosphere (TOA) where it passes into the cosmos. First consider the flux (radiant enery per unit area per unit time) correspondin to a sinle wavelenth ( ) and proceedin alon a line that makes the anle = arccos with the vertical. Later we will sum over the wavelenths in and over the directions of an upward hemisphere. This flux is partially absorbed by the CO (this absorption is characterized by an absorption coefficient, k, by the amount of carbon dioxide in the path, P, and by the lenth of the path, dz/ (chane in heiht CO divided by cos )). Furthermore additional radiation, all directions by CO at z., is emitted in

31 These thouhts lead to the basic equation relatin the chane in flux, df of CO, P flux, F, CO,, the chane in path lenth, dz/, the partial pressure, and the absorption coefficient, k, to the, and the flux oriinatin from the atmosphere, B, all at heiht z as the radiation passes throuh the atmosphere. df (F )(k P )(dz / ),, CO

32 The chane in absorbance is: du k P dz CO so the basic equation is, df / du F B.,,

33 In what follows fluxes (w m ) at 0 (round level) and at TOA (the top of the atmosphere) will be desinated F(0) and F(TOA), respectively. The oriin of the flux will be desinated by a superscript, e.. F (0) desinates the flux at the round level from the earth and F or F atm atm atm (TOA) (0) are fluxes oriinatin in the atmosphere. A subscript, as in F iven wavelenth or, in the case of (0), will desinate the flux at the i i, a small interval of wavelenths centered on. A subscript,, as in F (0), will indicate a flux that has been interated or summed over the whole interval.

34 Multiplyin throuh the basic equation by e is the linear absorbance : u u u,, e df / du e F e B. 1 u, 1 u, where u Note that the left hand side is d[e F ] / du. Therefore, interatin from the round level at which: z=0, u u (0), F F (0) B (0),, to the top of the atmosphere, TOA, where: z =, F F (TOA), u 0,,, 1 u (0),, F (TOA) e F (0) 0 u (0) 1 u 1 B e du.,

35 Now, to et the total upward upward hemisphere; 1 1 u (0) 1 1 u (0) u, F (TOA) F (0) e d B,ze ddu. furthermore F atm = F (TOA) + F (TOA) flux at TOA, interate over the (0) - F (TOA) = F (0) + F (TOA) = F (0) atm atm atm 1 1 u (0) atm 0 thus F (0) F (0) e d = F (0) i.e. fhe flux that oriinates from the round but is scattered is twice the flux that returns to the earth because half the scattered flux returns to the earth and half leaves throuh the top of the atmosphere

36 Introducin the notation: u (0) (0) e d for the difuse transmissivity and interatin the quantity 1 1 u (0) 0 F (TOA) F (0) e d = F (0) (0) over to et the flux at TOA oriinatin from the round over the interval (9.3 to 19 m) yields: 19 m 9.3m F (TOA) F (0) (0) d where, as described previously, F (0) is the Planck flux with wavelenth at the round and oriinatin from the earth.

37 u(19m) 3 u u(9.3m) Recall Planck's law: F (0) = u / (e 1)du with u evaluated at the earth's averae temperature. Divide the interval into 11 smaller intervals labeled by i (runs from 1 to 11) where: m, m, m, etc until m. (the intervals are expressed in as in terms of wave numbers ( absorption data are published vs. i 1 i 1 i1 1 ) because the 1 u( ) 3 u [ u / (e 1)du] / F (0) u( 1 1 i1 i i to be that fraction of the flux in the interval between 5.5 and m (9.3 and 19 m) that falls between and ). Define F (0) / F (0) i 4 1

38 The slide before the last ended with: 18m F (TOA) F (0) (0) d 1m Replacin the interal over 11 F (TOA) F (0) (0) i1 i by a sum: and, introducin i F (0) / F (0), i 11 F (TOA) F (0) (0) i i i1 i

39 11 i1 i i (0) is the Planck weihted difuse transmissivity for the interval from the round to TOA. The value for F (TOA) is the product of this quantity with F (0). Thus the problem of relatin the flux leavin the earth to that oriinatin from the earth is solved when (0) is known. 11 i1 i i

40 Values for the Planck weihted transmissivities are obtained below usin the in-line transmissivity values reported by Stull, Wyatt, and Plass, Applied Optics, Vol.3, No., p43 (1964) and iven for i 1 four i values in the interval in the table on the next slide. Values for the interal over a hemisphere and for values 1 u i (0) < (0)> = e d i 0 1 were calculated by interation usin the iven u i

41 1 Transmissivities cm 1000 atm-cm 500 atm-cm 00 atm-cm 100 atm-cm , i

42 The products i i (0)> and the sums of these over 1 the interval are then iven for the four i on the followin slide.

43 (0) i 1 cm 1000 atm-cm 500 atm-cm 00 atm-cm 100 atm-cm i

44 Our interest here is to find the broadband difuse transmitivity of CO : 11 i i i1 (0) (0)> appropriate to the intermediate values characteristic of the current (385 ppm) and 100 years hence (555 ppm) CO concentrations in the total atmosphere. To find these we interpolate a plot of the four values obtained for the broadband difffuse transmissivity (shown on the next slide) to obt ain values of u/k appropriate to the CO in the atmosphere as found on the slide that follows the next one.

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46 I obtained the optical depths for the atmosphere as follows. By definition: z 4 For CO in the current atmosphere P = Pair , air u(z) = kpdz. and P =P(0) e e Mz /(8.314T) 0,03395z / T barometric formula. Thus, takin k(t,p) constant: z / T u(0) = k P air (0) e dz 0 air K km from, accordin to the Takin P (0) = 1 atm and T = 88 K at z=0 and decreasin by 0 to 10 km, constant from 10 to 15 km and 1 increasin by 1.1 K km from 15 to 50 km interation yields: u(0)/k = 3.5 atm-m (or 35 atm-cm), For 555 ppm u(0)/k = 469 atm-cm.

47 The interpolated values found for: are: 11 i i i1 (0) > at 385 ppm < (0) at 555 ppm < (0) 0.454

48 Accordin to what was found above the averae contribution of CO to the blue arrow flux is: atm F tot (0) (1 F (0) 1/ (0) ) This equation says that one half the radiation scattered by CO returns to the earth.accordinly, usin the Planck weihted transmissivity values, the blue arrow effects of CO and F (0) = F F atm,385,385 tot tot atm,555,555 tot tot at 385 and 555 ppm are: (0), (0) = F ( 0).

49 The increase in the steady state flux to the round level from the atmosphere when the CO increases from 385 to 555 ppm is matched by an increased Planck flux from the round when a steady state is reached at each partial pressure: Therefore atm F (0) F tot (0). F (0) F (0) F F,555,385,555,385 tot tot tot tot Dividin by F or F,555 tot / F,385 tot,385 tot,555,385 tot tot (0) and collectin terms: ( ) ( ) F / F

50 Usin the Stefan-Boltzmann law F (0) / F (0) (T T) / T,555, tot tot (T 4T T) / T, F (0) / F (0) = 1+4T/T,,555,385 tot tot and, 4T/T = , T = 0.7 K.

51 7. Comments on the result: A. Accordin to the IPCC the averae surface temperature of o the earth increased by C between 1905 and 005. The 100 year increase found here (0.7 K) shows that there is solid science (of which the climatoloists are thorouhly aware) supportin their conclusion that human production of CO plays an important role in lobal warmin.

52 B. The increase in temperature found here is a lower bound because: 1. it is based upon the assumption that the rate of production of CO will not increase. As has been widely discussed substantial increases are expected in China, India, and in the developin world.. feedbacks have been inored e.. i. meltin of arctic ice, ii. releaseof CO from the earth and ocean due to increased temperature, iii.release of CH 4 from permafrost and ocean floor, iii. chanes in the concentration of water in the atmosphere

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