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OCEANS: Mass:1.4 x 10 21 kg Area:3.6 x 10 8 km 2 Average depth:~4 km How much ice needs to melt to raise the level by 1 cm? Enough.

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Presentation on theme: "OCEANS: Mass:1.4 x 10 21 kg Area:3.6 x 10 8 km 2 Average depth:~4 km How much ice needs to melt to raise the level by 1 cm? Enough."— Presentation transcript:

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12 OCEANS: Mass:1.4 x 10 21 kg Area:3.6 x 10 8 km 2 Average depth:~4 km How much ice needs to melt to raise the level by 1 cm? Enough to add 3.6 x 10 8 km 2 x 10 -5 km = 3.6 x 10 3 km 3 of water = ~ 4 x 10 3 km 3 of ice

13 ~3 x 10 6 km 3 ~3 x 10 7 km 3  ~10 meter  ~100 meter

14 OCEANS: Mass:1.4 x 10 21 kg Area:3.6 x 10 8 km 2 Average depth:~4 km What temperature change will raise the level by 1 cm? expansion coefficient: between 0 (at 3 C) to 2 x 10 -4 K -1 (at 20 C)

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25 Are we Warming up? HOAX NO YES Not All Natural? HOAX NO YES dangerous? HOAX NO YES

26 Are we Warming up? HOAX NO YES Not All Natural? HOAX NO YES dangerous? HOAX NO YES

27 Are we Warming up? HOAX NO YES Not All Natural? HOAX NO YES dangerous? HOAX NO YES Now What?

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29 Limiting the global temperature rise at 2%, considered as a high risk level by the Stockholm Environmental Institute, demands 75% decline in carbon emissions in the industrial countries by 2050

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32 ~ 2 TerraWatts = 2 x 10 12 Watts (2000 million K) Per Year: 2 x 10 12 x (60x60x24x365) = ~ 60 x 10 18 Joules = 60 ExaJoules

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35 Covalent Bond Energy Covalent bond energy is measured by the energy required to break the bond. The bond enthalpy, Δ(X-Y) is the average  H for breaking one mole of X-Y bonds in the gas phase: When one mole of X-Y bonds is formed, the enthalpy change is   H(X-Y).  H =   (bonds broken) -   (bonds formed)

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37 Estimate  H for the combustion of CH 4 :  H = 4  (C–H) + 2  (O=O) - 2  (C=O)- 4  (O–H) = [ 4(413) + 2(495) - 2(799)- 4(463) ] kJ = -808 kJ +  + H C H H H 2 O O =  O C O ==  2 H O H 

38 Comparing fuels Natural gas: CH 4 + 2O 2 → CO 2 + 2 H 2 O ΔH=-808 kJ/mol Coal: C + O 2 → CO 2 Oil: C 20 H 42 + 30½O 2 → 20CO 2 + 21 H 2 O

39 Comparing fuels Natural gas: CH 4 + 2O 2 → CO 2 + 2 H 2 O ΔH=-808 kJ/mol Coal: C + O 2 → CO 2 ΔH=-393.5 kJ/mol Oil: C 20 H 42 + 30½O 2 → 20CO 2 + 21 H 2 O

40 Comparing fuels Natural gas: CH 4 + 2O 2 → CO 2 + 2 H 2 O ΔH=-808 kJ/mol Coal: C + O 2 → CO 2 ΔH=-393.5 kJ/mol Oil: C 20 H 42 + 30½O 2 → 20CO 2 + 21 H 2 O ΔH=-13315 kJ/mol ΔH=-666 kJ/mol.CO 2

41 Comparing fuels Production of 1 GigaJoule of energy releases: Natural gas: (10 9 J ÷ 808,000 J/mol ) x 0.044 kg/mol = 54.5 kg CO 2

42 Comparing fuels Production of 1 GigaJoule of energy releases: Natural gas: (10 9 J ÷ 808,000 J/mol ) x 0.044 kg/mol = 54.5 kg CO 2 Coal: Oil:

43 Comparing fuels Production of 1 GigaJoule of energy releases: Natural gas: (10 9 J ÷ 808,000 J/mol ) x 0.044 kg/mol = 54.5 kg CO 2 Coal: 112 kg CO 2 Oil: 66 kg CO 2 To produce 500 exaJoule with gas releases ……. (500 x 10 18 J) x (54 kg / 10 9 J) = 27 x 10 12 kg = 27 Gton = 4 ppm

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49 October 12, 1999: 6 billion! Now doubling every 61 years December, 2012: 7 billion!

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55 1000 kilowatt-hours of electricity is equivalent to the average amount of electricity consumed per month by: 1 resident of the United States. 2.3 residents of Europe. 7.6 residents of Mexico. 7.4 residents of South America. 12.35 residents of the Far East 26.3 residents of Africa..

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57 1000 kilowatt-hours of electricity is equivalent to the energy stored in each of the following: 574 fast-food meals. 34 pieces of firewood. 28.5 gallons of gasoline. 274pounds of coal. 34 therms of natural gas. lead-acid battery weighing 61110 pounds.

58 1000 kilowatt-hours of electricity production emits the following pollutants: KgPing pong balls CO 2 782.514960000 SO 2 1.935900 NO X 1.630530 And require 422 gallons of cooling water

59 Sunlight coming in: 1365 (1-A) J/m 2. sec 1 year = 3 x 10 7 sec π R 2 = 1.27 x 10 14 m 2 ~4 x 10 24 J/year or ~ 4 x 10 20 J/hr or ~ 10 17 W

60 Sunlight coming in: 1365 (1-A) J/m 2. sec 1 year = 3 x 10 7 sec π R 2 = 1.27 x 10 14 m 2 ~4 x 10 24 J/year or ~ 4 x 10 20 J/hr or ~ 10 17 W 1 hour of sunlight Equals world energy use In a year!

61 1 hour of sunlight Equals world energy use In a year!

62 Total surface area of earth  5.1  10 14 m 2 Land  25%  1.3  10 14 m 2  13 Gha (1 hm = 10 2 m; 1 ha = 1 hm 2 = 10 4 m 2 ) Need a fraction of 1 hr/ I year = 1/(24x365) = ~ 10 -4 This is ~ 5 x10 10 m 2 = 50,000 km 2 Or~ 200 x200 km NB: India: 3.3 x 10 6 km 2

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65 2 photosystems: 8 photons / cycle If QY for each cycle 100%.... 8 mol of photons  1400 kJ Reduce 1 mole of CO 2 : 480 kJ  Maximum efficiency: 34 % Realistic efficiencies: Sugar cane: ~1 % Rice, Potatoes, Soy beans: 0.1 % PAR = photosyntehtically active radiation 400-700 nm --- 43%

66 Approximate present land use: Million hectares Total Land13,000 Forest & savannah4,000 Pasture & Range3,100 Cropland1,500 Total Food4,600 ~11% of earth’s surface produces food

67 At 0.1 % efficiency on 10% of earth surface that is ~ 10 18 kJ of free energy / year in PhSynth equivalent to removing ~ 10 15 gram C/year Sunlight coming in: 1365 (1-A) J/m 2. sec 1 year = 3 x 10 7 sec π R 2 = 1.27 x 10 14 m 2 ~4 x 10 24 J/year or ~ 4 x 10 20 J/hr or ~ 10 17 W

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70 Fracking

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83 Are we Warming up? HOAX NO YES Not All Natural? HOAX NO YES dangerous? HOAX NO YES

84 Are we Warming up? HOAX NO YES Not All Natural? HOAX NO YES dangerous? HOAX NO YES

85 Are we Warming up? HOAX NO YES Not All Natural? HOAX NO YES dangerous? HOAX NO YES Now What?

86 Available alternative energy estimates

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