oceans: mass:1.4 x 10 21 kg area:3.6 x 10 8 km 2 average depth:~4 km
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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 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. ~3 x 10 6 km 3. ~10 meter. ~3 x 10 7 km 3. - PowerPoint PPT PresentationTRANSCRIPT
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OCEANS:
Mass: 1.4 x 1021 kgArea: 3.6 x 108 km2 Average depth: ~4 km
How much ice needs to meltto raise the level by 1 cm?
Enough to add 3.6 x 108 km2 x 10-5 km =3.6 x 103 km3 of water= ~ 4 x 103 km3 of ice
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~3 x 106 km3
~3 x 107 km3
~10 meter
~100 meter
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OCEANS:
Mass: 1.4 x 1021 kgArea: 3.6 x 108 km2 Average depth: ~4 km
What temperature changewill 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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Are weWarming
up?
HOAX
NO
YES NotAll
Natural?
HOAX
NO
YESdangerous?
HOAX
NO
YES
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Are weWarming
up?
HOAX
NO
YES NotAll
Natural?
HOAX
NO
YESdangerous?
HOAX
NO
YES
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Are weWarming
up?
HOAX
NO
YES NotAll
Natural?
HOAX
NO
YESdangerous?
HOAX
NO
YES NowWhat?
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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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~ 2 TerraWatts = 2 x 1012 Watts (2000 million K)
Per Year: 2 x 1012 x (60x60x24x365) = ~ 60 x 1018 Joules = 60 ExaJoules
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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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13a–36
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Estimate H for the combustion of CH4:
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 HH
H2 O O=
·· ·
· ·
·
·· O C O= =
·· ·
· ·
·
·· 2 H O H
·· ·
·
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Comparing fuels• Natural gas:
CH4 + 2O2 → CO2 + 2 H2O ΔH=-808 kJ/mol
• Coal: C + O2 → CO2
• Oil: C20H42 + 30½O2 → 20CO2 + 21 H2O
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Comparing fuels• Natural gas:
CH4 + 2O2 → CO2 + 2 H2O ΔH=-808 kJ/mol
• Coal: C + O2 → CO2 ΔH=-393.5 kJ/mol
• Oil: C20H42 + 30½O2 → 20CO2 + 21 H2O
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Comparing fuels• Natural gas:
CH4 + 2O2 → CO2 + 2 H2O ΔH=-808 kJ/mol
• Coal: C + O2 → CO2 ΔH=-393.5 kJ/mol
• Oil: C20H42 + 30½O2 → 20CO2 + 21 H2O ΔH=-13315 kJ/molΔH=-666 kJ/mol.CO2
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Comparing fuelsProduction of 1 GigaJoule of energy releases: • Natural gas:
(109 J ÷ 808,000 J/mol ) x 0.044 kg/mol = 54.5 kg CO2
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Comparing fuelsProduction of 1 GigaJoule of energy releases: • Natural gas:
(109 J ÷ 808,000 J/mol ) x 0.044 kg/mol =54.5 kg CO2
• Coal:
• Oil:
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Comparing fuelsProduction of 1 GigaJoule of energy releases: • Natural gas:
(109 J ÷ 808,000 J/mol ) x 0.044 kg/mol =54.5 kg CO2
• Coal: 112 kg CO2
• Oil: 66 kg CO2
To produce 500 exaJoule with gas releases …….(500 x 1018 J) x (54 kg / 109 J) = 27 x 1012 kg
= 27 Gton = 4 ppm
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October 12, 1999: 6 billion!
Now doubling every 61 years
December, 2012: 7 billion!
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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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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.
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1000 kilowatt-hours of electricity production emits the following pollutants:
Kg Ping pong balls
CO2 782.5 14960000
SO2 1.9 35900
NOX 1.6 30530
And require 422 gallons of cooling water
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Sunlight coming in:1365 (1-A) J/m2 . sec1 year = 3 x 107 secπ R2 = 1.27 x 1014 m2
~4 x 1024 J/year or ~ 4 x 1020 J/hr or ~ 1017 W
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Sunlight coming in:1365 (1-A) J/m2 . sec1 year = 3 x 107 secπ R2 = 1.27 x 1014 m2
~4 x 1024 J/year or ~ 4 x 1020 J/hr or ~ 1017 W
1 hour of sunlightEquals world energy use In a year!
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1 hour of sunlightEquals world energy use In a year!
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Total surface area of earth 5.1 1014 m2
Land 25% 1.3 1014 m2 13 Gha
(1 hm = 102 m; 1 ha = 1 hm2 = 104 m2)
Need a fraction of 1 hr/ I year = 1/(24x365) = ~ 10-4
This is ~ 5 x1010 m2 = 50,000 km2 Or~ 200 x200 km
NB: India: 3.3 x 106 km2
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2 photosystems:8 photons / cycleIf QY for each cycle 100%....8 mol of photons 1400 kJ
Reduce 1 mole of CO2: 480 kJ Maximum efficiency: 34 %
Realistic efficiencies: Sugar cane: ~1 %Rice, Potatoes, Soy beans: 0.1 %
PAR = photosyntehtically active radiation
400-700 nm --- 43%
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Approximate present land use:
Million hectares
Total Land 13,000
Forest & savannah 4,000
Pasture & Range 3,100
Cropland 1,500
Total Food 4,600
~11% of earth’s surface produces food
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At 0.1 % efficiency on 10% of earth surfacethat is ~ 1018 kJ of free energy / year in PhSynthequivalent to removing ~ 1015 gram C/year
Sunlight coming in:1365 (1-A) J/m2 . sec1 year = 3 x 107 secπ R2 = 1.27 x 1014 m2
~4 x 1024 J/year or ~ 4 x 1020 J/hr or ~ 1017 W
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Fracking
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Are weWarming
up?
HOAX
NO
YES NotAll
Natural?
HOAX
NO
YESdangerous?
HOAX
NO
YES
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Are weWarming
up?
HOAX
NO
YES NotAll
Natural?
HOAX
NO
YESdangerous?
HOAX
NO
YES
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Are weWarming
up?
HOAX
NO
YES NotAll
Natural?
HOAX
NO
YESdangerous?
HOAX
NO
YES NowWhat?
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Available alternativeenergy estimates
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