energy is derived from fossil fuels?

When we use fuels to liberate heat, we liberate CO

earth's atmosphere. How much more heat? Let's estimate it:

Suppose you use a small space heater (1000 Watts, W) for 1 hour. A Watt is 1 Joule (J)

per second, so over the hour you used 3.6 million J (3.6x10

is a lot easier to use for big numbers; read it as "3 point 6 times 10 to the 6th", with the

last phase meaning 10 raised to the 6th power, or multiplied by itself 6 times, to give a

million). For now, we'll assume that's all the fuel used- this was a heater directly using fossil

fuel, say, kerosene.

How much CO

retain at the earth's surface while that CO

The amount of CO

about 40,000 J per gram (4x10

Taking that first number, we have that

Mass of fuel used = (energy used)

---------------

(energy delivered per mass of fuel used)

= (3.6x10

= 90 g (OK, about 3.2 ounces, for the metrically challenged)

How much CO

composition of liquid fuels is close to CH

Chemists work in units of moles,; 1 mole is a fixed number of molecules (Avogadro's

number, 6.02x10

chemical elements, multiplied by their atomic masses. Hydrogen (H) has a mass of 1

(OK, 1.008, really, with some normal isotopes present), and carbon (C) has a mass of 12.

So, CH

So, our sample of 90 g of fuel contains

n = number of moles of C = (mass of fuel used) / (mass of fuel per mole)

= 90 / 14

= 6.4, with appropriate rounding

Oops - only about 40% of the CO

stays in the air. Right now, the rest is absorbed by the ocean or by plants on the

land. So, only about 0.4 x 6.4 = 2.56 moles of CO

when we used that heater.

How much solar energy does that help trap at the earth's surface? Every mole or every

molecule of CO2 acts the same, so let's compute the total amount of CO

human activities since the Industrial Revolution, and relate it to the change in the earth's

temperature. We can then convert the change in temperature to the amount of heat

that has to be radiated back to the earth from our layer of greenhouse gases.

How much has the CO

it has gone from 280 parts per million (ppm; our best estimate, from ice cores in glaciers

that have preserved samples of the ancient air) to about 370 ppm now (measured by

a worldwide network of samplers, with the one on Mauna Loa, Hawaii, keeping the

longest record, since about 1957). So, we've added about 90 ppm. In the same time,

the average increase in temperature worldwide is about 0.5 degrees Celsius (0.5oC),

or about 0.9 degrees Fahrenheit). This is the best estimate of the part of climate change

due to our adding CO

An increase of 90 ppm must be mulitplied by the total number of moles of air to see

how many moles of CO

about 10,000 kg of air above it (you can work this out from the air pressure and the

gravitational acceleration constant). Now, the earth's surface is 4 pi times the square of

the earth's radius, which is about 6400 km (about 4000 miles). To cut to the chase, the

area is about 53 billion hectares or 5.3x10

atmosphere contains about (10,000 kg / m2 ) x (5.3 x 10

The average molecular weight of air is between that of nitrogen gas (N

mole) and that of oxygen gas (O

O

put this in kg, as 0.029 kg per mole. Then, the number of moles of air is about

n

= (5.3x10

= 183x10

= 1.83x10

How much of this is our added CO

90x10

n

= 1.65x1016 mol as CO

How much solar energy did each mole of CO

that's radiated is related to temperature. All bodies above absolute zero in temperature

radiate energy as thermal infrared. There's a lot of information about this, but for us, the

important point is that the amount of radiation increases as the fourth power of the

body's temperature. So, if the absolute temperature doubles, the amount of radiation

increases by a factor of 24 = 16. Absolute temperature is that temperature measured

relative to absolute zero, which is -273oC or -459oF. So, an object at 20oC is at an

absolute temperature of 293, in units we call Kelvin to distinguish them from Celsius on

a different scale (from Fahrenheit, the change in name is to "Rankine"). There is a

fixed constant of proportionality, the Stefan-Boltzman constant, symbolized by the

character sigma (which I can't readily get in this composer package). Then, the

rate (I) at which thermal infrared energy is radiated from a given area of an object at

absolute temperature Tabs is

I = (sigma) T

We can add a small factor, the emissivity, to account for the fact that real objects don't

act as perfectly black bodies ("black"has almost nothing to do with color in visible light

note). For the earth's surface, the factor is close to 1 (about 0.96), so we'll ignore

it for now. To a first approximation, then, the earth's surface has an average temperature

of 15oC or 288 K; it then radiates thermal infrared at a rate

I = 5.67x10

= 390 W / m

An increase of 0.5oC changes this to the same expression, but with 288 replaced by 288.5.

The increase in I is computed as 2.7 W / m

as the calculation of radiative forcing, and it uses our transient increase in global temperature,

not our final or equilibrium increase, which is likely to be twice as much after accounting

for the greater absorption of sunlight when glaciers and sea ice melt, etc.)

Over the whole surface of the earth, our gazillions of moles of CO

caused an additional heat retention that resulted in the radiation of

E = 2.7 W / m

= 2.7 W / m

= 14.3x1014 W - a lot, of course

This was contributed by 1.6x10

an amount of radiation

E' = (total radiation increment) / (number of moles of CO

= (14.3x10

= 0.09 W / mol

Our piddling 2.56 mol added 2.56 x 0.09 = 0.23 W to the earth's surface radiation....

but, as the difference in units tells us (Joules, vs. Watts = Joules per second), we

are comparing apples and oranges. How long did the 0.23 W addition persist?

About as long as the CO

CO

seconds per hour = 31 million seconds (3.1x10

trapped by the amount of CO

H = (0.23 Joules / s) x (3.1x10

= 3.6 x 10

This is 100 times larger that the original heat liberated in burning the fuel!

The story doesn't quite end there. If the burner were electric, the fuel was burned

at the power plant, with only about 35% having been converted into electric power.

Only about 90% of that got to you, after losses in the power lines. So, the net

efficiency is about 31.5%. The amount of fuel actually burned, and thus, the

amount of CO

about 3.2. The fuel did not even get to the power plant itself without fuel being

burned to mine and transport the fuel, and to make the equipment an the

buildings. These efforts inflate the amount of fuel used by another 20% or so,

so we're up to 3.2 x 1.2 = 3.84 times as much CO

heat trapped by all the fuels we used is almost 400 times as large as the heat we

got from the heater! We can't use fuels without looking at the long-term

consequences.

Some other interesting aspects:

The amount of extra thermal radiation emitted - at the top of the atmosphere - from fuel use

and CO

(climatologists and climate modellers). Radiative forcing does not come only from adding CO

we also change cloudiness, as well as the amount of aerosols in the air. Aerosols, such as smog

as well as ordinary soot, can add to or subtract from the final radiative forcing.

The best estimates of total radiative forcing from human-induced change (fuel

burning, forest clearing, etc.) is about 1.2 W / m2. This is somewhat less than half

of what we computed above.