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# Chemical Potential Energy

## Practice

### practice problem 1

How much coal needs to be burned to keep a 60 W light bulb glowing continuously for a full year?

#### solution

Power is the rate at which work is done (or energy is transformed).
 P̅ = ΔW = ΔE Δt Δt

The energy needed to power a 60 W light bulb for a year is then…

 E = Pt E = (60 W)(365.25 × 24 × 60 × 60 s) E = 1,893,456,000 J = 1890 MJ

Coal is a naturally occurring mineral composed of compressed, fossilized plant material. It's physical characteristics and chemical composition varies from location to location. High quality coal is black, hard, and lustrous. Low quality coal is soft, brown and dull. The primary ingredient in coal is carbon (60~90%) followed by water and volatile organic compounds with traces of sulfur, arsenic, selenium and heavy metals. High quality coal is high in carbon and low in water.

Energy density of coal Source: ASTM International and Chemistry and Technology of Coal
type physical
characteristics
density
(kg/m3)
carbon
(%)
water
(%)
specific energy
(MJ/kg)
anthracite hard black lustrous 800–929 81–86 02–16 > 32.6
bituminous hard black dull 673–913 45–78 02–16 24.4–32.6
sub‑bituminous soft black dull       19.3–24.4
lignite soft brown dull 641–865 31 39 14.7–19.3

Anthracite may have the highest energy density of all the types of coal, but it is rare and expensive. Some people burn it in home furnaces, because it burns relatively cleanly, but probably no power company burns it to generate electricity. Lignite is notoriously dirty when burned and, because it is so low in energy content, you need a lot more of it. Assuming you buy your electricity from a power plant that burns coal, chances are high that, unless it's near a "brown coal" mine, they aren't burning lignite. Bituminous and sub-bituminous coals are the types mostly likely used to generate electricity.

 e = E ⇒ m = E m e
 mmin = E = 1890 MJ = 58 kg emax 32.6 MJ/kg mmax = E = 1890 MJ = 98 kg emin 19.3 MJ/kg

Trade associations, government agencies, and nongovernmental organizations dealing with economics will sometimes state energy values in terms of the amount of fuel burned to produce that energy.

fuel unit energy
(MJ)
source
tce
ton of coal equivalent
(1,000 kg)
29,308 International Energy Agency
29,310 Center for Energy Efficiency
29,290 Carbon Dioxide Information Analysis Center
Fuel-based energy units

While not identical, these three sources agree with one another to three significant digits. It seems like a reasonable average value to work with. Let's set the problem up this time as a ratio — for no reason other than to demonstrate a different problem solving technique.

 eave = E = 29,300 MJ = 1,890 MJ mave 1,000 kg mave

mave = 64.5 kg

Let's use this last value and the average density for bituminous coal quoted in the first table to determine the volume instead of the mass.

 ρ = m ⇒ V = m V ρ
 V = m ρ V = 0.0645 kg 793 kg/m3 V = 0.081 m3 = 81 liters

I don't visualize sizes well when stated in cubic units and 81 liters doesn't help me much either. Let's take the cube root of this number to get the length of one side of a cube with this volume.

s = ∛V = ∛(0.081 m3) = 0.43 m = 43 cm

Technically, that's called a lump of coal. I'd like to propose this as a new unit.

lump
Brit. /lʌmp/, US /ləmp/, noun,
1. a unit of energy equivalent to 1,893,456,000 joules; or the amount of electric energy neeed to operate a standard 60 watt incandescent light bulb for one year; or the amount of chemical energy released by the complete combustion of a cube of bituminous coal, 43 cm on a side, in air.
2. a unit of volume equal to 0.081 cubic meters, used to measure coal.

### practice problem 2

Write something else.

### practice problem 3

Write something different.