practice problem 1
The energy needed to power a 60 W light bulb for a year is then…
|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.
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|
|mmin =||E||=||1890 MJ||= 58 kg|
|mmax =||E||=||1890 MJ||= 98 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.
ton of coal equivalent
|29,308||International Energy Agency|
|29,310||Center for Energy Efficiency|
|29,290||Carbon Dioxide Information Analysis Center|
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 = 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 =||0.0645 kg|
|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.
- 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.
|A coal seam||Lumps of coal (non-standard)|
practice problem 2
practice problem 3
practice problem 4