I found this article at howstuffworks interesting:
I wondered how big a pile of coal that was.
The density of coal is 1.1 - 1.5 g/cm^3 (average 1.3g/cm^3).
323000 g / 1.3 g/cm^3 = 248461 cm^3 (a cube 63cm or 25in on a side)
= 0.25 cubic meters = 8.8 cubic feet = 248 L = 66 gallons (55 imperial)
That's per year. And per day it's
248 L/yr = .68 L/day = .18 gallons = .72 quarts = 2.9 cups = 1.45 pints
Here are the figures per kWh:
.28 L = .075 gallons = .30 quarts = .6 pints = 1.2 cups
So, on the order of a cup (or quarter liter) per kWh.
[BTW, is there a list of useable tags?]
Here is a summary (my words & format) of their results (with a small correction):How much coal is required to run a 100-watt light bulb 24 hours a day for a year?
http://science.howstuffworks.com/question481.htm
So they end up with an answer of 712 lbs or 323 kg.Wattage x Hours in a year = Total energy consumed
0.1 kW x 8,760 hours = 876 kWh
Generator efficiency x Thermal energy content of coal = Energy per ton of coal
0.4 x 6,150 kWh = 2,460 kWh/ton
Total energy / Energy per ton of coal = Coal mass
876 kWh / 2,460 kWh/ton = 0.356 tons = 712 lbs = 323 kg
I wondered how big a pile of coal that was.
The density of coal is 1.1 - 1.5 g/cm^3 (average 1.3g/cm^3).
323000 g / 1.3 g/cm^3 = 248461 cm^3 (a cube 63cm or 25in on a side)
= 0.25 cubic meters = 8.8 cubic feet = 248 L = 66 gallons (55 imperial)
That's per year. And per day it's
248 L/yr = .68 L/day = .18 gallons = .72 quarts = 2.9 cups = 1.45 pints
Here are the figures per kWh:
.28 L = .075 gallons = .30 quarts = .6 pints = 1.2 cups
So, on the order of a cup (or quarter liter) per kWh.
[BTW, is there a list of useable tags?]