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Name: Jessica
Status: Student
Grade: 6-8
Location: MN
Country: United States
Date: February 2008

Hello, I am an 8th grader in Laporte MN. I am studying phantom electrical loads. I am trying to figure out how much you can save on energy bills a year just by unplugging the common household items. Along with trying to figure how much coal or carbon dioxide is being burned/produced per watt. I have an electric meter that tells me watts but not KWH. So I am wondering if you could help me with how to calculate the amount of coal/carbon dioxide is being put out per watt.

Hi Jessica,

Unfortunately there are far too many unknowns to give you an answer to your question. It is easy to figure out how many Kilowatt-hours you are saving, but to figure out how much carbon dioxide is created, is impossible without a lot more information.

If you can measure how many watts the appliances you are unplugging use, using your wattmeter, then to get the number of KW-H you are saving, simply multiply the number of watts they consume, by the number of hours they remain unplugged, then divide by 1000 (there are 1000 watts in a Kilowatt).

But to know how much CO2 is saved from being emitted, you need to know the efficiency of your power plant, and the exact composition of fuel it is burning, among other things. The other thing needed is the to know if the power station is even burning coal in the first place, or is it nuclear or hydro powered, or even an on-demand gas turbine system powered by natural gas? The calculation would be extremely complex.

I assume your interest in carbon dioxide is because you have an interest in the claims of global warming many claim is caused by the greenhouse effect. Even this is a vastly more complex issue than many believe. For example, the worst greenhouse gas is not even CO2. Amazingly, water vapor is much worse as a greenhouse gas than carbon dioxide is. In fact, the total warming caused by manmade emissions is only a small fraction of the total natural greenhouse effect, most of which is caused by water vapor in the atmosphere. Like your original question, this is an issue with no simple answers.


Bob Wilson

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