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Is it right as xkcd states:

You can use heat flow to come up with simple rule of thumb: If an unused charger isn’t warm to the touch, it’s using less than a penny of electricity a day.

Or, more importantly, because one has usually a lot of devices like that at home:

For a small smartphone charger, if it’s not warm to the touch, it’s using less than a penny a year.

Can one estimate by some joule per time-unit calculation how much electricity is turned into heat to warm up a wall-plug cellphone-charger to, say, a maximum of 25°C?

From there I would will calculate the wasted money by using .25€/kWh -- a calculation which does not involve physics, which I can therefore do myself ;-)

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+1 for doing calculations for yourself ;) – Michael Brown Mar 5 '13 at 10:42
@MichaelBrown Usually I can only calculate with bits, bytes, apples and bananas. But in this case I can handle €. :-) -- but kW and J elude me completely. – towi Mar 5 '13 at 10:46
It depends on whether radiation, conduction, or convection dominates the heat transfer. I'm not sure off hand which it is, but assuming radiation will probably give an ok estimate at first. If radiation dominates then it's easy, since the radiated power is proportional to the surface area and you can use that the human body constantly burns about $100\ \mathrm{W}$ to maintain temperature (under normal conditions). – Michael Brown Mar 5 '13 at 10:47
A watt is a joule per second. And the k in kW means 1000, not 1024. ;) – Michael Brown Mar 5 '13 at 10:49
@MichaelBrown See? That's Why. ;-) – towi Mar 5 '13 at 10:50

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