Visualizing Power in Watts Tuesday, Jul 31 2007
physics 2:18 am
Below is a list containing various power values and the quantity of water they can boil. Click for the larger version. For the water, the initial temperature is approximately that of the ocean’s surface, 20 °C. The text version may be downloaded here.
July 31st, 2007 at 10:17 am
this would be more legible if the scientific notation were used throughout. you can always have the friendly name in ()
July 31st, 2007 at 3:01 pm
Try formatting it with tabs.
July 31st, 2007 at 5:13 pm
I think the second half of the table might be off.
I used the formula: heat = (heat of vaporization)*(density)*(volume)/time and derived the formula:
v = (t/(h*rho))*Q
where t = 60 sec, h = 2.261*10^6 J/kg, rho = 1000kg/m^3, v = volume in m^3, and Q = heat in watts. (t/(h*rho)) = a constant = 2.654*10^-8 m^3/watt.
For example: Inserting 1000 watts into Q and solving for V gives a volume of 2.654*10^-5 m^3 or a cube of water 2.983 cm on a side boiled by 1 kw in 1 minute. This assumes the (latent) heat of vaporization of water is measured at standard atmospheric conditions and at its boiling point of 100 degrees celcius with no temperature change.
Perhaps I made a mistake but the numbers looked a bit low to me considering that water holds a lot of heat. I think boiling a cube of water 1 meter on a side should take ten 100 watt light bulbs more than 1 minute, even if the water starts at 100 celcius.
http://en.wikipedia.org/wiki/Heat_of_vaporization
July 31st, 2007 at 7:53 pm
”
I think the second half of the table might be off.”
Aye. The last few numbers don’t even scale properly- three orders of magnitude increase in length is nine OOM increase in volume, and yet it only gives three OOM increase in power. To give you an idea of how off they are, a cube of water 10^13 km on a side is a full light-year across. Such a volume of water would mass 10^48 metric tons (10^51 kg); trying to boil it in one minute with 10^33 watts is… er.. around twenty full orders of magnitude off of the proper figure. Also notice how all the figures match up; eg. 1 kW-1 meter-1 minute, 1 MW-10 meters-1 minute, which is strange considering the amount of energy needed to boil a kilo of water is some messy number around 2.26 MJ.
July 31st, 2007 at 8:51 pm
Well, I’m no physicist, and the point of posting this here was to get feedback, so thanks.
I just assumed that the energy to heat 1 kg of water at 20 degrees up to 100 degrees would be 80 kilocalories, and just went from there.
There is a mistake on the last week numbers that Tom pointed out, I will fix that shortly. As for the numbers matching up, it actually turned out roughly that way (maybe off by 5%-10% but this is handwaving anyway) according to my calculations based on the previously stated assumption… you can do the math on it if you like.
July 31st, 2007 at 9:27 pm
“I just assumed that the energy to heat 1 kg of water at 20 degrees up to 100 degrees would be 80 kilocalories, and just went from there. ”
It is, but “boiling” water is commonly understood to mean “vaporizing” it, which requires a lot more energy. 80 kcal roughly equals 250 KJ/60 seconds = ~4 KW for the cube .1 meters on a side (one liter).
July 31st, 2007 at 9:29 pm
Oops, arithmetic error- please replace “250″ with “330″ and “4″ with “5.5″.