- May 15, 2009
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I got into an argument with a business major yesterday about a product that could boil water via a usb port. Below is the math I did but at the end is a product that seems to violate the formulas. Did I make a mistake? (other than arguing a science topic with a non science person)
g * (temp change needed °c) * (specific heat of water 4.186J/g °C) = J
10oz of water (1 fluid oz = 29.6ml && 1ml water = 1g) * 29.6g = 296g water
296g * (100c-30c) * 4.186J/g °C = 86733.9 J
USB port = 5v at a max of .5amp
w = VI so USB port max power is 5*.5 or 2.5watts
J = ws
s = J/w
86733.9/2.5w = 34693s = 9.64 hours
This is in line with what I thought it would be but then I found this online
http://www.vat19.com/dvds/usb-warmer-cooler.cfmhttp://www.vat19.com/dvds/usb-warmer-cooler.cfm
Which claims to heat up to 140f. Doing the math again with that I come up with just over 2 hours to heat it.
So what am I missing? The product isn't just bogus is it?
edit: I also am assuming ideal conditions. I don't think the race is close enough to bother with the added hassle.
g * (temp change needed °c) * (specific heat of water 4.186J/g °C) = J
10oz of water (1 fluid oz = 29.6ml && 1ml water = 1g) * 29.6g = 296g water
296g * (100c-30c) * 4.186J/g °C = 86733.9 J
USB port = 5v at a max of .5amp
w = VI so USB port max power is 5*.5 or 2.5watts
J = ws
s = J/w
86733.9/2.5w = 34693s = 9.64 hours
This is in line with what I thought it would be but then I found this online
http://www.vat19.com/dvds/usb-warmer-cooler.cfmhttp://www.vat19.com/dvds/usb-warmer-cooler.cfm
Which claims to heat up to 140f. Doing the math again with that I come up with just over 2 hours to heat it.
So what am I missing? The product isn't just bogus is it?
edit: I also am assuming ideal conditions. I don't think the race is close enough to bother with the added hassle.