Posted by Lydia on Thursday, February 21, 2008 at 5:44pm.
You need the heat transfer coefficient of styrofoam. It is about .01 watts meters/deg
call area of foam A = .016 m^2
call temperature difference across styrofoam = 35 - 0 = 35 deg C
call thickness of foam t = 2*10^-3 m
then heat gained through foam = (k A/t)((35-0)
heat gain rate watts = (.01*.016/.002)(35)
That watts * time in seconds = Joules
so that times 3600 = Joules gained to melt ice in one hour. Now kg melted = Joules gained/ heat of fusion in Joules/ kilogram
By the way, I think heat of fusion is about 3.33 * 10^5 Joules/kg. You have a typo.
I don't really get the last part...so that times 3600 = Joules gained to melt ice in one hour. Now kg melted = Joules gained/ heat of fusion in Joules/ kilogram
Please explain.
(.01*.016/.002)(35)joules/s *3600 s = X kg ice * 3.33*10^5 Joules/kg ice melt
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