Posted by Lydia on .
At a picnic, a Styrofoam cup contains lemonade and ice at 0 degree C. The thickness of the cup is 2.0*10^3m, and the area is 0.016 m^2. The temperature at the outside surface of the cup is 35 degree C. The latent heat of fusion for ice is 3.35*1065 J/kg. What mass of ice melts in one hour?
Please give some hints to do it!Thanks!

PHYSICS!!!HELP！！ 
Damon,
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)((350)
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 
PHYSICS!!!HELP！！ 
Damon,
By the way, I think heat of fusion is about 3.33 * 10^5 Joules/kg. You have a typo.

PHYSICS!!!HELP！！ 
Lydia,
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. 
PHYSICS!!!HELP！！ 
Damon,
(.01*.016/.002)(35)joules/s *3600 s = X kg ice * 3.33*10^5 Joules/kg ice melt