Question

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!

Answer 1

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

Answer 2

By the way, I think heat of fusion is about 3.33 * 10^5 Joules/kg. You have a typo.

Answer 3

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.

Answer 4

(.01*.016/.002)(35)joules/s *3600 s = X kg ice * 3.33*10^5 Joules/kg ice melt