PHYSICS!!!HELP!!

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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!! -

    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

  • PHYSICS!!!HELP!! -

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

  • PHYSICS!!!HELP!! -

    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!! -

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

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