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July 25, 2014

July 25, 2014

Posted by **Lydia** on Thursday, February 21, 2008 at 8:17pm.

I had posted this question several hours ago, and someone gave me such solution:

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

However, I don't quite get the last part, can someone please explain? Thanks!

- physics -
**Damon**, Thursday, February 21, 2008 at 8:24pmwe got the heat gain in watts

watts * time = Joules

(.01*.016/.002)(35) was our watts

so

(.01*.016/.002)(35) Joules/s * 3600 seconds = X * 3.35*10^5 Joules/kg

Solve for X kilograms of ice

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