Thursday
May 5, 2016

# Homework Help: physics

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

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*10^5 J/kg. What mass of ice melts in one hour?

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 8:24pm

we 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