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castrodisastro
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Homework Statement
An ice cube at 0.00 °C measures 14.9 cm on a side. It sits on top of a copper block with a square cross section 14.9 cm on a side and a height of 18.1 cm. The bottom of the copper block is in thermal contact with a large pool of water at 92.5 °C. How long does it take the ice cube to melt? Assume that only the part in contact with the copper liquefies, that is, the cube gets shorter as it melts. The density of ice is 0.917 g/cm3.
Tice = 273 K
Twater = 365.5 K
LHeat of fusion = 334 kJ/kg
Vice = (14.9cm)3 = 3,307.95 cm3
h = 0.181 m
κCu = Thermal conductivity of Copper = 385 W/(m*K)
A = face area of Cu = (0.149m)2 = 0.0222 m2
Homework Equations
ρ = m/V
Q = mLHeat of fusion
PCond = Q/t = Aκ(ΔT/L)
The Attempt at a Solution
I did this is 3 steps.
1)
Used density to obtain the volume of the ice cube
ρi = mi/Vi
mi = ρiVi
mi = (0.917 g/cm3)(3,307.95 cm3)
mi = 3,033.39 g
2)
Used the mass to obtain amount of energy required to melt the ice, Qi.
Qi = miLHeat of fusion
Qi = (3,033.39g)(334 J/g)
Qi = 1,013,152.3 J
3)
Used Qi and the conduction rate equation to calculate the time t.
PCond = Qi/t = ACuκCu(ΔT/LCu)
QiLCu = (ACuκCuΔT)t
QiLCu/(ACuκCuΔT) = t
t = (1,013,152.3 J)(0.181 m)/(0.0222 m2)(385 W/(m*K))(92.5 K)
t = 231.952 J/W = 231.952 (W*s)/W = 231.952 s
t [itex]\approx[/itex] 232 s
I submitted this answer to my online homework and it told me I was incorrect. I need to make sure that I am correctly analyzing the situation and accounting for everything I need to.
Any help is appreciated.
Please don't be rude. I will gladly provide more info on my calculations. Thanks.