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 Defrosting the Cone (Posted on 2014-10-02)
You have two identical ideal cones of radius R at the base and height H. The bases are open and they are made of a material of zero thickness which conducts heat perfectly.

One cone is filled with ice and the other is filled with water. Lower the frozen cone into the water-filled cone to a height that optimizes the transfer of heat and minimizes the time it takes to thaw the ice.

At this point, the vertical distance from the vertex of the frozen cone to the water level is "h".

Assume that the portion of the frozen cone that is above the water level is insulated and loses no heat there; the only heat transfer is where the water touches the frozen cone.

(a) You are told to use a model which assumes that the heat transfer is proportional to the volume of the water remaining in the lower cone TIMES the surface area of the frozen cone which is touching water.

(b) If you have a different model employing a more accurate assumption with respect to the Physics involved, please state your assumption and solve, or if you prefer, submit a new problem with your model.

The Question: What is h (in terms of H)?

 No Solution Yet Submitted by Larry No Rating

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 Most calculations not shown Comment 1 of 1
I did this on paper and didn't feel like typing everything in.

The Volume remaining in the bottom cone
V=(pi*R²H-pi*r²h)/3
The Surface of contact
S=pi*r√(r²+h²)
The conversion from r to h
r = h(R/H)
The Transfer formula (for part a)
T=VS
putting this together gives a big formula for T in terms of h (and parameters R and H)
T(h) is a fifth degree polynomial which I won't enter here.
The derivative dT/Dh is a fourth degree polynomial with only two terms: the leading and a linear term.
This can then be factored and solved for h.
A lot of simplifying later I got
h = H*³√(2/5) ≈ .632 H

So this is how far the frozen cone should be dunked.

 Posted by Jer on 2014-10-03 11:29:41
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