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Here is my procedure for doing this problem and lets see what happens:
First we must find the area over which the can will be cooled. I will ignore the top and bottom areas since they will most likely be stacked so therefore natural convection will not occur.
A=pi*D*L=3.14*.06*.125 = .02355 m^2
We can now find the thermal resistances:
I am ignoring the thermal resistance of the can since it is very thin and made of aluminum so it will cool much faster than the beer.
R from convection = 1/h*A 1/ [10 w/m^2*C * .02355 m^2] = 4.246 C/W
We can now calculate the steady rate of heat transfer to the beer
Q = T2 –T1 / R = 22-4 /4.246 = 4.239 W or J/s
Now it is time for some more assumptions. I am going to assume that beer has the same thermal properties as water.
Density = 998kg / m^3
Specific Heat = 4182 J/kg * C
Density * Volume gives a mass of .3527 Kg
Finally the amount of energy dissipated by cooling the can 18 C will be 26549 J
26549/ 4.239 = 6263 Seconds = 104 minutes = 1.74 Hours
My final answer for the cooling of the beer cans is 1.74 hours. Does that sound like a reasonable answer?
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