In an aluminum pot, 0.15 kg of water at 100°C boils away in five minutes. The bottom of the pot is . . .?

In an aluminum pot, 0.15 kg of water at 100°C boils away in five minutes. The bottom of the pot is 2.5 10-3 m thick and has a surface area of 0.016 m2. To prevent the water from boiling too rapidly, a stainless steel plate has been placed between the pot and the heating element. The plate is 1.1 10-3 m thick, and its area matches that of the pot. Assuming that heat is conducted into the water only through the bottom of the pot, find the temperature at each of the following:





a) the aluminum-steel interface (in Celsius)





b) the steel surface in contact with the heating element (in Celsius)

In an aluminum pot, 0.15 kg of water at 100°C boils away in five minutes. The bottom of the pot is . . .?
Well, first we need to figure out the rate heat has been transferred:





To boil water at 100C, 2260 J/g are needed.





q = 150g(2260J/g)/300seconds = 1130 Watts





And now...the part I am having trouble with...you need to find the the convective heat transfer coefficent, h, between water and aluminum boiling water, so we can find the temperature of the aluminum touching the water.





This needs to be given or has to be estimated via a correlation to properly solve this. I have found some decent info here.





http://www.wlv.com/products/databook/db3...





See pages 9-9 thru 9-11 and used the Gorenflo correlation for the nuclation boiling of water. From this I found h ~ 5960 W/m^2/K





Now we can use this equation to find the temperature of the aluminum on the water side.





q = hA(Tsurface - Twater)





1130W = (5960 W/m^2/K)(0.016m^2)(Tsurface - 373.15K)


Tsurface = 385K = 111.85 deg C





Now we can easily solve the problem using equations for conduction:





q = kA(T2-T1)/thickness


for aluminum, wikipedia says k = 237 W/K/m





the same 1130 W are transferred:


1130 W = (237 W/K/m)(0.016m^2)(T2 - 385K)/0.0025m


T2 = 385.745K =112.595 deg C





For the steel plate, the same 1130 Watts are transferred:


steel is primarily iron:


k for iron = 80.4 W/K/m


1130 W = (80.4 W/K/m)(0.016m^2)(T3 - 385.745K)/0.0011m


T3 = 386.711K = 113.591 deg C





Note, depending on what class this question was asked in...I have a feeling you were supposed to assume that the edge of the aluminum pot was 100C...if that would be the case, just subtract all the answer by 11.85 deg C. If this is an engineering class on heat transfer, then you shouldn't ignore the calculation of the pot temperature touching the water (the most ambiguous) step. Also if your textbook has other reference data...you should appropriately substitute those numbers in.

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