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Thermal Conductivity and Boundry Layer 
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#1
Apr911, 09:03 PM

P: 380

Thermal Conductivity , k , for water is .1455 cal/sec/ meter * C ( converted watts to cal/sec)
Q/t = k* A delta T/ d A= area M^{2} d = thickness of water/ice boundary layer Suppose I want to calculate time for 100 grams of ice to melt ( 8000 calories absorbed from surroundings by conduction. I have delta T and initial surface area of ice/water interface . Two questions : What would be the boundary layer thickness. d And in the case that the surface area, A, is decreasing as the ice melts is it correct to use just the initial surface area in the calculation ? My reasoning on the area is that it would actually be : Integral A(initial)  A(final) and A(final ) = 0 


#2
Nov2411, 02:26 AM

P: 380

so final temp 22000 cal /1100 gram water = 20 C , melting time was 6 minutes In close agreement with Q/t = k*A delta T/ x x = thickness of boundary layer between ice and water = .002 M Reference Journal Physical Oceanography A= surface area 100 cc ice cylinder = .01224 M^2 delta t = 30 deg k, thermal conductivity water = .1455 cal/sec/Meter * Kelvin So, Q/t = .1455 cal/sec/M*T ( 30 C) (.01224 M^2) / .002 M = 26.5 cal/sec 8000 cal/ 26.5cal/sec = 301 sec. = 5 min 


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