8 c b we expect and this is supported by the result

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Unformatted text preview: r rate of conductive heat transfer. (c) Repeating the calculation above with the new value for k2 , we have = which leads to ∆ T2 = 13.8 °C. This is less than our part (a) result which implies that the temperature gradients across layers 1 and 3 (the ones where the parameters did not change) are greater than in part (a); those larger temperature gradients lead to larger conductive heat currents (which is basically a statement of “Ohm’s law as applied to heat conduction”). _______________________________________________________________________ _ Problem 18-65 ((My solution)) kice=0.0040 cal/s cm C, ρice = 0.92 g/cm3 TC = -10°C, TH = 0°C. Lf = 3.33 x 105 J/kg = 79.50 cal/g ∆Q = L f ( Adx) ρice ℘= dQ dT ∆T ∆T = −kice A = − kice A = kice A dt dx x x From these equations, we have ℘ = Lf ( A dx ∆T ) ρice = kice A , dt x where ∆T = 0 − (−10) = 10C dx k ∆T = ice dt ρice L f x At x = 5 cm, we have dx = 1.09 × 10− 4 cm / s = 0.4cm / h dt ((Note)) dx k ∆T = ice , dt...
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This note was uploaded on 05/14/2013 for the course MATH 346 taught by Professor Professormiguelarcones during the Spring '08 term at Binghamton University.

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