# 09 10 4 cm s 04cm h dt note dx k t ice dt ice

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Unformatted text preview: ρice L f x or ∫ xdx = ρ kice ∆T ∫ dt ice L f or 12 k 2 ( x − x0 ) = ice ∆Tt 2 ρice L f At t = 0, x = 0. So we have x0 = 0. x= 2kice ∆T t = 2.18 × 10− 4 t (sec) = 2.18 × 10− 4 × 3600t (h) = 0.89 t (h) ρice L f In other words, the thickness of the ice grows as t with increasing time. ((WileyPlus)) 65. Let h be the thickness of the slab and A be its area. Then, the rate of heat flow through the slab is kA ( TH − TC ) Pcond = h where k is the thermal conductivity of ice, TH is the temperature of the water (0°C), and TC is the temperature of the air above the ice (–10°C). The heat leaving the water freezes it, the heat required to freeze mass m of water being Q = LFm, where LF is the heat of fusion for water. Differentiate with respect to time and recognize that dQ/dt = Pcond to obtain Pcond = LF dm . dt Now, the mass of the ice is given by m = ρAh, where ρ is the density of ice and h is the thickness of the ice slab, so dm/dt = ρA(dh/dt) and Pcond = LF ρ A dh . dt We equate...
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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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