University Physics with Modern Physics with Mastering Physics (11th Edition)

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17.113: a) See Figure 17.11. As the temperature approaches C, 0 . 0 ° the coldest water rises to the top and begins to freeze while the slightly warmer water, which is more dense, will be beneath the surface. b) (As in part (c), a constant temperature difference is assumed.) Let the thickness of the sheet be , x and the amount the ice thickens in time . be dx dt The mass of ice added per unit area is then , e ic dx ρ meaning a heat transfer of . f e ic dx L ρ This must be the product of the heat flow per unit area times the time, ( 29 ( 29 . dt x T k dt A H = Equating these expressions, . or f ice f e ic dt L ρ T k xdx dt x T k dx L ρ
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Unformatted text preview: ∆ = ∆ = This is a separable differential equation; integrating both sides, setting , at = = t x gives . 2 f ice 2 t L ρ T k x ∆ = The square of the thickness is propotional to the time, so the thickness is propotional to the square root of the time. c) Solving for the time in the above expression, ( 29 ( 29 ( 29 ( 29 ( 29 s. 10 . 6 m 25 . C 10 K mol J 6 . 1 2 kg J 10 334 m kg 920 5 2 3 3 × = ° ⋅ × = t d) Using m 40 = x in the above calculation gives s, 10 5 . 1 10 × = t about 500 y, a very long cold spell....
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  • Heat, constant temperature difference, Lf dx, ρic e Lf, slightly warmer water

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