ch18-p036 - m ice = Q c ice T = 88828 (2220)(20) = 2.0 kg ....

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36. (a) Eq. 18-14 (in absolute value) gives | Q | = (4190 J/ kg C ⋅° )(0.530 kg)(40 °C) = 88828 J. Since d Q d t is assumed constant (we will call it P ) then we have P = 88828 J 40 min = 88828 J 2400 s = 37 W . (b) During that same time (used in part (a)) the ice warms by 20 C°. Using Table 18-3 and Eq. 18-14 again we have
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Unformatted text preview: m ice = Q c ice T = 88828 (2220)(20) = 2.0 kg . (c) To find the ice produced (by freezing the water that has already reached 0C so we concerned with the 40 min < t < 60 min time span), we use Table 18-4 and Eq. 18-16: m water becoming ice = Q 20 min L F = 44414 333000 = 0.13 kg....
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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