P21_053 - . 3 27 . 2 = 3 . 2 J/K. (Note: these calculations...

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53. (a) Starting from Q = 0 (for calorimetry problems) we can derive (when no phase changes are involved) T f = c 1 m 1 T 1 + c 2 m 2 T 2 c 1 m 1 + c 2 m 2 =40 . 9 C , which is equivalent to 314 K. From Eq. 21-1, we have S copper = Z 314 353 cmdT T = (386)(0 . 6) ln µ 314 353 = 27 . 2J / K . (b) Also, S water = Z 314 283 cmdT T = (4190)(0 . 07) ln µ 314 283 =30 . 4J / K . (c) The net result for the system is 30
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Unformatted text preview: . 3 27 . 2 = 3 . 2 J/K. (Note: these calculations are fairly sensitive to round-o errors. To arrive at this nal answers, the value 273 . 15 was used to convert to Kelvins, and all intermediate steps were retained to full calculator accuracy.)...
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