P21_066 - = mc I ln T 4/T 3 = mc I ln T 1/T 2 = − mc I ln...

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66. We consider a three-step reversible process as follows: the supercooled water drop (of mass m )s tar ts at state 1 ( T 1 = 268 K), moves on to state 2 (still in liquid form but at T 2 = 273 K), freezes to state 3 ( T 3 = T 2 ), and then cools down to state 4 (in solid form, with T 4 = T 1 ). The change in entropy for each of the stages is given as follows: ∆ S 12 = mc w ln( T 2 /T 1 ), ∆ S 23 = mL F /T 2 , and
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Unformatted text preview: = mc I ln( T 4 /T 3 ) = mc I ln( T 1 /T 2 ) = − mc I ln( T 2 /T 1 ). Thus the net entropy change for the water drop is ∆ S = ∆ S 12 + ∆ S 23 + ∆ S 34 = m ( c w − c I ) ln µ T 2 T 1 ¶ − mL F T 2 = (1 . 00 g)(4 . 19 J / g · K − 2 . 22 J / g · K) ln µ 273 K 268 K ¶ − (1 . 00 g)(333 J / g) 273 K = − 1 . 18 J / K ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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