Problem 4.2 - = = 1 2 1 2 ln T T Vc s s m S v kcal J 4190...

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Problem 4.2 [Difficulty: 2] Given: An ice-cube tray with water at 15 o C is frozen at –5 o C. Find: Change in internal energy and entropy Solution: Apply the Tds and internal energy equations Governing equations: pdv du Tds + = dT c du v = Assumption: Neglect volume change Liquid properties similar to water The given or available data is: () K 288 K 273 15 1 = + = T ( ) K 268 K 273 5 2 = + = T K kg kcal 1 = v c 3 m kg 999 = ρ Then with the assumption: dT c du pdv du Tds v = = + = or T dT c ds v = Integrating = 1 2 1 2 ln T T c s s v or ()
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Unformatted text preview: = = 1 2 1 2 ln T T Vc s s m S v kcal J 4190 288 268 ln K kg kcal 1 mL m 10 mL 250 m kg 999 3 6 3 = S K kJ 0753 . = S Also ( ) 1 2 1 2 T T c u u v = or ( ) T Vc T T mc U v v = = 1 2 ( ) kcal J 4190 K 288 268 K kg kcal 1 mL m 10 mL 250 m kg 999 3 6 3 = U kJ 9 . 20 = U...
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