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Entropy Changes in the System and Surroundings Quiz 10

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?Scold = mcold ccold dT / T = (25.0 g) (1.00 cal / g K)) ln ( 343.2 K / 303.2 K ) ... ?Scold = 3.10. ?Shot = - 2.83. ?Stotal = 0.27. 0 (isolated system) ... – PowerPoint PPT presentation

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Title: Entropy Changes in the System and Surroundings Quiz 10


1
  • Entropy Changes in the System
    and Surroundings Quiz 10
  • 50.0 grams of hot water at 90.0 C is mixed at
    constant pressure with 25.0 grams of cold water
    at 30.0 C in an insulated dewar and allowed to
    come to thermal equilibrium. The only heat
    exchange is between the hot and cold water. What
    is the entropy change in cal/K for the system,
    the surroundings, and the total entropy change.
    You are free to choose the system and
    surroundings as you like.
  • - heat lost by hot water heat gained by cold
    water
  • mhot chot ?Thot mcold ccold ?Tcold
  • (50.0 g) (T 90.0 C) (25.0 g) (T 30.0
    C)
  • T 70.0 C

2
?Shot ? mhot chot dT / T (50.0
g) (1.00 cal / g K)) ln ( 343.2 K / 363.2 K )
- 2.83 cal / K ?Scold ? mcold ccold
dT / T (25.0 g) (1.00 cal / g K)) ln (
343.2 K / 303.2 K ) 3.10 cal / K
?Ssystem (cal/K) ?Ssurroundings (cal/K) ?Stotal (cal/K)
?Shot - 2.83 ?Scold 3.10 ?Stotal 0.27
?Scold 3.10 ?Shot - 2.83 ?Stotal 0.27
(-2.83) (3.10) 0.27 0 (isolated system) ?Stotal 0.27
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