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Change in Entropy in steam

  1. Dec 7, 2011 #1
    1. The problem statement, all variables and given/known data

    Suppose that 1 kg of water, initially at 350 K, is turned into steam at 373 K. What is the change in entropy? (Lv = 2.26 x 106 J/kg; cwater = 4186 J/(kg Co)

    a. 6060 J/K
    b. 6070 J/K
    c. 6320 J/K
    d. 6330 J/K

    2. Relevant equations

    ΔS=ΔQ/T
    Q=mcΔT
    Q=mL

    3. The attempt at a solution

    heat is needed to get the water to the boiling point 100C or 373K and then heat is needed
    to vaporize the water

    ΔS=(mcΔT+mL)/T
    ΔS=(1kg*4186J/kg*23K+1kg*2.26x10^6)/373
    ΔS=6320

    and the answer key says it is supposed to be 6070
    what am I doing wrong?
     
  2. jcsd
  3. Dec 8, 2011 #2

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    Hi physgrl! :smile:

    The temperature changes while heating.
    This means that your formula ΔS=ΔQ/T is not right.
    You need to integrate for that part of the process: ΔS=∫dQ/T
     
  4. Dec 8, 2011 #3
    What is the right formula?
     
  5. Dec 8, 2011 #4

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    For the part where the water is heating up, you should use:

    ΔS=∫dQ/T

    substitute dQ=m c dT

    and integrate from T1=350 K to T2=373 K


    Do you know how to do that?
     
  6. Dec 8, 2011 #5
    No sory idk how to integrate yet
     
  7. Dec 8, 2011 #6

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    All right. In that case you should have the following formula as a given.
    [tex]\Delta S=m c \ln({T_2 \over T_1})[/tex]

    Do you have that somewhere on a formula sheet or something?
     
  8. Dec 8, 2011 #7
    Nop...i didnt have it. But with that formula how would I include the energy from the vaporization?
     
  9. Dec 8, 2011 #8

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    Since T is constant for vaporization at T2=373 K, the change in entropy is what you already had:
    [tex]\Delta S={m L \over T_2}[/tex]
     
  10. Dec 8, 2011 #9
    Ohh ok so the total would be the sum of the mcln(t1/t2)+mL/t2
     
  11. Dec 8, 2011 #10

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    Almost.
    You seem to have reversed the temperatures in the formula.
     
  12. Dec 8, 2011 #11
    Ohh ok...thanks :)
     
  13. Dec 8, 2011 #12

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    So... did you get the right answer?
     
  14. Dec 8, 2011 #13
    I did it and got 6325 but the answer key says 6070 still
     
  15. Dec 8, 2011 #14

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    I just calculated it myself and I also get 6325.
    So I suspect your answer key is wrong.

    Let me check with the specialists and I'll get back to you.
     
  16. Dec 8, 2011 #15
    :) ok
     
  17. Dec 8, 2011 #16

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    I just got a response back from my esteemed colleague and specialist in the field, vela, who says, and I quote:
     
  18. Dec 8, 2011 #17
    Hehe okay thank you :)
     
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