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Help with Entrophy

  1. Feb 9, 2008 #1
    One kg of ice at 0°C is added to one kg of boiling water. The mixture comes to
    equilibrium. What is the change in entropy in cal/K of the system? (Lf = 80cal/g)

    a. 300
    b. 100
    c. 200
    d. 50
    e. 25

    Qcold = -Qhot

    Mw*Cw*dT + Lf*Mw = -Mw*Cw*dT + Mw*Lv

    All has Mw in common, therefore it should cancel out

    Cw*dT + Lf*Mw = -Cw*dT + Mw*Lv

    (1cal/g)(Tf - 0) + 80cal/g = -(1cal/g)(Tf-100) + 540cal/g;
    I converted the latent heat of vaporation (2.26x10^6J/kg)*(1/4186cal*kg/J*g) ~
    540cal/g

    after equating the two sides, I got 280C for Tf

    My answer for Tf doesn't seem reasonable because the Final Temp should be in between
    0-100C

    Anyways, My other problem is getting the dQ

    I have no idea how to get it...or maybe I do..when it doubt I just ask people out!

    dS = dQ/T

    if my T is right then just finding dQ problem

    Thanks in advance folks.
     
  2. jcsd
  3. Feb 9, 2008 #2

    HallsofIvy

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    Science Advisor

    Why didn't you cancel the "Mw" out in each term?
    Dividing each term in your original equation by mW gives
    Cw*dT+ Lf= -Cw*dT+ Lv

     
  4. Feb 9, 2008 #3
    I did cancel! ...
     
  5. Feb 9, 2008 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Then you must mean something different by "cancel" than I do.

    You start with
    "Mw*Cw*dT + Lf*Mw = -Mw*Cw*dT + Mw*Lv"

    and conclude
    "Cw*dT + Lf*Mw = -Cw*dT + Mw*Lv"

    What exactly did you do in order to "cancel"? Did you divide both sides of the equation by something? Did you add or subtract something from both sides?
     
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