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Entropy of Liquid water Calculation.

  1. Dec 14, 2015 #1
    1. The problem statement, all variables and given/known data
    Calculate the ΔS° when 0.5 mole of liquid water at 0°C is mixed with 0.5 mole of liquid water at 100°C.Assume Cp=18cal/deg mole over the whole range of temperatures.

    2. Relevant equations
    ΔS=∫ Cp/T dT
     
  2. jcsd
  3. Dec 14, 2015 #2

    BvU

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    Hello Joseph, :welcome:

    You erased part of the template. That's a nono in PF -- let's hope the spirits that watch over us don't delete your thread altogether. PF requires effort on your part, so work out the equation !
     
  4. Dec 14, 2015 #3

    Andrew Mason

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    Welcome to PF Joseph!

    The assumption is that the only heat flow into the cold water comes from the hot water, not from the surroundings. So there is no change in entropy of the surroundings.

    Step 1: find the end state of each quantity of water (ie. the final temperature - easy).
    Step 2: find a reversible path between the initial and final states for each quantity of water
    Step 3: calculate ##\Delta S = \int dQ/T## over that path for each quantity of water
    Step 4: add the two results to find the total change in entropy.

    AM
     
  5. Dec 14, 2015 #4
    I am new in here.Forgive me.
     
  6. Dec 14, 2015 #5
    Step 1: Heat Loss=Heat Gain

    m.Cwater.ΔT1=m.Cwater.ΔT2 (Moles are equal so the mass)
    (100°C-T)=(T-0°C)
    T=50°C

    Step 2: Calculations for 0.5 mole liquid water at 100°C
    Ti=100°C----> 100 +273.3=373.3°K
    Tf=50°C------> 50+273.3=323.3°K

    ΔS=∫dQ/T pressure is constant therefore ΔS=∫CpdT/T from 373.3°K to 323.3°K

    ΔS=Cp.ln(323.3/373.3)=18 cal/deg (-0.144) = - 2.59 cal /deg

    Calculations for 0.5 mole liquid water at 0°C
    Ti=0°C------> 0 + 273.3 = 273.3°K
    Tf=50°C-----> 50+ 273.3 = 323.3°K

    ΔS=∫CpdT/T from 273.3°K to 323.3°K

    ΔS=Cp.ln(323.3/273.3)= 18 cal/deg (0.168) =3.02 cal/deg

    Step 3: ΔS=3.02 cal/deg - 2.59 cal/deg =0.43 cal/deg

    Is it correct ?
     
  7. Dec 14, 2015 #6

    Andrew Mason

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    If you have done the math right, that should be the right answer.

    AM
     
  8. Dec 14, 2015 #7
    What if we mix 1 mole of liquid water at 0°C with 2 mole of liquid water at 100°C.How will the result change?
    Because in this equation ΔS=∫dQ/T entropy doesn't depend on quantity of substance.So can we say entropy doesn't depend on mole of the substance for the whole concept?
     
  9. Dec 14, 2015 #8
    In post # 5, you should have multiplied by the number of moles of each liquid before adding the changes.
     
  10. Dec 14, 2015 #9
    ΔS=-2.59 X 1/2 cal/deg mol =-1.295
    ΔS= 3.02 X 1/2 cal/deg mol =1.51

    ΔS°=1.51-1.295=0.215 cal/deg mol I think this is the correct answer now.Is there any mistake in the unit of ΔS° ?
     
  11. Dec 14, 2015 #10
    Yes. It should be cal/deg.

    Do you know what to do now when you have unequal amounts?
     
  12. Dec 14, 2015 #11
    We should multiply entropy by the number of moles of each liquid before finding the net entropy of the system.Thus,

    If we mix 1 mole of liquid water at 0°C with 2 mole of liquid water at 100°C so the result should be:

    ΔS=1x3.02 cal/deg= 3.02 cal/deg
    ΔS=2×(-2.59 cal/deg)= -5.18 cal/deg

    ΔS°=3.02 cal/deg - 5.18 cal/deg
    =-2.16 cal/deg (We can say that entropy of the system is decreasing )
     
  13. Dec 14, 2015 #12
    You forgot to calculate and use the new equilibrium temperature.
     
  14. Dec 14, 2015 #13
    Ohh.You are right .I forget it.I am going to calculate again.
     
  15. Dec 14, 2015 #14
    Qloss=Qgain

    m1Water.Cwater.ΔT1=m2water.Cwater.ΔT2

    2 mole × 18g/mole × 1 cal/deg(100°C-T)=1 mole × 18g/mole × 1 cal/deg (T-0°C)

    2(100°C-T)=T , 200°C=3T T≅66.7°C

    Calculation for 2 mole liquid water at 100°C

    Ti=100°C ------> 100+273.3=373.3°K
    Tf=66.7°C-------> 66.7 + 273.3=340°K

    ΔS=∫CpdT/T from 373.3°K to 340°K

    ΔS=Cp.ln(340/373.3)=18 cal/deg .(-0.093)=-1.68 x 2=-3.36 cal/deg

    Calculations for 1 mole liquid water at 0°C

    Ti=0 + 273.3=273.3 °K
    Tf=66.7+ 273.3=340°K

    ΔS=∫CpdT/T from 273.3°K to 340°K

    ΔS=18 cal/deg x 0.021
    = 3.93 x 1=3.93 cal/deg

    ΔS°=3.93 cal/deg - 3.36 cal/deg=0.57 cal/deg

    I hope it is correct.
     
  16. Dec 14, 2015 #15
    I'm not going to check your arithmetic, but your methodology is definitely correct now.

    Chet
     
  17. Dec 14, 2015 #16
    Thank you very much sir !
     
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