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Heat of dilution

  1. Jun 7, 2006 #1
    I have a problem here that troubles me for weeks. I was asked to calculate the final temperature if we dilute sodium hydroxide from initial concentration of 50% to final concentration of 2.1%. And also from intial 50% to final 20%. I got 2 answer that diffenrent a lot! 1 is 25.04 oC and the other is 182.36 oC! Is this answer correct? I even do the calculation in between which are 2.1%, 5%, 10%, 15%, 20%. And i got an exponential graph. Is this correct? Can some please tell me? I tried this for weeks and weeks already! Attached here is my graph from calculation. Thank you!
     

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  3. Jun 8, 2006 #2

    GCT

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    Remember, some people can't open your zip files, might want to scan your stuff and then save them as jpeg.
     
  4. Jun 8, 2006 #3

    Bystander

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    You've plotted the results of some calculation you've done. You haven't stated what that calculation is. The plot is meaningless without such a statement.

    What calculation?
     
  5. Jun 8, 2006 #4
    I...

    I also want to show the calculations, but I just dont know why cant I upload the excel sheel, the system doesnt support. That's why! I also dont want to zip the file too, but the file size is too big to upload! I really ahve no idea how can i show my calculations! Please help me!
     
  6. Jun 9, 2006 #5

    Bystander

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    A spreadsheet is NOT a calculation; it's an exercise in baffling onlookers.

    Describe the calculation: 1) name a source for the values you're using, International Critical Tables, JANAF, NBS 500, something; 2) state the numbers you're actually using, x kcal or kJ per mole for what concentration at what temperature; 3) state how you're using that number to determine a "final temperature" (you're doing adiabatic calorimetry, essentially); 4) state what assumptions you're making about the heat capacities of the various solution concentrations.
     
  7. Jun 9, 2006 #6
    Information

    For density and heat capacity, I obtained the value from Perry's Chemical Enginneers' Handbook according to % concentration and interpolation if I need to get the value in between.

    Value for delta_Hsolution I got it from Elementary Principles of Chemical Process, 3rd Edition.

    I used Energy Balance,
    [m*(Cp)*(delta_T)]in=m*(delta_Hsolution)+[m*(Cp)*(delta_T)]out
    where (delta_T)in = Tinitial - Tref
    (delta_T)out = Tref - Tfinal
    and Tinitial = 25oC and Tref = 25oC
    So, the equation reduced to
    - m*(delta_Hsolution) = [m*(Cp)*(delta_T)]out

    The value I used in calculation, from 50% sodium hydroxide to 20% sodium hydroxide, outlet flowrate = 10.2m3/hr
    Density Water = 1000kg/m3
    Density NaOH = 1216.30kg/m3
    Cp Water = 4.18kJ/kg.K
    Cp NaOH = 3.33kJ/kg.K
    Delta_Hs = 1511.47kJ/kg

    The result is
    Delta_T = 157.36 oC
    Final_T = 182.36 oC (Which I think its impossible!)
     
  8. Jun 9, 2006 #7

    Bystander

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    No.

    Going from 50% to 20%: initial state is m(50%NaOH) + 1.5mH2O, and enthalpy is 0.5m x del Hsoln(50%NaOH) + 0; final state is 2.5m(20%NaOH), and enthalpy is 0.5m x del Hsoln(20%NaOH) --- the quantity of NaOH has NOT changed.

    No.

    (mNaOH = 0.5m(50%NaOH) x (delH(20%) - delH(50%)) = 2.5mCp(20%NaOH)delT

    This is heat capacity for 50%. You're not raising the temperature of a 50% solution after you've diluted it to 20%, the enthalpy of dilution is raising the temperature of a 20% solution.

     
  9. Jun 11, 2006 #8
    The Cp Value I used is 20% not 50%.

    About the del Hsoln, I didnt use del H 20% - del H 50% but instead, I used del H 20% - del H infinite, like what i learn from the book Elementary Principles of Chemical Process. Does this cause a difference?
     
  10. Jun 11, 2006 #9
    confused

    The Cp Value I used is 20% not 50%.

    About the del Hsoln, I didnt use del H 20% - del H 50% but instead, I used del H 20% - del H infinite, like what i learn from the book Elementary Principles of Chemical Process. Does this cause a difference?

    By the way, why you use this ratio 0.5 and 2.5?
    (mNaOH = 0.5m(50%NaOH) x (delH(20%) - delH(50%)) = 2.5mCp(20%NaOH)delT
     
  11. Jun 12, 2006 #10

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    Just looked at Perry's (5th) --- it gives you enthalpy, mass fraction, and temperature. You don't need to redo calculations that're done for you --- just go ahead and use them.

    If you're working between 20 and 50%, you use 20 and 50%, not infinite dilution values --- yes, it makes a difference.

    Mass of solution changes with concentration if you're holding the amount of solute constant in a dilution process --- that's the "0.5 and 2.5." Those aren't correct numbers for using the data in Perry's, since Perry's is written in "Chem. E.-speak" (BTU per lbm of solution rather than kJ/mol solute like normal people use).
     
  12. Jun 14, 2006 #11
    1 more question

    I solve the dilution of NaOH already. Thank you Bystander. 1 more question to ask, is this method suitable for all types of dilution? Like if I want dilute 96% sulphuric acid to 30% sulphuric acid? But I cant get a similiar answer to what I got from this website http://www.chm.davidson.edu/chemistryapplets/calorimetry/HeatOFSolutionOfSulfuricAcid.html. I am using the same method as NaOH, but the temperature I got is 27.06 oC but from the website, it is 40.94 oC.

    What happen this time? Thank you for your help!
     
  13. Jun 19, 2006 #12
    Heat of Diluition

    Hello pikkie, I also would like to know out what is the final temp for diluting 98% conc sulfuric acid to 10% conc.
    I am trying to dilute a conc acid stream at 220 litersperhour to a water stream at 2000 litersperhour. both solution are initially at ambient temperature of 28degC.
    Based on your experience derived from the forum, can you help to advise me what is the final temp of the diluted stream will be.
    I am not good in chemistry, therefore really need someone to advise. Thank you.
     
  14. Jun 20, 2006 #13
    Confused

    I would like to know the answer too. But from my previous calculation, I got 32.20 oC (If using del_H(96%) - del_H(infinite)), and 25.08 oC (If using del_H (96%) - del_H(10%)). From this website , the answer is 33.70 oC. And I even run a test myself, and the temperature I got is 46 oC. So I am so confused now, which is correct? Can this help you? I think not as it cant even help me, just want to share with you.
     
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