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Specific heat of solution

  1. Mar 2, 2016 #1
    It's possible to calculate the latent heats of fusion and the specific heat of a solution of H2O and NaCl?

    fig6_1_1.gif

    In some cases, it would have two latent heats of fusion? For example, at 10% NaCl, one at -21ºC and another at -5ºC?

    Thanks!
     
  2. jcsd
  3. Mar 3, 2016 #2

    DrDu

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    Yes, the heat of fusion will be a function of the concentration of the brine.
     
  4. Mar 3, 2016 #3
    Thank you DrDu. I guess that heat of fusion will be at -21ºC, but If I have 10% of NaCl, would I have another different heat of fusion at -5ºC, or it would be the same?
     
  5. Mar 3, 2016 #4

    BvU

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    With a 10% solution there is no NaCl formation at either of these two temperatures. The heat of fusion comes into the calculation only on the line between brine and brine + NaCl .
     
  6. Mar 3, 2016 #5
    Well, thank you BvU. But for me it's interested what happend on the left of eutectic point. I could calculate the latent heat of fusion of ice+brine?
     
  7. Mar 4, 2016 #6

    BvU

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    You form ice and the composition moves towards the eutectic point when cooling further.
     
  8. Mar 4, 2016 #7

    DrDu

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    You can do so, at least for 0% concentration. For very small concentrations, the chemical potential of water in the liquid phase follows Raoult's law ##\mu_l(T,x)=\mu_{l0}(T)+RT \ln(x)## where x is the molar fraction of water in brine. Pressure is constant, so I don't mention it any further. The chemical potential of the solid phase, which consists of pure ice, is ##\mu_s(T)## and depends only on temperature. Now on the melting curve, ##\Delta G=\mu_l(T,x)-\mu_s(T)=0##. We can now use a Taylor expansion for ##\mu(T,x)## around ##\mu(T_0,1)##, where ##T_0## is the melting point of pure ice. We know that ##\partial \Delta G/\partial T|_{P,x}=\Delta S=\Delta H/T_0## and ## \partial \Delta G/\partial x|_{T_0,P}=-RT_0 (1-x)##. So ##0=\Delta G=(\Delta H/T_0)\cdot \Delta T -RT(1-x)## or
    ##\Delta H=RT_0^2(1-x)/\Delta T##. In the calculation of x, you have to take into account that NaCl will dissociate into sodium and chloride ions.
     
  9. Mar 4, 2016 #8
    Ok, thank you very much!
     
  10. Mar 4, 2016 #9
    I have been thinking about this all day, but I'm not able to resolve the problem. If you don't mind, could you show me an example?

    Than you.
     
    Last edited: Mar 4, 2016
  11. Mar 6, 2016 #10

    DrDu

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    Basically, the final result I derived is the formula for freezing point depression. Every book on physical chemistry should contain a discussion.
     
  12. Mar 6, 2016 #11
    Thank you, but I'm going to do with an practical experiment. Even so I'll try to resolve the formula.
     
  13. Mar 6, 2016 #12

    DrDu

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    A first step would be to convert weight percent to molar fraction.
     
  14. Apr 19, 2016 #13
    Well I'm back! I would like to resolve this exercise so...

    For example, applying this formula to a solution with 10 wt% NaCl (0.17 mol NaCl), which has a T0 = 268 K, ΔT = 1 K and R = 8.31 J/(molK); I have ΔH = 495392 J

    Now with n = 1 mol, and ΔT = 1 K:
    Cp = ΔH/(nΔT) = 495392 J/(molK) → 8470*103 J/(KgK)

    So this would be the specific heat of a solution with 10 wt% NaCl in water at -5 ºC. But this huge number has no sense... something I'm doing wrong, could you help me?
     
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