Specific heat of solution

1. Mar 2, 2016

joaquinjbs

It's possible to calculate the latent heats of fusion and the specific heat of a solution of H2O and NaCl?

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. Mar 3, 2016

DrDu

Yes, the heat of fusion will be a function of the concentration of the brine.

3. Mar 3, 2016

joaquinjbs

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?

4. Mar 3, 2016

BvU

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 .

5. Mar 3, 2016

joaquinjbs

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?

6. Mar 4, 2016

BvU

You form ice and the composition moves towards the eutectic point when cooling further.

7. Mar 4, 2016

DrDu

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.

8. Mar 4, 2016

joaquinjbs

Ok, thank you very much!

9. Mar 4, 2016

joaquinjbs

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
10. Mar 6, 2016

DrDu

Basically, the final result I derived is the formula for freezing point depression. Every book on physical chemistry should contain a discussion.

11. Mar 6, 2016

joaquinjbs

Thank you, but I'm going to do with an practical experiment. Even so I'll try to resolve the formula.

12. Mar 6, 2016

DrDu

A first step would be to convert weight percent to molar fraction.

13. Apr 19, 2016

joaquinjbs

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?