# Effect of impurities on enthalpy of fusion of ice

When I freeze water with some impurities (1% NaCl by mass), the mixture seems to have a much lower enthalpy of fusion that pure water. The impurities cause the freezing point to depress to 0.6°C. I only have a simple calorimeter available but it looks like the enthalpy of fusion is dropping by 25%+ compared to pure water (tested by melting the ice into water and measuring the temperature drop).

Additionally, the ice does not have a nice crystal structure. It is "mushy" and easily deformed.

This effect is surprising to me since colligative theory tells us that most properties changes in a pretty small linear way at low concentrations (like the freezing point).

Am I observing a real phenomenon or is there some problem with my testing?

If its a real phenomenon, is there a way to calculate/estimate the effect?

• are you mesuring change in freezing point, $\Delta T_f$ or change in enthalpy of fusion $\Delta H_f$ ? Apr 23 '15 at 16:48
• @ADG Both. The change in freeze point matches theory nicely- $\Delta T_f = i*K_b*m = 0.6°C$. I back-calculated the enthalpy of fusion from $m_{water}*c_p*(T_f-T_i) = m_{ice}*h_f$ (where the final temp is 0C so the ice has no sensible energy change.) Apr 23 '15 at 17:43
• ok the change in freezing point is matching but not the heat of fusion? Apr 23 '15 at 17:44
• Thats the question. I don't know of a way to calculate the enthalpy of fusion of a salt/water mixture to compare the experimental value against. Is there a formula for this? Apr 23 '15 at 17:45
• Maybe 1% isn't "low" concentration Apr 24 '15 at 8:07

According to the article An Empirical Correlation between the Enthalpy of Solution of Aqueous Salts and Their Ability to Form Hydrates (Pandelov et al. 2010), your observation of a lower enthalpy is correct:

We have found a correlation between the enthalpy of solution of the salt and its ability to form a hydrate, namely, that the salt’s enthalpy of solution is lower than the standard enthalpy of fusion of ice (6 kJ/mol)

However, the drop in enthalpy you observed of 25%+ seems excessive, as can be seen in the graph below: Latent heat of fusion comparison between water, $\ce{NaCl}$ solution and ethanol solution.

Image source: Colligative properties of water cited from the original article Formulation of the latent heat of fusion of ice in aqueous solution (Kumano et al. 2009)

KUmano et al. 2009's methodology centred around the use of

Differential scanning calorimetry analysis was performed and the measurements and calculated values compared with respect to the specific enthalpy of the ice/solution mixture.

As for the 'mushy ice' that you observed, it is likely to be a mixture of ice and brine, as per the diagram below: Image source: Wikipedia

• Thanks for the references, these are very helpful. Update for anyone else who might ever be reading this: I put the mushy stuff in a chest freezer for a week and it is a nice solid block. The issue must have been in the way I was making the ice. May 4 '15 at 22:24

I think your methodology is a bit wrong. When the frozen brine melts into the water, the water won't be 100% water anymore. So, the specific heat capacity of the water changes, making the change in temperature different. Maybe that's why you have a 25% drop when compared to water. You can follow the steps below only if you have a heat source that generates a known amount of heat(energy), like an immersion heater with known voltage and wattage.

1. Weigh an empty container

2. Clamp a filter funnel to a retort stand (or anything that can hold it off the table surface)

3. Place the container below the filter funnel

4. Dry the ice and crush it

5. Place the crushed ice into the filter funnel

6. Set a timer for a specific amount of time, place the heater into the crushed ice and turn the power on, start the timer at the same time.

7. Once the time is up, switch off the power to the heater and remove the container

8. Now, weigh the container with the melted ice

9. Use the same set up and follow the same steps in the steps above, but this time, don't turn on the heater (but still leave it in the ice) (this is to find the amount of ice that melted due to the room temperature)

10. Find the mass of ice that melted due to the heater by subtracting the mass of the container and melted ice without the heater on from the mass of the container and melted ice with the heater on

11. Find the amount of heat generated from the heater by taking the power rating of the heater (in Watts) and multiply it by the time the ice melted for.

The main equation you'll use is ML = Pt, where M is the mass of ice that melted due to the heater, L is the specific latent heat of fusion, P is the power rating of the heater in Watts and t is the time that you let the ice melt for. From there I think you can figure out L. All this is done with the assumption that the heat from the heater that is lost to the surroundings is 0 or negligible.

As for the "mushy" ice, you probably didn't let it freeze long enough.

IDK if any of this will help now though, cus it's been more than 5 years, and I'm pretty sure you're done with the experiment.LOL