I am in my last year of high school completing an assignment hot and cold packs, and particular the enthalpy behind them.

My major lack of understanding is why my calculations are working for heat of water but not for heat of ethanol.

In the past week, I have tested $0.1$–$0.4~\mathrm{M}$ of calcium chloride, magnesium sulphate and potassium chloride and ammonium chloride, in $100~\mathrm{ml}$ of distilled water.

Before doing these experiments I did some basic calculations Using $\Delta H= mc\Delta T$ for each solute and its respective concentration, I assumed water temperature to be $25~\mathrm{^\circ C}$ to just get a rough understanding of the temperature I could expect and got some calculation which looked like this:

Example results

I got the heat of solution from a PDF I found.

I then did the experiments and got results which were very close to what these calculations showed.

However, this week I decided to test another solvent, ethanol. I found its specific heat to be $2.44\ \mathrm{J/g}$ and did the calculations again. I saw that almost all the estimated temperature increased/decreased drastically depending on if it was an exothermic reaction or endothermic reaction. However, after doing this experiment the results I got was very different. I decided to just test $0.2~\mathrm{M}$ of each solution and my calculations estimated these temperatures

Ethanol temperatures

But my calcium chloride temperature ended up only reaching $42\ \mathrm{ \circ C}$ whilst the potassium chloride and ammonium chloride barely changed at all, if anything they went up by $0.1~\mathrm{^\circ C}$.

So where is the problem with my thinking of what I have done incorrectly. I'm guessing ethanol can't be modelled the same as water but I cant find any research supporting that.

  • 3
    $\begingroup$ Ethanol can be modeled like water all right (the same formula linking $\Delta\rm H$ to $\Delta\rm T$, etc.), but ethanol is not water. The enthalpies of dissolution of salts (all of them!) will be very different from those in water, to the point that knowing the latter gives you absolutely no clue as to the former. To add more to the confusion, your ethanol is most likely a water/ethanol mixture, but that's another story. $\endgroup$ – Ivan Neretin Jun 6 '16 at 12:08

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