# Thermodynamics adding salt to water changes the temperature

$5\:\mathrm{g}$ of an unknown salt are dissolved in $325\:\mathrm{g}$ of water. Both the water and the salt are initially the same temperature. The water's temperature falls by $11.4\:\mathrm{^\circ{}C}$. Explain how it is possible for the salt and water to change temperature even though both substances are initially at the same temperature.

My approach:

We know that $$q = mc \Delta T$$

Since $q, m,$ and $\Delta T$ aren't changed by this action, this process must result in the raising of water's specific heat. Is this correct?

-

The keyword is that the salt dissolves. Dissolution entails at least two steps:

1) Overcoming solvent-solvent interactions and bonds. An extreme example: your dinner plate doesn't dissolve in your kitchen table. One reason is that there is an extremely high energy barrier to overcoming the hypothetical solvent's (in this case, the kitchen table's) intermolecular forces. You'd have to take an ax to the table to overcome the strong intermolecular forces that hold the molecules of your dinner table together.

On the other hand breaking these solvent-solvent "bonds" or intermolecular forces in water is easy; you can jump into a swimming pool just fine. I wouldn't suggest jumping into a table.

Also, as this step's suggests, this is an energy intensive process or an endothermic process. Energy must be consumed to break these solvent-solvent bonds or interactions.

2) Breaking solute-solute interactions and bonds. This however isn't required. For example, some unionized $\ce{NaCl}$ may be solvated by water. You'll find this more true with some of the less soluble salts. But most of the sodium chloride you toss in water will have dissolved (we'll get to why later), so what is solvated generally isn't $\ce{NaCl_{(s)}}$ or $\ce{NaCl_{(aq)}}$ (this is somewhat misleading terminology) but rather $\ce{Na^+}$ and $\ce{Cl^-}$.

This again is an endothermic process as bonds are being broken.

3) Solute-solvent stabilization (solvation). This steps entails the formation of solute-solvent intermolecular forces. For example, the sodium ion, with its positive charge, may form a hydration shell of water molecules.

This is an exothermic process, as "bonds" or more accurately, stabilizing intermolecular forces are formed.

Couple all three of these processes together and you have the actual dissolution process of $\ce{NaCl}$.

What's interesting about $\ce{NaCl}$ is that even with the third, exothermic step, the dissolution (which comprises all three steps) is still a tiny bit endothermic. So that's what the question is premised on - the addition of sodium chloride to water kicks off a spontaneous process which is endothermic. So one might ask why might this endothermic process still occur. The answer lies with entropy; the dissolution of sodium chloride is not enthalpically favorable (dissolution consumes energy - specifically - from the water) - yet the dissolution is entropically favorable.

Remember that all processes tend toward greater disorder over time, and the dissolution of sodium chloride is a perfect way to increase disorder; we go from unionized sodium chloride to two ions! The two ions are definitely going to have a lot more degrees of freedom than a single, unified molecule.

$\ce{NaCl_(s) + H_2O ->Na^+_(aq) + Cl^{-}_(aq)}$

-
In terms of dissolution processes as in sodium chloride it is better to speak of lattice energy instead of bonding energy. –  Martin May 29 at 9:29
Good point. Will edit. –  Dissenter May 30 at 2:28

You are starting on the right track, but the specific heat of water won't change (at least not by a measurable amount).

You know that the flow of heat is defined by $q=mC{\Delta}T$.

But, you also know that energy is conserved. So if heat is flowing out of the water, where is it going?

This looks like a homework question (if it is, please tag it as one!), so I am reluctant to just give you the answer. I will give you a hint though - the problem states that the salt dissolves - I think that will be important.

-- Edit --

Since it's not a homework question, but rather review, I'll explain in some detail. As @Dissenter said, dissolution of any salt (ionic compound) can be either exothermic or endothermic, depending on the salt and on the solution. In this case, it must be endothermic.

We know this because ${\Delta}T < 0$, which means $q < 0$ for the water. The heat must have gone somewhere - the only place it could have gone (assuming that it wasn't just lost to the environment) is into the molecules as stored energy (enthalpy). Since the heat transferred from the water is negative (exothermic), that means the enthalpy change for dissolving the salt must be positive (endothermic).

-
This isn't a homework question. It is review for a test. If the water and the salt are the same temperature, I don't see where the heat would be going. –  okarin May 28 at 18:34
Heat = Energy. When something dissolves it may have positive or negative enthalpy - energy change. See? –  Swedish Architect May 28 at 19:23
@Swedish Architect That's right - in this case the enthalpy of solution is positive (endothermic), and so heat is withdrawn from the water. Ammonium nitrate is a common salt with positive enthalpy of solution. Since this isn't a homework problem I will expand my answer. –  thomij May 28 at 19:59
@Okarin Even if it is a review for a test it is well within the scope of the homework as defined in the homework policy. thomij, you might want to suggest an edit to adjust the tag for the next time. –  Martin May 29 at 9:36