Does dissolving something in water change the specific heat of the solution?

Question

I’m a little confused on this topic. If I have a solid that I dissolve in water and that undergoes an exothermic reaction, can I assume that the specific heat $c$ of the solution instantaneously after the dissolution occurs is equal to $\pu{4.18 J g-1 °C-1}$? Does dissolving something in water change the specific heat of the solution? Or can I just say that the specific heat of the solution (solid plus water) equals the specific heat of water?

My first instinct was to think that yes, dissolving a solid in water does change the specific heat of the solution. If this is true, then I thought I might determine the overall specific heat of the solution by taking the specific heat of water and multiplying it by the percentage of the solution that is water, then adding that to the specific heat of the solid times the percent of the solution that is ions from that solid.

I’ve overthought this quite a bit, and I’ve tried to find an explanation on the internet. I’m pretty sure I’m thinking about this incorrectly, and I apologize if this is a bad question. Any clarity you could provide would be much appreciated.

Answer 1

It depends on the solute concentration. If the solution is dilute in terms of the concentration of solid you added, then the heat capacity will be dominated by that of the solvent, or, the heat capacity of the solution will be little altered by addition of a minute amount of solute relative to the value for the pure solvent.

To borrow an expression from a different answer, we can write

$$c_p=x_\mathrm{solv}c_{p,\mathrm{solv}}+(1−x_\mathrm{solv})c_{p,\mathrm{solute}}$$

where $c_p$ is the (molar) heat capacity of the mixture, $c_{p,\mathrm{solv}}$ and $c_{p,\mathrm{solute}}$ are the (molar) heat capacities of solvent and solute, respectively, and $x_\mathrm{solv}$ is the mole fraction of solvent in the mixture. Since for dilute solution we can often make the approximation $x_\mathrm{solv}\approx 1$, the second term on the right is approximately zero and can be ignored. Then

$$c_p\approx c_{p,\mathrm{solv}}$$

So for many examples, in particular those involving dilute solutions, you can safely ignore the effect of the added solute and assume that the heat capacity of the mixture is equivalent to that of the pure solvent.

To give a slightly more quantitative example, consider the molar heat capacity of solutions of sodium chloride in water, data from the Dortmund Data Bank:

A 1 molal solution corresponds to an NaCl mole fraction of 0.0177. At this concentration of NaCl the heat capacity is ~1.4% smaller than that of pure water. An error of this size is often acceptable.

Answer 2

If you were adding two liquids ideally, that is, assuming that the mixing occurs without any change in enthalpy - which occurs for chemically similar substances, then yes, the heat capacity of the mixture is a weighed average of the heat capacity of the components ($A$, $B$):

$$c_p = x_Ac_{pA}+(1-x_A)c_{pB}$$

However, when we are talking about dissolving solids, the situation is a little different. That is because infinite-dilution properties of a solute that derives from a solid source do not mimic the properties of the (crystalline) solid. For example, table salt ($\ce{NaCl}$) has a certain density as a powder, but this value has nothing to do with the density it acquires in solution once the ions dissociate and live freely in solution. The same goes for heat capacity, which is also a thermodynamic property.

So the answer is yes, the heat capacity of the solution does change by adding a solute and, in fact, it depends on the amount that is added. Only experimentally can you measure the exact function of concentration that defines the heat capacity of a particular solution. From a molecular perspective, it will be highly dependent on how the phenomenon of solvation develops.

As always, you can always look up the scientific literature to get the data of the thermodynamic functions of certain solutions including heat capacity.