# Confusion on when to use mass of entire solution when performing q=mc∆T

I am studying for the AP Chemistry exam in May 2020. While studying, I encountered a doubt.

In Problem 1 (the question on the top), the educator saw that the problem was referring to the solution as a whole (the problem states ∆Hsoln), and used 100 g + 10 g = 110g grams for the mass in Q = mc∆T and then, on the next part, did ∆H = -Q/n to find the enthalpy of solution.

However, in Problem 2 (the question on the bottom), the educator uses the total mass of the solution (200 grams) even if the problem refers to the reaction (the problem states ∆Hrxn) and not the solution. When I saw the question, I thought I had to convert the 1 mol Ag+ and 1 mol Na+ to grams, and add those masses for (m) in Q=mc∆T since the problem referred to the reaction and not the solution.

As a result, I am confused as to when exactly I should use the mass of the entire solution instead of the mass of the reactants in my Q = mc∆T and ∆H = ±Q/n calculations.

• In short you are saying that mass of water has not been taken into account in the second question while it has been in the first. - is that what you are trying to say? May 1, 2020 at 2:05
• Are you saying because one is a dissolution process and the other is an acid base reaction, you thought they should be treated differently? But it is always the entire solution (i.e. mostly water) that changes temperature.
– Karsten
May 1, 2020 at 2:31
• Karsten, an acquaintance recently showed me a similar first year undergraduate problem and he was in a very good university. It was hopeless to see how this was being taught to him (online classes have made it worse). In some online solutions to such problems, 10 g was being considered and in other online solutions, 10 g salt was ignored. This exact crazy problem is circulating here and there. My opinion was that it must be included in the calculation. Someone should address this problem in detail here so that the confusion ends. May 1, 2020 at 3:55
• M. Farooq, in my experience as an educator, there are often mistakes in solutions manuals. The total mass of the solution must be used. Another way to verify this is that heat capacities tend to be given per mass solution, not mass solvent May 1, 2020 at 4:40

I am confused as to when exactly I should use the mass of the entire solution instead of the mass of the reactants in my Q = mc∆T and ∆H = ±Q/n calculations.

Ideally, you always take the mass of the entire solution, and the heat capacity of the product mixture. The two questions tell you what assumptions to make with respect to the heat capacity. There are other assumptions you also have to make, even if they are not stated.

[top question] ... assuming that the specific heat capacity of the solution is $$\pu{4.18 J/(g ^{\circ}C)}$$.

The instructions are straightforward. You know the mass of the product mixture (conservation of mass) and the assumed specific heat capacity. This is a dissolution process, indicated by the subscript "soln", i.e. $$\Delta H_\mathrm{soln}$$. The first sentence in the problem description should read "enthalpy of dissolution" instead of "enthalpy of solution" to be less confusing.

[bottom question] Assume the density of the solution is $$\pu{1 g/mL}$$ and the and the specific heat capacity is the same as that of water.

Here, solution refers to the product mixture. You have to make an additional assumption, namely that the volume of the product mixture is equal to the sum of the volumes of the reactants. For the given situation, you are not making a big systematic error assuming that. With this assumption, you can calculate $$Q$$.

### What $$Q$$ are we calculating?

The reactants are at a certain starting temperature, and the product mixture are at a different temperature. In a thought experiment, you could heat up or cool down the product mixture back to the starting temperature. That is the magnitude $$Q$$ you are calculating (opposite sign, though). Because the entire mixture is "too" hot or cold, the mass (and specific heat capacity) of the entire product mixture needs to be considered.

### Approximations in other problem sets

In other problems, you are instructed to consider the mass of water instead of the mass of the entire product mixture. This is a fine approximation for diluted solutions. In rare cases, problems such as this give you a more realistic value for the specific heat capacity of the product mixture instead of asking you to use the value for water. Again, using the value for pure water is fine for diluted solutions.

• Just a quick question though; why do we usually use the heat capacity of the solution after mixing, and the mass after mixing? Basically that would be like a thought experiment "dissolving" the salt without enthalpy change and then adding the heat with the salt solutions' heat capacity and mass after mixing Sep 14 at 7:54
• Is it because enthalpy is a state function and we can model it after this thought? Sep 14 at 7:54
• @Mäßige Yes. It's a cheap setup. Instead, you could have the reaction vessel in a calorimeter heat bath, and either keep the bath temperature constant (isothermal calorimetry) or almost constant (traditional large heat bath). In the first case, the reaction mixture would have the same temperature before and after (enthalpy is reflected in cooling or heating necessary to keep bath at constant temperature), and in the second case, the temperature change is tiny (and the enthalpy would almost be equal to the heat transferred to the bath because it has the higher thermal mass).
– Karsten
Sep 14 at 13:24

In mcΔT, you must use the total mass of the system that is changing temperature. In this case, that is the solution.

The same thing is occurring in the second problem. The temperature of the solution changes, and therefore the solution mass is the mass you should use.

It wouldn't make sense if the solute changes temperature but the solution it is dissolved in doesn't change temperature, does it? I find it always helps me to think about the physical process that is happening for questions like this. Always identify what your system of interest is, and the initial and final states of the system. Doing so can clear up a lot of these problems, and will also help you get through more difficult problems if you ever take a thermodynamics course.