# How does this method of calculating the water equivalent of a calorimeter work?

I am doing a AP Chemistry lab in which we determine the heat capacity of a calorimeter (a concept that seems like it should depend on a lot more than just the calorimeter.) We have to calorimeters with water of different temperatures (which is tracked the whole time) but equal mass. We then mix them. We then track temperature again. We then extrapolate to the time of mixing (with a linear regression for some reason.) We look at how much energy was lost by the hot water and gained by the cold water based on all 3 lines (each calorimeter before, and the mixed calorimeter after.) Next we take the difference of the energy gained and energy lost at a single instant divide by the temperature change of the cold calorimeter (I have no idea why the cold one) and this is the calorimeters heat capacity.

I am having trouble understanding this, since at a single instance, the energy gained by the hot water and lost by the cold should be equal, and any difference would be error. It also seems completely random to divide by the temperature of the cold water and not the hot. What is going on here?

P.S. Later in the lab, extrapolating to the time of mixing is used to find the energy change due to a reaction, which makes more sense.

P.S.S One Calorimeter was fully mixed into the other (I forget which one, but either way does not make sense, and calorimeters themselves (like the material it was made from) had negligible heat capacity.))

P.S.S.S It is the teacher's first year doing this lab I think.

• Maybe it's the cold one because it can't be the hot one. The hot one loses heat in many ways, and that makes the calculations more "approximate". – M.A.R. Jan 12 '15 at 12:31
• @PyRulez "One Calorimeter was fully mixed into the other (I forget which one, but either way does not make sense, and calorimeters themselves (like the material it was made from) had negligible heat capacity.))" What do you mean by it ?You are finding water equivalent and saying calorimeter heat capacity is negligible.It's so contradicting. – DSinghvi Jan 12 '15 at 17:18

So you're mixing hot and cold water in a calorimeter, and from the temperature change read from a time versus temperature curve, you're finding the heat capacity of the calorimeter. You're not understanding why the heat capacity is computed the way it is.

First, identify all of your heat flows. You've got heat coming from the hot water, heat absorbed by the cold water, and heat absorbed by the calorimeter. If energy is conserved, $$q_{hot} + q_{cold} + q_{calorimeter} = 0$$

One thing that's confusing you is that last term, $q_{calorimeter}$. It isn't zero. It's very small, since your calorimeter is Styrofoam, but it will be measureable. Measuring it is a central goal of this experiment!

Now, you have a time versus temperature curve for the contents of the calorimeter. You're extrapolating the temperature plateaus before and after mixing to the time of mixing to figure out the initial and final temperatures. Since you had the cold water sitting in the calorimeter initially, and you dumped the hot water into it, the final temperature minus the initial temperature gives you the temperature change for the cold water. It also gives the approximate temperature change for the calorimeter if you can assume that the temperature of the calorimeter is the same as the temperature of its contents.

The heat capacity of the calorimeter is $$C = \frac{q_{calorimeter}}{\Delta T_{calorimeter}}$$ You know $\Delta T_{calorimeter}$ from your time vs temperature graph. You know $q_{calorimeter}$ from the energy conservation equation above; $q_{calorimeter} = -q_{hot}-q_{cold}$. Therefore you know $C$. You should know how to get $q_{hot}$ and $q_{cold}$ from the masses and initial and final temperatures of the hot and cold water (and the specific heat of water). Can you take it from there?

The point is that heat is not just stored in the water, but also in the calorimeter sturcture. Heat that is unaccounted for just considering the temperature and amount of water must have been taken up by the calorimeter itself.

• This should really be a comment. Any way, it has now been addressed in the question. – PyRulez Jan 12 '15 at 12:22
• @PyRulez Old men like me say, "you can lead them to water but you can't make them drink". When you say "calorimeters themselves...had neglibile heat capacity", you are contradicting your earlier statement "we determine the heat capacity of the calorimeter". The point is that heat is not just stored in the water, but also in the calorimeter sturcture. Heat that is unaccounted for just considering the temperature and amount of water must have been taken up (or released) by the calorimeter itself. – DavePhD Jan 12 '15 at 13:01
• When one talks about the heat equivalent of a calorimeter, you also take heat lost intro account. These calirimeters mostly leak their heat, and does not contain it (Styrofoam.) – PyRulez Jan 12 '15 at 16:07
• @PyRulez The phrase "heat capacity of the calorimeter" refers only to the heat stored in the material of the calorimeter per degree. It does not include heat transfer in or out of the system over time. See chm.davidson.edu/vce/calorimetry/heatcapacityofcalorimeter.html For someone to give a more helpful answer, I think you need to give more information in question. As far as heat leaking into or out of the calorimeter, the rate is proportional to the difference between the inside and outside temperatures, in accordance with Newtons law of cooling. – DavePhD Jan 12 '15 at 16:26
• @PyRulez, it sounds like you're accounting for heat lost into the surroundings with your time vs. temperature curve. By extrapolating the initial and final temperatures at the time of mixing, you are compensating for temperature drift. – Fred Senese Jan 12 '15 at 21:45