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In explaining how to evaluate data from a coffee cup calorimeter, textbooks often claim that $$ \Delta H = q $$

This makes sense for an isothermic calorimetry experiment, where the temperature of the reaction mixture is kept constant by heating or cooling. It also makes sense for a bomb calorimeter, where there is heat exchange between the bomb and the surrounding water bath. How do you rationalize this for a coffee cup calorimeter, where ideally there is no heat exchange between inside and outside of the cup?

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    $\begingroup$ $\Delta H = q = 0$ and $\Delta H = n \cdot \Delta H_\mathrm{r} + C_{p, \mathrm{fin}} \Delta T$ $\endgroup$
    – Poutnik
    Commented Jun 13, 2022 at 14:29
  • $\begingroup$ @Poutnik How is the sign of $\Delta T$ defined with respect to an exothermic reaction, for example? Does it refer the the change in temperature during the reaction, or the change in temperature to return to the original temperature? $\endgroup$
    – Karsten ♦
    Commented Jun 13, 2022 at 14:41
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    $\begingroup$ For exothermic reaction, the 1st term in negative and the 2nd one is positive, the content warms up and $\Delta T$ is obviously positive. I.e. the change from initial to final temperature. $\endgroup$
    – Poutnik
    Commented Jun 13, 2022 at 14:44
  • $\begingroup$ @Poutnik So this is the view that the product mixture is the environment at the same time as it is the system (the reactants turn into product, giving off heat to the solvent and "to itself"). Or in your mind, you let the system cool down (it releases all of the enthalpy in the form of heat, and you need $q = C_{p, \mathrm{fin}} \Delta T$ to heat it up again, see my answer. $\endgroup$
    – Karsten ♦
    Commented Jun 14, 2022 at 16:33
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    $\begingroup$ If I got it correctly and the scenario is an isolated system, there is no cooling down. If reaction enthalpy is zero, temperature does not charge. With nonzero enthalpy, temperature changes. Negatively taken reaction enthalpy, properly scaled, is then equal to heat, that would be needed to warm up the final system by the same T difference. $\endgroup$
    – Poutnik
    Commented Jun 14, 2022 at 16:38

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First, a remark about notation and dimensions. $\Delta H$ has dimensions of energy. I will use $\Delta_r H$ for the enthalpy of reaction, and it has dimensions of energy per amount of substance. The enthalpy of reaction is the enthalpy change of a system (where initial and final temperature are equal) when a reaction happens, divided by the amount of reaction. This means it refers to a specific reaction equation, and its value changes when you multiply coefficients by a constant factor.

Back to the question.

How do you rationalize this for a coffee cup calorimeter, where ideally there is no heat exchange between inside and outside of the cup?

You want to compare the state before the reaction to the state after the reaction at the same temperature. In the scheme below, this would be state 1 and state 3. However, experimentally we went from state 1 to state 2.

enter image description here

Enthalpy is a state function, so we can complete the path to state 3 by going from state 2 to state 3 computationally, and adding up the changes along the path. To get back to room temperature, we have to change the temperature by negative $\Delta T$. If we have good estimates of the specific heat capacity and mass of the cup's contents (and assume the heat capacity of the container is negligible), we can estimate the heat exchange necessary to get back to room temperature, as shown in the image. If the cup is insulated well, we are done because the enthalpy change between state 1 and 2 is zero, ideally. If not, we have to somehow estimate the heat loss or gain in that step.

The entire calculation assumes the absence of non-PV work and constant pressure.

There is a nice discussion of the conceptual difficulties in calculating the enthalpy change from state 2 to state 3 here. They conclude that using the heat capacity of water as an estimate for the heat capacity of the solution often tricks students into using the mass of water instead of the mass of the solution (which will typically be larger for the same volume). They also point out that the specific heat capacity of 1 mol/L saline (the product in one of the common GenChem1 reactions explored by calorimetry) is significantly smaller than that of water, so it might make sense to provide that value, both for accuracy and for better understanding. Relevant data is here and there. For example, the heat capacity of 1 mol/L aqueous sodium chloride is about 3.8 J/(g K) compared to 4.182 J/(g K) for pure water.

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