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It is my understanding that enthalpy differs from internal energy only in that the change in enthalpy takes into account any pressure-volume work done on the system (or by the system) in a reaction, and thus it is equal to the heat transferred to the system, at least in constant pressure.

Many sources say that in a constant pressure, "coffee cup" calorimeter, $q_p = \Delta H$. I understand that. However, they also say that the $q_v$ in a constant-volume calorimeter is equal to the change in internal energy, and not necessarily the change in enthalpy. My thought is that since $\Delta H = \Delta U + P\Delta V$, with $\Delta V$ being zero (it is by definition constant volume), is not $\Delta H = \Delta U = q_v$?

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The definition of enthalpy is $$H = U + pV$$

which gives

$$\begin{align}H_2 - H_1 &= (U_2 + p_2V_2) - (U_1 + p_1V_1) \\ \Delta H &= \Delta U + \Delta(pV)\end{align}$$

The equation you cited is only true at constant pressure: if $p_1 = p_2 = p$,

$$\begin{align} H_2 - H_1 &= (U_2 + pV_2) - (U_1 + pV_1) \\ \Delta H &= \Delta U + p(V_2 - V_1) \\ &= \Delta U + p\Delta V \end{align}$$

The condition for that equation is that the pressure is constant. That equation is not applicable for constant-volume (isochoric) processes unless the process is also constant-pressure (isobaric).

However, if you do have a process that is both isobaric and isochoric, you are right to say that $\Delta H = \Delta U$.

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A coffee cup calorimeter is typically used for liquid phase reactions and, as such, generally involves little or no volume change. Under these circumstances $\Delta (PV)=P\Delta V$ is very close to zero, and the change in enthalpy of the reaction is very close to the change in internal energy (particularly at low pressures like atmospheric pressure). Also, since liquids are nearly incompressible, the heat capacity of the reaction mixture at constant pressure is virtually the same as the heat capacity of the mixture at constant volume.

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