Why don't we add energy wasted as work to the heat measured by calorimeter in calculation of $\Delta H$?

Question

Lets say a exothermic reaction is happening in a constant pressure calorimeter. This reaction releases energy as heat($q$) and work($w$) (done to expand its volume against atmospheric pressure). The energy for work comes from kinetic energy of molecules i.e. from heat released.

So effectively "the energy used to raise the temperature of water the calorimeter"($q_o$) is less then the actually heat released($q$) as some of the energy from it is "wasted" in expanding($w$), hence the calorimeter measures this new energy which is $q_o=q-w$.

Now $\Delta H$ at constant pressure is equal to actual heat released($q$), hence if we want to measure $\Delta H$ from calorimeter we must add the wasted energy($w$) to energy released measured by calorimeter($q_O$) i.e. $\Delta H=q_o +w= q-w+w=q$.

But in all the books I read$^1$ it is written that $\Delta H=q_o$, that is change in enthalpy equals to heat measured by the calorimeter, but this clearly is wrong according to me. But since it is written in many books that means they are correct hence I am wrong but I cant see my mistake, so help me. Thanks.

1: Some internet refrence:

Enthalpy , Calorimeters

Answer 1

Well, to heat a liter (10^-3 m) of water 1°C around room temperature takes about 4180 J. The corresponding thermal expansion coefficient is about 2.1×10^-4 1/°C, for a volume change of 2.1×10^-7 m. Thus, the P–V work is PΔV = 0.02 J, which is minuscule compared to 4180 J.

Put another way, energy and enthalpy changes (ΔU and ΔU + PΔV, respectively) can often be used as surrogates for each when condensed matter is concerned, because their volume is relatively constant (compared to gases).

Put yet another way, for condensed matter, the constant-pressure heat capacity $C_P$ is close to the constant-volume heat capacity $C_V+VT\frac{\alpha^2}{\beta}$, with volume $V$, temperature $T$, thermal expansion coefficient $\alpha$, and compressibility $\beta$, because $\alpha$ is typically quite small.

(But actually, I don't see any approximation being made in the link you provide. They calculate the temperature change from the enthalpy and the constant-pressure heat capacity, as you'd expect, right?)