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

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:

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?)

• Ohhh! Just to make sure youre saying that we approximate the change enthalpy and change in internal energy of solids and liquids, since their volume change is negligible but we dont(actually cant) that in gases since their volume change is significantly big? Commented Jul 10, 2021 at 3:23
• Thank you very much for your answer. But I don't understand what you mean in last paragraph, do you mean to say that their derivation is diffrent from mine? If yes, then yes they indeed are different. THANKS again. Commented Jul 10, 2021 at 6:20
• PS: Do you have recommendation for books on thermodynamics for beginners? Commented Jul 10, 2021 at 6:21
• Could you give an example of a textbook you’ve read that has $\Delta H=q_0$? I just want to check if I’m reading your question correctly. The term “heat” is often used to refer to a change in enthalpy, corresponding to a change in internal energy plus P–V work. Commented Jul 10, 2021 at 15:38
• It is not directly written that $\Delta H=q_o$, but actually it is written $\Delta H=q$ for constant pressure but the book uses the first formula while discussing the measurement of $\Delta H$ with calorimeter. The book is Peter Atkins Chemical principles. Here is screenshot of certain relevant portions of the book collaged together: i.sstatic.net/LOgwT.jpg Commented Jul 11, 2021 at 1:16