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I am reading a book (Thermodynamics an Engineering approach), I am reading over the section of steady-flow systems with chemical reactions. The book just introduced the following formula for the enthalpy (per mole) of a component as:

$$\bar{H} = \bar{h_f^o} + (\bar{h} - \bar{h^o})$$

Where $\bar{h_f^o}$ is the standard enthalpy of formation at the reference state of 25 ºC and 1 atm, and $\bar{h^o}$ is the sensible enthalpy at the standard reference state of 25 ºC and 1 atm.

How are these two quantities different? In my book they give an example where they calculate the enthalpy for oxygen at 7 ºC, they express it as:

$$\bar{H} = (0 +\bar{h}_{280K} - \bar{h}_{298K} ) $$ where $\bar{h}_{280K} $ = 8150 kJ/kmol

and $\bar{h}_{298K} $ = 8669 kJ/mol

When using enthalpy values to calculate the work done by a turbine at 25 ºC, say : $W = \dot{m}(h_{1,@1atm} -h_{2,@50kPa})$. I can get the corresponding enthalpy values from a pressure table and estimate the work. However, I do not quite get why the enthalpy value of oxygen at 7 ºC needs to be given with respect to $\bar{h^o}$.

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    $\begingroup$ When you calculate the work, with respect to what are the enthalpies $h_{1,@1atm}$ and $h_{2,@50kPa}$? $\endgroup$
    – Tomáš Létal
    Commented Aug 1 at 20:30
  • $\begingroup$ @TomášLétal I normally just look at the enthalpy tables on the textbook that just list "h" values at diff temperatures and pressures. In most examples in the book so far I had only needed to considered the "h" value provided in the tables, no need to account for a reference. I have tried to find what is the reference for the reported values in the pressure and temperature tables but without luck. I am also uncertain if the values given are just $h$ or $h-h_0$. $\endgroup$
    – STOI
    Commented Aug 1 at 22:24

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The thing with enthalpies is that you rarely need the "absolute value". Instead, like in your example, you need difference between enthalpies. But for the difference to be correct, both enthalpies have to have the same "reference zero" state.

Something similar is with temperatures, where degree Celsius and Kelvin have the same "width", i.e., the temperature differences have the same magnitudes in both units, but if you take one temperature in degree Celsius and the other in Kelvins, the difference will be wrong: $$\frac{\Delta T}{[\text{°C}]} = \frac{\Delta T}{[\text{K}]} = \frac{T_2}{[\text{°C}]}-\frac{T_1}{[\text{°C}]} = \frac{T_2}{[\text{K}]}-\frac{T_1}{[\text{K}]} \neq \frac{T_2}{[\text{°C}]}-\frac{T_1}{[\text{K}]} \text{ or } \frac{T_2}{[\text{K}]}-\frac{T_1}{[\text{°C}]}$$

Importance of the same reference state for correct difference calculation should be clear from this. So just use the same source or approach for estimating both enthalpies when you need to calculate their difference and you should be fine.

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  • $\begingroup$ Thank you very much for your response. I am I guess trying to have more intuition for the given formula. When I did similar problems considering water and steam, I often just took enthalpy values at the given initial and final states and took their difference: $h_{f}-h{i}$. In the example above, the enthalpy of oxygen at 7º should just be $H = h_{7 ºC}$, and on the example they write it as $H = h^o_f +h_{7 ºC} - h_{25 ºC} $, the quantity $h_{7 ºC} - h_{25 ºC}$ , from my understanding that's the thermal energy change of oxygen changing from 25 ºC to 7 ºC, not the actual enthalpy value at 7 ºC. $\endgroup$
    – STOI
    Commented Sep 23 at 18:14
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    $\begingroup$ It seem like in that book, the reference state might be $\bar{h^o}$, which is at 25 °C. Other enthalpies are then defined with refence to this state, so you are right that quantity $h_{7\text{ ºC}} - h_{25\text{ ºC}}$ is the thermal energy change between these 2 temperatures, but it also represents the enthalpy with reference to the zero state at 25 °C. $\endgroup$
    – Tomáš Létal
    Commented Sep 23 at 19:25

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