# How do you calculate the Enthalpy of a geothermal resource?

Enthalpy seems to be commonly used to characterize geothermal reservoirs, but how is it calculated? I have an example here where the enthalpy is cited as 800 $kJ kg^{-1}$ for a reservoir with a temperature of 190 °C. In that case the enthalpy was apparently just taken to be the specific heat of water (4200 $J kg^{-1} °C^{-1}$) multiplied with the temperature, something like $H = c\Delta T$ (meaning that the reference temperature must be 0 °C).

Is that the correct way to do it? And if yes, is this really the enthalpy in the thermodynamical sense and why is the reference temperature 0 °C?

• They made the assumption of the reference temperature perhaps due to the location of where they were working or what they were trying to achieve. It may or may not suit you, in which case you may need to adjust the reference according to your assumptions. Jun 11, 2017 at 18:19
• @SolarMike This was from a textbook under "Classifications of geothermal systems", so it sounded rather general. Could you elaborate "due to [...] what they were trying to achieve"? What could that be? Jun 26, 2017 at 13:34
• Hi @ye-ti-800 is this still open? Did you find an answer (if yes it would be great if you would post it)? I agree that it is not obvious why one would use the specific heat of water to calculate the enthalpy, on the other hand: if all you can use to extract the heat would be water, then this is the heat capacity of your process fluid. Jan 14, 2018 at 10:41
• @rul30 Hi, yes it is still open and I don't have a really good answer to it yet. To clarify: my main question is not, why they use the specific heat of water, but rather why the reference temperature is 0 °C. My best guess is that this is some kind of maximum possible value, a potential enthalpy so to say. But I never saw this being defined anywhere Jan 15, 2018 at 15:39

${\displaystyle \mathrm{d}H=\mathrm{d}U+\mathrm{d}\left(pV\right)}$