High Temperature Heat Capacity of Aqueous Species

The thermodynamic data of aqueous species have been extensively investigated in the high temperature geology and related fields. During this course, a new equation of representing the high temperature heat capacity of aqueous sepcies has been proposed and widely used in those fields. Unfortunately, this equation is not well consist with the MALT representaion equation.

(1) Literature

271. Shock, EL., Sassani, DC., Willis, M., Sverjensky, DA., Geochimica Cosmochimia Acta 61(5),907-950 1997.
272. Accornero, M., Marini, L., Lelli, M., Applied Geochem. 25, 242-260 2010.
273. (a) Marini, L., Accornero, M., Environ. Geol. 52, 1343-1363 2007.
        (b) Marini, L., Accornero, M., Environ. Earth Sci.59, 1601-1606 2010.
332. Haas, JR., Shock, EL., Sassani, DC., Geochim. Cosmochim.Acta 59(21), 4329-4350 1995.
333. Sverjensky, DA., Shock, EL., Helgeson, HC., Geochim. Cosmochim. Acta 61 (7), 1359-1413 1997.
334. Sassani, D., Shock, EL., Geochim. CosmoChim. Acta 62(15), 2643-2671, 1998.

(2) Method of Conversion

1. Proposed heat capacity equation
   
   When pressure is given as the standard pressure, the high temperature heat capacity of aqueous spceis can be represented as follows;
   
   C_p = c₁ + c₂/(T-Θ)²
   Here, c₁ and c₂ are the constant of the constant term and the special term introduced for the aqueous species in which Θ, constant, is used to specify the temperature dependence.
   
   
   
2. MALT adopted equation
   
   MALT adopts the following equation.
   
   C_p = C_a + C_b/10^-3 * T/K + C_c/10⁵ /(T/K)² + C_d/10^-6 * (T/K)² + C_e/10⁸ /(T/K)³
   Here, normal temperature dependences are assumed.
   
   
3. Conversion scheme
   
   Since it will be troublesome to include the above special temperature dependence term, it is adopted to convert this term into the normal terms by least square method. The adopted results are given as follows;
   For c₂=1000 and Θ = 228
   the following coefficients were obtained:
   C_a = 22.232;     C_b/10^-3 = -27.174;     C_c/10⁵ = -37.699;     C_d/10^-6 = 0;     C_e/10⁸ = 8.025
   
   
   Comparison between two values are made in the following graph. Agreement is sutisfactorily good.
   
   Fig. 1 Comparison between two Cp equations.
   
   

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