HTE Coefficients for sFe2O3

HTE = A + B x T + 0.001 x C x T^2 + 100,000 x D / T

Using the data for Hematite, Ref. B672158, A = neg 20748.867 B = 46.15174 C = neg 3.87510 D = 21.94615

then a = 1 x A b = 1 x B c = 0.001 x C d = 100,000 x D

Coeffs = a b c d Powers = 0 1 2 neg1 where " = " means APL "gets", "neg" means APL "Alt + 2"

and HTE(T) = + / Coeffs x T * Powers where " * " is the APL "power" which is of course " * "

From USGS Bulletin 1259, data for HEMATITE, page 111, data is provided for T values of 298.15, 400, 500 ... where ... is increasing increments of 100 to 1800, except there are some additional points (950, 1050) to provide more detail at points around crystal phase changes.

We will test a selection of T values, T = 298.15 800 1400 1800 and find HTE = 27.6 16,435.8 37,836 50,988.2

which matches the published data for HT-H298 as shown below: T HT-H298 HTE K KCAL cal 298.15 0 27.6 800 16.130 16,436 1400 37.650 37,836 1800 51.880 50,988

Cheers, Jim