= Solution
With water <fugacity coefficient> $\phi_f=0.8$, use $f=\phi_fP=4800$ bar. Retain the same ideal mineral-site activities, giving
$$
Q=\frac{4800}{64}=75.
$$
Hence
$$
\boxed{T=\frac{127442.28}{170.7-8.3145\ln75}=945.40\ \mathrm K=672.25^\circ\mathrm C.}
$$
The lower fluid <chemical fugacity> stabilizes the water-bearing products of the forward dehydration relative to the ideal-gas calculation and lowers the required <temperature> by about $13.19$ K. Here the stated <fugacity coefficient> is treated as a given constant for evaluating the equilibrium.
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