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[tex] g=\left( \frac{\tau_F^2}{\tau_1\tau_2}\right)^{\frac{mC_V}{k_B}}[/tex]

where the system is two identical copper blocks at fundamental temperatures [itex]\tau_1, \tau_2[/itex] brought into equilibrium through thermal contact, both with mass [itex]m[/itex] and heat capacity [itex]C_V[/itex]. I've calculated the final temperature through

[tex]\Delta U = C_V(\tau_F-\tau_1)=C_V(\tau_2-\tau_F) \quad \Rightarrow \quad \tau_F=\frac{1}{2}(\tau_1+\tau_2)[/tex]

I tried then to use that (with [itex]\sigma=\log g[/itex] as entropy)

[tex] \Delta \sigma = \left(\frac{\partial \sigma_1}{\partial U_1}\right)(\Delta U_1) + \left( \frac{\partial \sigma_2}{\partial U_2}\right)(-\Delta U_2)[/tex]

But with this I get a result of

[tex]\Delta \sigma = \log g = 2C_V\left(\frac{\tau_F^2}{\tau_1\tau_2}-1\right)[/tex]

But raising this with an exponential to find the change in [itex]g[/itex] would obviously not give me the desired result. Any help on where I'm going wrong with this would be greatly appreciated.