Q3(b). Two phases in thermal equilibrium have the same temperature

1. Take phases α and β in contact, the pair isolated from the surroundings. Let their volumes and amounts be fixed (rigid, closed), so only energy (heat) can pass between them.
2. Total energy constant: U_α+U_β= constant, so dU_β=-dU_α.
3. At constant V and n, dU=T dS for each phase (fundamental equation with dV=0), so dS_α=(dU_α)/(T_α) and dS_β=(dU_β)/(T_β)=-(dU_α)/(T_β)
4. Total entropy change: dS=dS_α+dS_β=((1)/(T_α)-(1)/(T_β))dU_α
5. In an isolated system the entropy is a maximum at equilibrium (second law): dS=0 for any small virtual transfer dU_α, positive or negative. That needs 3pt] (1)/(T_α)-(1)/(T_β)=0 T_α=T_β
6. If T_α≠ T_β, a spontaneous change must give dS>0. With T_α>T_β the bracket is negative, so dU_α<0: energy flows from the hotter to the colder phase until the temperatures are equal. This is the thermodynamic basis of the zeroth law.
7. The same argument with volume exchange gives p_α=p_β (mechanical equilibrium), and with matter exchange μ_α=μ_β (material equilibrium): together the conditions of phase equilibrium.

Application. the condition T_α=T_β, p_α=p_β, μ_α=μ_β for every component is the starting point of the Clapeyron equation and of the phase rule, which describe ice–water, liquid–vapour and alloy equilibria.