Question

the standard molar enthalpy and entropy of the denaturation of a certain protein are 512 kj/mol and 1.60 kj/K. mol, respectively . comment on the sign s and magnitudes of these quantities, and calculate the temperature at which the process favors the denatured state

Answer 1

I suppose you are speaking of a reaction something like this.
protein.H2O ==> protein + H2O

I also suppose that 512 kJ represents the dHrxn and 1.6 represents dSrxn. Then
dGrxn = dH - TdS
dG will be zero at the point where the sign changes. The + sign for dH favors the initial state while the + sign for dS favors the denatured state. The overall dG is 512-298*1.6 = 35 kJ at 25C so the overall dG favors the protein in the hydrated state.Set dG = 0 and solve for T (which will be in kelvin) and that will be the T at which the equilibrium changes from one direction to the other.