The von Neumann entropy $-\mathrm{Tr}(\rho\ \log\rho)$ of a quantum thermal state with $\rho=\frac{1}{Z}e^{-\beta H}$ gives the thermal entropy, see e.g. this question.
The von Neumann entropy is a fine-grained entropy, does that mean thermal entropy is also a fine-grained entropy? To my understanding, the thermal entropy (in classical sense) is a coarse-grained entropy. Is there a difference between "classical" and "quantum" thermal entropy? Do the thermal and von Neumann entropies follow the second law?
Moreover, we could purify a thermal state to a thermal field double (TFD). The thermal entropy is given by the entanglement entropy of the TFD. Is this only a trick, or is there any relationship between thermal entropy and entanglement entropy?
Edit: For coarse-grained entropy, I use the definition in section 4 of The entropy of Hawking radiation (2020). When we could only observe some simple observables $A_i$, the coarse-grained entropy is given by the maximum von Neumann entropy of all possible $\tilde{\rho}$ with $-\mathrm{Tr}(\rho A_i)=\mathrm{Tr}(\tilde{\rho} A_i)$.