How can work be generated from quantum correlations? There are several articles (like this or this one) concerning extracting work from (quantum and/or classical) correlations. Nevertheless, these articles appear to be quite dense and without showcasing a simplest toy model, so I have no intuition behind why correlations (defined as mutual information) can be transformed to some degree into useful work.
Can someone provide a reference or hopefully explain with the simplest example they can think of how quantum correlations (for example, from a Bell state) can be transformed into useful work? 
It would be great if emphasis is made on the role information theory plays into this (i.e., how this work is bounded by the mutual information).
My background is late undergraduate physics level.
 A: The maximum average amount of work that can be extracted from a quantum system in state (i.e., density operator) $X_S$, with Hamiltonian $H_S$, using an arbitrary thermal environment at temperature $T$, is given by $kT \log Z_S$ plus the non-equilibrium quantum free energy:
\begin{equation}
F(X_S,H_S) = \mathrm{Tr}[H_S X_S] - kT S(X_S),
\end{equation}
with $S$ the von Neumann entropy: $S(X_S) = - \mathrm{Tr}[ X_S \log X_S]$, and $\mathrm{Tr}$ the trace. Here $Z_S = \mathrm{Tr}[e^{H_S/kT}]$ is the partition function and k is Boltzmann's constant. Note that F has the standard form  "average energy" $- kT$ "entropy", but it is defined out of equilibrum. The above has been known for a long time; for a simple proof that the average extracted work cannot be larger than $F(X_S,H_S) + kT \log Z_S$ see the appendix of arXiv:1705.05397. Saturation is more complicated, and is related e.g. to the results in New J. Phys. 16, 103011 (2014).
Now, suppose S is a bipartite system AB, i.e. $X_S = X_{AB}$, with Hamiltonian $H_S = H_A \otimes I_B + I_A \otimes H_B$ ($I_X$ is the identity). $F$ then admits a decomposition into "local parts plus correlations"
\begin{equation}
F(X_{AB}) = F(X_A) + F(X_B) + kT I(A:B)
\end{equation}
where $I(A:B)$ is the quantum mutual information: $I(A:B) = S(X_A) + S(X_B) - S(X_{AB})$ (this can be verified by summing and subtracting the local entropy terms $S(X_A)$, $S(X_B)$ to the expression for $F(X_{AB})$. Here $X_A$ ($X_B$) is the partial trace over $B$ ($A$) of $X_{AB}$. Hence correlations, as measured by $I(A:B)$, contribute directly to the non-equilibrium free energy of the state, and hence increase the extractable work. For a Bell state, $I(A:B) = 2 \log 2$. Note, interestingly, that $I(A:B) \leq \log 2$ for qubit states that are not entangled: the fact that you can go above $\log 2$ is a quantum feature. 
A: An explanation that I found useful for understanding the general idea here is from the community blog lesswrong: "The Second Law of Thermodynamics, and Engines of Cognition".

So (again ignoring quantum effects for the moment), if you know the states of all the molecules in a glass of hot water, it is cold in a genuinely thermodynamic sense: you can take electricity out of it and leave behind an ice cube.
This doesn't violate Liouville's Theorem, because if Y is the water, and you are Maxwell's Demon (denoted M), the physical process behaves as:

M1,Y1 -> M1,Y1
M2,Y2 -> M2,Y1
M3,Y3 -> M3,Y1
M4,Y4 -> M4,Y1

Because Maxwell's demon knows the exact state of Y, this is mutual information between M and Y.  The mutual information decreases the joint entropy of (M,Y):  H(M,Y) = H(M) + H(Y) - I(M;Y).  M has 2 bits of entropy, Y has two bits of entropy, and their mutual information is 2 bits, so (M,Y) has a total of 2 + 2 - 2 = 2 bits of entropy.  The physical process just transforms the "coldness" (negentropy) of the mutual information to make the actual water cold - afterward, M has 2 bits of entropy, Y has 0 bits of entropy, and the mutual information is 0.

See also: Szilard's engine on wikipedia
