Work performed by ramping of magnetic field (non-interacting Ising model) [closed]

Consider the following hamiltonian

$$H=-h\sum_{i=1}^N\sigma_i$$

where $\sigma_i=\pm1$ and $h$ is the magnetization.

Let us assume that the system is equilibrated with a bath at temperature $T$ with $h=0$. Then the field is ramped up to $h=h_0$, and the system equilibrate again with a bath at temperature $T$.

what is the work performed in this process?

closed as off-topic by By Symmetry, Jon Custer, Kyle Kanos, AccidentalFourierTransform, John RennieAug 4 '18 at 5:34

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – By Symmetry, Jon Custer, Kyle Kanos, AccidentalFourierTransform, John Rennie
If this question can be reworded to fit the rules in the help center, please edit the question.

• Your use of the word “equilibrate” is very unclear to me. Do you perhaps mean: “put into thermal equilibrium with...” and “the system in equilibrium” resp. instead? – Antaios Jul 29 '18 at 7:50
• @Antaios. the system equilibrate again with a bath at temperture $T$. corrected in the post. – jarhead Jul 29 '18 at 7:53

By work, I presume you mean macroscopic work. This is the work done by an external agent in sustaining a magnetic field $h_0$ until a Magnetisation $M$ is induced.
Macroscopic work done in changing the magnetisation from $0$ to $M$ is given by a by the expression(See Heat and Thermodynamics, Zemansky and Dittman, Seventh Edition, Eq 3.11): $$W=\int_0^M hdM$$ Assuming the field is flipped to $h_0$ instantaneously $$W=h_0\int_0^M dM=h_0 M$$ This is numerically equal to the average energy with a negative sign. This is a standard exercise in statistical mechanics that requires the definition of a partition function: $$Z=2^Ncosh^N \beta h_0$$ Where $\beta=\frac{1}{KT}$.
Average Energy is given by: $$U=-\frac{\partial \log Z}{\partial \beta}$$