= Solution
Immediately after the reversal the state populations have not changed, but the new Hamiltonian satisfies
$$
H(-B)=-H(B).
$$
Thus
$$
e^{-\beta H(B)}
=e^{-(-\beta)H(-B)},
$$
so the same state is canonical for the reversed field with
$$
\boxed{\beta'=-\beta}.
$$
This is the <magnetic-field reversal and negative temperature> construction. Since
$$
\beta'\mu B'=(-\beta)\mu(-B)=\beta\mu B,
$$
the partition function is unchanged. Therefore
$$
\boxed{F'=-F},
\qquad
\boxed{C'=C}.
$$
A gas has kinetic energy unbounded above. Its partition function diverges for $\beta<0$, so it cannot have a normalizable <negative temperature> canonical state and cannot satisfy the zeroth-law equilibrium condition $\beta_{\mathrm{gas}}=\beta'<0$.
If an amount $\delta Q>0$ flows from the spins into a positive-temperature gas, the total entropy change is
$$
\delta S_{\mathrm{total}}
=\delta Q(\beta_{\mathrm{gas}}-\beta')
>0.
$$
The second law therefore selects heat flow
$$
\boxed{\text{from the negative-temperature spin system to the gas}}.
$$
This is the general ordering expressed by <heat flow from negative to positive temperature>.
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