= Solution
With inflow and overflow both zero and fixed reservoir volume $V$, the transient balances are $V\dot T=-AF_T$ and $V\dot S=-AF_S$. Therefore
$$
\frac{d\eta}{d\theta}=\frac{R_F}{R_0}=q,\qquad
\eta=\eta_s+q(\theta-\theta_s)=q\theta.
$$
The last equality uses the initial steady state. Define $k_c=Ab(\alpha T_0)^{1/3}/(V R_0^2)=CQ/V$, where the latter $Q$ is the pre-shutdown flow. The <post-shutdown double-diffusive reservoir cooling> equation is
$$
\dot\theta=k_c\frac{(1-\theta)^{10/3}}{(1-q\theta)^2}.
$$
Separate variables. An explicit primitive of $(1-q\theta)^2/(1-\theta)^{10/3}$ is
$$
G(\theta)=\frac{3(1-q)^2}{7}(1-\theta)^{-7/3}
+\frac{3q(1-q)}2(1-\theta)^{-4/3}+3q^2(1-\theta)^{-1/3}.
$$
Differentiate it to verify the three powers. Hence
$$
\boxed{\eta=q\theta,\qquad t=t_0+\frac{G(\theta)-G(\theta_s)}{k_c}.}
$$
For $q<1$, $\theta$ increases monotonically to one only at infinite time, with $1-\theta\sim[3(1-q)^2/(7k_ct)]^{3/7}$ up to the choice of time origin. The reservoir approaches $T_\infty$, but its salinity tends to $S_i-qS_0$, leaving the positive salt contrast $(1-q)S_0$. Thus $R_\rho\to\infty$ and the double-diffusive flux shuts down asymptotically. The assumption of a stationary interface and continued applicability of this flux parameterization is part of the reservoir model, not a claim about every actual hydrothermal pool.
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