= Solution
Let $\theta=T-T_0>0$ and $\theta_1=T_1-T_0$. The aperture spans the enclosure's depth $H$, so the incoming <volume flux>, equal to the outgoing <volume flux>, is
$$
q=\frac{Hh}{2}\,c\sqrt{g\beta h\theta}.
$$
Only one of these equal exchange fluxes multiplies the <temperature> difference in the <heat> budget: incoming fluid brings $qT_0$, outgoing fluid removes $qT$. With constant <mass density> and <specific heat capacity>, those factors cancel and the <well-mixed ventilation temperature balance> is
$$
H^3\dot\theta=-q\theta,\qquad
\dot\theta=-k\theta^{3/2},\qquad
k=\frac{c h^{3/2}\sqrt{g\beta}}{2H^2}.
$$
Integrating from $\theta_1$ gives
$$
\boxed{T(t)=T_0+\frac{T_1-T_0}{\left[1+\dfrac{c h^{3/2}\sqrt{g\beta(T_1-T_0)}}{4H^2}\,t\right]^2}.}
$$
The cooling is algebraic rather than exponential because the buoyancy-driven <single-opening exchange flow> weakens as its driving <reduced gravity> decreases. The formula assumes negligible wall <heat capacity>, no heat input, the specified aperture geometry, and a uniform interior <temperature> maintained by mixing.
With a low-level opening, the dense incoming cold fluid tends to spread along the floor instead of falling through the warm interior. A cold lower layer and warm upper layer develop, giving <stable density stratification>. The incoming fluid can short-circuit back out through the low opening after that layer occupies it, leaving warm fluid above poorly ventilated. Thus the uniform-temperature assumption is generally inappropriate: cooling becomes vertically nonuniform and substantially slower for the upper warm fluid. If mixing were artificially maintained, the same lumped budget would still apply; it is the loss of that assumption, rather than a reversal of <buoyancy>, that changes the physical outcome.
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