= Solution
The shallow-water equations cease to apply inside the narrow nose, so a <gravity-current front condition> is needed to relate its speed to the depth immediately behind it. Use
$$
\boxed{\dot R=Fr\sqrt{g'h}=Fr\,c},
$$
where $Fr$ is the front <Froude number>; the ideal deep-ambient von Kármán condition gives $Fr=\sqrt2$.
In a uniform axisymmetric box model, conservation of contaminated volume gives
$$
h=\frac{V}{\pi R^2}.
$$
The total nondimensional heat content is $V\theta$, while cooling acts over area $\pi R^2$, so
$$
\boxed{
\dot\theta=-\frac{K\pi R^2}{V}\theta,
\qquad
\dot R=\frac{Fr}{R}
\sqrt{\frac{g'_0V}{\pi}}\,\theta
}.
$$
Eliminating time and taking the initial release radius as negligible gives
$$
\theta(R)=1-\frac{K\pi R^4}
{4Fr\,V\sqrt{g'_0V/\pi}}.
$$
Spreading stops as $\theta\to0$, at
$$
R_{\max}^4
=\frac{4Fr\,V}{K\pi}
\sqrt{\frac{g'_0V}{\pi}}.
$$
Hence the maximum covered area is
$$
\boxed{
A_{\max}=\pi R_{\max}^2
=2\pi
\left[
\frac{Fr\,V}{K\pi}
\sqrt{\frac{g'_0V}{\pi}}
\right]^{1/2}
}.
$$
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