5. In the acute triangle $\triangle A B C$, the side lengths $a, b, c$ and the area $S$ satisfy $S=\frac{c^{2}-(a-b)^{2}}{k}$, and $\angle C$ is neither the largest nor the smallest interior angle of $\triangle A B C$. Then the range of the real number $k$ is ( ).
(A) $(0,4)$
(B) $(4(\sqrt{2}-1), 4)$
(C) $(0,4(\sqrt{2}-1))$
(D) $(4,+\infty)$
5. B.

Assume $0<\angle A \leqslant \angle C \leqslant \angle B<\frac{\pi}{2}$, then
$$
\frac{\pi}{2}<\angle A+\angle C \leqslant 2 \angle C \leqslant \angle C+\angle B<\pi \text {. }
$$

Thus, $\frac{\pi}{4}<\angle C<\frac{\pi}{2}$.
Also, $k=\frac{c^{2}-(a-b)^{2}}{S}=\frac{2 a b-\left(a^{2}+b^{2}-c^{2}\right)}{\frac{1}{2} a b \sin C}$
$$
=\frac{2 a b(1-\cos C)}{\frac{1}{2} a b \sin C}=4 \tan \frac{C}{2}
$$

is an increasing function on $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$, so,
$4 \tan \frac{\pi}{8}<k<4 \tan \frac{\pi}{4}$.
Therefore, $4(\sqrt{2}-1)<k<4$.