5. In the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, it is known that the edge length is 1, point $E$ is on $A_{1} D_{1}$, point $F$ is on $C D$, $A_{1} E$ $=2 E D_{1}, D F=2 F C$. Then the volume of the tetrahedron $B-F E C_{1}$ is $\qquad$ .
5. $\frac{5}{27}$.

As shown in Figure 2, draw $F F_{1} \perp C_{1} D_{1}$, connect $B_{1} F_{1}$, which intersects $E C_{1}$ at point $K$.
It is easy to see that $B_{1} F_{1} \perp E C_{1}$, and $E C_{1} \perp$ plane $B F K$.
Since $B F$ and $E C_{1}$ are skew perpendicular lines with a distance of 1, and $B F=E C_{1}=\frac{\sqrt{10}}{3}$, therefore,
$$
\begin{array}{l}
V_{\text {tetrahedron } B-F E C_{1}}=\frac{1}{3} E C_{1} \cdot S_{\triangle B F K} \\
=\frac{1}{3}\left(\frac{\sqrt{10}}{3}\right)^{2} \times \frac{1}{2}=\frac{5}{27} .
\end{array}
$$