11. Given $a+b+c=0, a^{2}+b^{2}+c^{2}=4$. Then, the value of $a^{4}+b^{4}+c^{4}$ is $\qquad$ .
11.8.

From the known equations, we get $a+b=-c, a^{2}+b^{2}=4-c^{2}$.
$$
\begin{array}{l}
\text { Also, } a b=\frac{1}{2}\left[(a+b)^{2}-\left(a^{2}+b^{2}\right)\right] \\
=\frac{1}{2}\left[(-c)^{2}-\left(4-c^{2}\right)\right]=c^{2}-2 .
\end{array}
$$

Therefore, $a^{4}+b^{4}=\left(a^{2}+b^{2}\right)^{2}-2 a^{2} b^{2}$
$$
=\left(4-c^{2}\right)^{2}-2\left(c^{2}-2\right)^{2}=8-c^{4} \text {. }
$$

Thus, $a^{4}+b^{4}+c^{4}=8$.