The disintegration energy Q for the nuclear fission of ${ }^{235} \mathrm{U} \rightarrow{ }^{140} \mathrm{Ce}+{ }^{94} \mathrm{Zr}+\mathrm{n}$ is $\_\_\_\_$ MeV .
Given atomic masses of ${ }^{235} \mathrm{U}: 235.0439 \mathrm{u},{ }^{140} \mathrm{Ce} ; 139.9054 \mathrm{u}$, ${ }^{94} \mathrm{Zr}: 93.9063 \mathrm{u} ; \mathrm{n}: 1.0086 \mathrm{u}$, Value of $c^2=931 \mathrm{MeV} / \mathrm{u}$.