How it does look like a basis of $L^2(0,T;X)$?

Question

The space $L^2((0,a)\times(0,b))$ admits an orthogonal basis of the form $(\phi_n^1,\phi_m^2)_{n,m\geq1}$ where $\phi_n^1(x)=sin(\frac{n\pi x}{a})$, $\phi_m^1(x)=sin(\frac{m\pi x}{b})$. My question is the following: What about the space $L^2(0,T;X)$?. It is clear that for $X=L^2(0,b)$ we fall in the first case. If $X$ a Hilbert space which admits $(w_n)_{n\geq1}$ as a basis, is it correct that the family $(sin(\frac{n\pi t}{a})(w_m))_{n,m\geq1}$ is an orthogonal basis in $L^2(0,T;X)$?