Why do I have to perform a second ReplaceAll in this InterpolatingFunction?

Question

I have an InterpolatingFunction that I'd like to plot, and I wonder why I have to perform a ReplaceAll on the following function:

Plot[Evaluate[IIz[r, t] /. soln] /. t -> 5, {r, 0, 490}, PlotRange -> {{0, 
 490}, {.000015, -.00035}}, AxesLabel -> Automatic]`

soln is the solution to a differential diffusion equation given by Mathematica (v. 9.0.1) as

{{IIz[r,t]->InterpolatingFunction[{{1.*10^-8,490.5},{0.,10.}},<>][r,t]}}[r, 
 t]}} 

When I plot the 3D solution, there is no problem in using

Plot3D[Evaluate[IIz[r, t] /. soln], {r, 0, 490}, {t, 0, 10}, 
 PlotRange -> 
 {{0, 490}, {0, 10}, {.000015, -.00035}}, 
 AxesLabel -> Automatic] 

and I would think that all I'd need to do to plot this in terms of a fixed t would be to replace t with a constant like so:

Plot[Evaluate[IIz[r, 5] /. soln] , {r, 0, 490}, 
 PlotRange -> {{0, 490}, 
 {.000015, -.00035}}, AxesLabel -> Automatic]`

However, when I input that, I get a blank plot… What gives?

For the original differential equation, see this question.

Answer 1

It will work as you expect if you get the solution in a form like

IIz ->InterpolatingFunction[...

and not like

IIz[r,t] -> ...

since in the second case you have to have the explicit t, not 5, in order for the match to succeed.