Do you feel the radiated heat of an explosion before you see it? Do the infra-red rays reach us before the visible light does. I understand they have different wavelength but I don't know much else. Please answer the question if the explosion was in air and if it was in a vaccum.
 A: The index of refraction for gases is commonly discussed using:
$$ \eta \equiv (n-1)\times 10^6 $$
The general value for air is $\eta=273$; however, if you need to consider dispersion, the accepted formula is:
$$(n-1) = \frac{0.05792105}{238.0185−\lambda^{−2}} + \frac{0.00167917}{57.362−\lambda^{−2}}$$
where $\lambda$ is the wavelength in microns. That looks like:

so the difference between visible (0.45 micron)and IR (> 1 micron) is negligble.
The propagation velocity is:
$$ v = \frac c n = \frac c {1+\eta/10^6} \approx c(1-\frac{\eta}{10^6}) $$
With $\delta\eta \approx 8$ between blue light and IR, the time delta is 1 part in 125,000, or one nanosecond per 24 miles.
One nanosecond is one foot of propagation. Given that the fireball is much larger than 1 foot at 24 miles (or any scale you choose), the rise time of the light/heat which spans at least the distance from the observer to each point on the surface, it's not measurable.
A: You would see the light and feel the heat from it at the same time, if it were an ordinary chemical explosion. This would be true in air as well as in a vacuum.
The situation is a little more complex in the case of a nuclear explosion. These exhibit two closely-timed flashes of light and heat. This is because after the initial explosion and flash of the bomb, the fireball that quickly expands outwards is so hot that it is opaque to light and the rest of the energy is trapped inside the fireball for a very short time. Once the fireball expands a bit more and cools off a bit, the trapped energy inside can freely radiate outwards again and the second flash (which is much longer in duration than the first one) occurs.     
