Prior to hearing this, I would have bet just about any amount of money that aluminized steel had the same thermal conductivity as regular steel.
So would I, and after further thought, I don't think I believe the thermal conductivity number in the AK Steel product data sheet.
Here is why I say that - we can think of the problem in terms of conductive heat transfer:
Fourier's Law of Heat Conduction: q = k A (ΔT/s) 
q =heat transfer (W)
k = thermal conductivity of the material (W/m°K)
A = heat transfer area (m2
ΔT = temperature difference across the material (K)
s = material thickness (m)
This formula can be extended to calculate heat transfer across multiple materials in series:
q = (ΔT) / ((s1
A)) + (s2
A)) + ... + (sn
We can use these thermal conductivity values:
Low carbon steel = 64 W/m°K
Aluminum = 205 W/m°K
And these thicknesses provided by AK Steel (the extremes that should give the highest k):
Low carbon steel = 0.15in (0.000381m) – the thinnest they offer
Aluminum = 1.2mil (0.00003048m) – the thickest they offer
We’re calculating a constant k for the clad metal, so it doesn’t matter what you use for A or ΔT and long as you use the same values throughout the steps (see  below).
Insert the values into  and solve for q (remember there are two layers of Al – same thing in  when you figure s for the Al-steel-Al clad).
Rearrange  to solve for k = (qs)/(AΔT) 
Insert q from  and the other values into  and solve for k = 70.7 W/m°K which is a lot lower than 89 W/m°K. I just don’t see how you can get to 89 with any reasonable set of assumptions. It almost looks like they did a weighted average k which would not be correct.
Inserting  into  and simplifying yields
+ … + sn
) / ((s1
) + (s2
) + … + (sn
which can be used to calculate the thermal conductivity for any n-layered clad material.
Back to the question at hand, if you use a steel thickness of 0.2in, and aluminum thickness of 1.2mil, the increase in thermal conductivity over plain steel is about 0.5 W/m°K which is completely insignificant. The thicker the steel, the less the increase.
I guess this brings back to where we started. Sorry Ron.