BZZZT! Reality check.
Radiation heat transfer q= (emissivity)(Stephan Boltzman constant)(T of hot item^4 minus T of cold item^4)(area)
The constant is 5.67E-8, so if you assume a 100C difference between the hot intake and the cold surroundings (and I think that is greatly exaggerated) and that the area is 1 meter^2, then q = around 80 W, depending on whether you pick polished or not.
Convective heat transfer (wind) q= (convective heat transfer coefficient)(area)(T of hot item - T of cold item cold)
The convective heat transfer coefficient of air is between 10 and 100. Take the same difference in temperatures and the same area, and the q = 10,000 W.
Note that my exaggerated temperature difference favors radiation, since it uses temperatures to the 4th power. Even so, the radiation heat transfer amount is 1% of the convective. Plus, we are talking about an engine part that is barely connected to the engine (not like carburetted manifolds were) and is probably gaining heat from radiation of the hot engine underneath it. In that case, a polished surface probably keeps it cooler.
All the formulae and constants are in the link provided earlier (engineeringtoolbox.com.) Unless there is a lab rat in your chemical plant that has some data, I think that for all practical purposes, a polished intake won't change power output.