If numbers make it obvious, where are they then? Show us if not peer-reviewed research, then at least something credible. Both wrt to dissipation into mb and amount of heat dissipated in the interconnect, which becomes separated by the cobalt gasket from the cooler (or are you [dis]missing the latter factor entirely?). Tell us, after all, something more specific about your own research you mentioned than very general words, as that would be interesting indeed.
What really makes it obvious though and indeed doesn't take measurements to realize -- as I have already pointed out at RWT -- is that motherboard is in fact a 2nd cooler attached to the cpu from the other side. Pretty lowsy, but still a cooler (the lower the power dissipated by the cpu, the higher the effect). It's enough to get hold of ir cam and take it on the backside of mb to see the picture quite clearly.
And what's probably more important here, is the combination of good thermal conductivity and thermal capacitance formed by Cu interconnect in the chip, its package and mb wires coming to the socket. It's this combo (with some contribution from silicon) which allows to absorb short bursts of Icc exceeding max average Icc (sustainable in long term). Use of poor thermal conductor in the IC violates function of both mechanisms.
Since -- as I said -- these thermal analyses were done on ASICs, which are by definition custom devices for a particular customer, it should be obvious that I'm not going to publish any detailed results. But as I said the overall numbers are pretty obvious, so let me give you a few clues...
Silicon substrate thermal conductivity is 200W/mK and area is 100% of chip area, so thermal resistance from circuits down through substrate to back of die is extremely low. Copper thermal conductivity is higher (400W/mK) but effective area is a tiny fraction, especially for vias, so thermal resistance to top metal is much higher than to die backside, especially when lateral flow through metal is also considered. Then there are solder bumps (60W/mk) which only cover ~20% of the chip. Then the heat has to get through the package substrate (small copper area of unfilled vias) to the package surface, then down through the solder bumps to the PCB, then get into the PCB planes and spread out through them. If a socket is used (most CPUs) then the thermal resistance of this is much higher than solder balls since the contact area to the balls is tiny.
Take all this into account together with thermal resistances of TIM1 (die to lid), lid/heat spreader, TIM2 (lid to heatsink) and compare the heat dissipated through the two paths, and invariably >90% goes out through the back of the die into the heatsink, as any thermal simulation (Icepak or similar) will show -- and the figure is typically >95% with a low thermal resistance package design like a CPU, especially if socketed. So the cobalt "thermal gasket" can only affect the ~5% that flows out of the top of the chip, and since all the thick metal layers are still Cu (and there are lots of other resistances in series) the difference is likely to be 1%-2%.
You're correct that an IR thermometer will show the back of the PCB under the chip is hot, but this shows temperature not heat flux -- it can be as hot as you want but this doesn't show any significant heat is being dissipated by it, it's like the difference between voltage and current.
So if you still don't believe me, go and build yourself a thermal model using realistic figures for all the above factors and see how ineffective heat dissipation up through the metal stack is -- and then you'll also see how Co makes almost no difference to overall thermal resistance compared to Cu, maybe 1 or 2 percent. I'm not going to do it for you, you're the one claiming this "Co thermal gasket effect" exists, so go and do the work to prove it.
[hint: it won't...]
As far as short bursts of high current are concerned the thermal mass of the Cu is negligible (a few percent) compared to the thermal mass of the silicon substrate, which in turn is much smaller than the thermal mass of the package lid, which in turn is much smaller than the thermal mass of the heatsink. In response to a high power pulse you get a series of cascaded time constants, but on a timescale of a second or so it's the heatsink that dominates, and it's the thermal mass of this that largely sets short-term overload capability.
Again, you can simulate all this if you don't believe me. I have on multiple occasions, which is why I can say that your ideas don't reflect reality.
By the way, you do realise that all advanced high-power chips are "flip-chips" with the circuits (and BEOL interconnect) facing down towards the PCB, and the back of the die facing up towards the package lid and heatsink, so your "thermal gasket" is under the chip (in the high thermal resistance path) not between the chip and the heatsink (the low thermal resistance path)?