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What does 29.3 w/m*k mean?
Assume you have a huge cube of this material that is 1 meter on a side.
Apply heat to one side of the cube, and have a heat sink at a fixed temperature
on the opposite side of the cube.  We assume no heat is otherwise lost to the environment.
If 29.3 watts of power is applied, that side of the cube will (eventually) be one degree Kelvin
hotter than the heat sink.   

A change of one degree Kelvin is the same as a change of one degree Celsius,
so I will just use Celsius from here on out.


Assume your insulator is 0.025 mm thick, or 0.000025 meters.  (About a thousandth of an inch.)
The surface area of the IRF510 from which we conduct the heat is around 1 cm by 1 cm, or 0.0001 square meters.
You are getting 10 Watts of RF out of that single IRF510 at 50% efficiency, so it is dissipating 10 Watts as heat.
How hot will the IRF510 tab be if the heatsink can be kept down at 30 degrees Celsius with a fan?

We divide the thickness into the area and reduce those dimensions to a single value of 0.0001/0.000025 = 4 meters.
Not terribly intuitive, but that's all we need to know about the geometry to do this heat conductivity calculation.
As the area increases, the amount of heat that can be transferred increases proportionally.
But as the thickness is increased, the heat has a proportionally harder time moving through it.


The material will have a 1 C temp rise with 29.3 Watts applied and a 1 meter figure of merit for the geometry.
Our 10 Watts is less than their 29.3 Watts,  so that temp rise to be (1 degree C) * 10/29.3 = 0.341 degrees C
Our geometry has a large area for the thickness, so the temperature rise is even less: 0.341 / 4 = 0.085 K

So the IRF510 side of the insulator is 0.085 degrees Celsius hotter than the heatsink side.
And is at 30.085 degrees C.
That is less of a temperature difference than I expected, perhaps someone will see an error in my math?


Inside the IRF510 it's a different story.
Page 2 of the datasheet says thermal resistance from silicon junction to the tab of the IRF510 case 
is 3.5 degrees C per Watt, plus an additional 0.5 degrees C per Watt from tab to a well greased heat sink.
    
So the silicon die inside the IRF510 is at (10 Watts * (3.5+0.5)) + 30.085 = 70.085 degrees C.
The datasheet also states that the IRF510 will work OK with a die as hot as 175 C
(though it likely won't last very long).


The interface from IRF510 tab to the insulating material and from the insulating material to the heat sink
is surprisingly problematic, given the 0.5 degrees C per Watt figure from the IRF510 datasheet.
There will likely be another 0.5 degrees C per Watt temp rise between the insulator and the heat sink.
Without heat sink grease, these temperature differences will be considerably greater.
Using appropriate heat sink grease appears to be far more consequential than choosing the correct insulating material.