LG, I can't believe an educated person like you with a Dr. degree would argue something that is factual...you think I'm as gullible as some uneducated person does...I'm so disappointed in you!
Okay, let's talk about thermal resistance and the law of conservation of energy!
So what is the thermal resistance that you're trying to base your argument on? It's an object's resistance to heat conductance. Yes, some materials conduct heat faster, some slower, just like the Raytheon graphite plate 6080 takes longer to warm up, and the regular plate 6080 warms up much faster. But have you considered the law of conservation of energy?!!!
As long as the same 2.5A heater current is drawn, and the same amount of plate current and grid current (yes, I intentionally list grid current here just in case someone will be nit-picky on the subject! I will elaborate later) are going through both the graphite-plate and regular-plate tubes, the same amount of total heat will be generated, which can be calculated by the following equations, Total power consumed = heater power consumption + plate power consumption + grid power consumption = I_heater x V_heater + Ip x Vp + Ig x Vg. The 6 parameters on the right hand side of the equation will be the same for both Ray graphite-plate 6080 and regular-plate 6080, hence the total power consumed will be the same for both tubes ( and total heat generated = total power consumed - power consumed by the load on tube, will also be same since load will be the same)! Yes, Ray graphite-plate 6080 will reach normal thermal operating temperature slower than regular 6080 does because of its higher thermal resistance, but it will eventually get there (some folks say graphite-plate 6080 needs 10-30mins to warm up), and what's more important, the total heater generated by both 6080 tubes will be the SAME, that's defined by the equations I listed above and the law of conservation of energy.
Now, let me elaborate why I intentionally listed grid current there previously. Yes, most tubes run on negative grid voltage on their grids, hence their grid current will be close to 0, but let's not forget that some tubes do run on positive grid voltages, so there will be measurable grid current on them! And even under the negative grid voltage mode, some electrons will go astray and hit the grid by chance, hence will generate grid current theoretically (though this current is almost 0 and is negligible). So if I did not list grid current there, some guy will be nit-picky on it. And even I did include the grid current there, someone is still trying to be nit-picky on it