Am I correct that the faster processor draws more power (and thus
dissipates more heat) under a computational load?
Not necessarily. There are two major components of power dissipation - static power (the power you burn when the chip is on) and dynamic/switching power (the power burned when the clock is running). While running the same chip at a higher frequency will result in more power dissipation, a chip may have a static power dissipation that is too high when combined with the faster clock rate to meet the bin requirements for the faster rating.
If so, is the power under computational load approximately
proportional to the rated/clocked frequency? In other words, inasmuch
as the one processor is clocked 8 percent faster than the other, does
it run about 8 percent hotter under load? Another way to ask the same
question is to ask: does each processor process about the same
quantity of data per unit of energy? or, if battery powered, can each
accomplish about as much before its battery dies?
For a given chip running identical calculations, the dynamic portion of the power consumption will be proportional to the clock frequency. The total power dissipation of the processor will increase a bit less than 8% for an 8% increase in clock frequency due to the static power dissipation.
When not under load, do the two processors idle equally cool, or are
there practical or theoretical factors that make the one idle cooler
than the other?
If you had two identical chips idling, the one with the lower clock frequency would dissipate less power. When the chips are idling, the static power becomes a much larger portion of the active power dissipation, so any differences there would be more noticeable.
Even if the processor's price were not determinate, might one prefer
the slower processor merely for the sake of cooler operation and
extended battery life?
Possibly, but again, you have a lot less of a guarantee that this would be the case. If you bought chips with different rated TDPs, then you could safely make this argument. Otherwise, you're at the mercy of the binning algorithm and the consistency of the manufacturer's process. Also, note that we're talking about power dissipation, not energy consumption. A faster processor may be able to complete a computationally heavy task faster, and switch to a low power idle mode sooner than a slower processor.
Would the answers differ for embedded processors?
Yes. The static power dissipation is most significant on the bleeding edge processes that Intel, TSMC, IBM, and Global Foundries use. Embedded processors are often optimized for low static power dissipation and use larger processes where static power dissipation is a much smaller portion of power dissipation. The variation at those larger process nodes is much less, so microcontrollers are much less susceptible to variation in power and frequency performance.
Best Answer
Internal Diodes can measure voltage by negative temp coefficient (NTC) and thus convert to a temperature. Many chips have overtemp protection (OTP), such as Smart switches and regulators.
Power dissipation is computed by Ohm's Law for static loads and switched charges with driver ESR losses for CMOS that vary with V and f and given in datsheets near the end of xxx pages.
Thermal design is spec'd with thermal resistance Rjc, junction to case and junction to ambient Rja by ['C/W] then heatink or force air cooling is like Ohm's law for thermal resistance. Heatsinks and interface thermal grease are then added in series, Rja (jcn to ambient) with convection air. Forced air can improve resistance from 10% to a few % proprotional to surface velocity.
An Intel CPU has a very low Rjc and then a forced air heatsink needs Rca of <0.1 to 0.2'C/W to keep cool, depending on budget $ and power [W] so that the sum of all thermal R's *Pd = temp rise above ambient.