If I have a 120V @ 50Hz AC heater rated at 750W, and I run it using 120V DC, will the DC heat it faster, and why? (Assuming I could get a clean 120V DC power source).
When you say 120V @ 50Hz AC you are implicitly saying 120Vrms.
The RMS voltage is qualitatively defined as the voltage which will give the same resistive heating (averaged out over time) as a DC voltage of the same number. Therefore, by the definition of RMS, the heating will be the same because the RMS voltages are the same.
If you said 120Vpeak or something different, then things would be different.
This is in reference to a heater modeled only as a resistor. No extra real-world components like motors for fans.
For heating, the only thing that matters is active power. If the load is a resistor, active power is \$ V^2/R \$ with V being the RMS voltage.
The RMS value of 120V DC is 120V RMS, which will give the same active power in a resistor than 120V AC RMS.
AC power will be pulsed, but a 750W resistor will have enough thermal mass to smooth it out, so there is no difference.
On a purely theoretical level, the resistor should get to working temperature a (really) tiny bit faster with DC power. The reason is the Stefan–Boltzmann law which state that the radiative output is proportional to the fourth power of the temperature.
During the beginning of the heating phase, during the power peak of the AC current, the resistor will get hotter than it's DC counterpart at the same time. But it will loose much more energy during the "low power" part of the AC cycle. Fourth power is a really step function.
The effect will be really small, because the ripple in the temperature will be microscopic 1/100th of second is infinitesimally small compared to the time constant of the resistor (typically 10th of seconds), but that doesn't means that it doesn't exist.
But... this is only about the temperature of the resistor itself. In fact, the thermal energy dissipated in the room during the heating time will be higher. Two reason to this :
The first one is that, as stated earlier, instantaneous power dissipation will be higher on average.
The second is that real resistors are not perfect and generally act as Negative Temperature Coefficient resistors. So, by being slightly cooler, the AC one will have lower resistance and drain more current, so more power.
But keep in mind that this calculation is purely theoretical and that in real live, even with perfectly equivalent power source, the effect will certainly not be measurable.