This was my 400 level thermo 2 class, we calculated everything, inefficiency, convection, conduction.
From a pure heat capacity viewpoint, heating cooler air is very slightly easier (on the order of a few thousandths at most for realistic temperatures) [0]. Don't think it'll be noticeable for the average heating system, though.
> we calculated everything, inefficiency, convection, conduction.
I'd be curious to how those calculations affect things. If the heating element is the same temperature, I'd think heating cooler air would be faster because heat conduction is faster across a larger temperature differential and you'd get "faster" convective mixing for air with bigger temperature differences. Have to admit I'm not sure what inefficiencies might come into play, and I haven't taken classes at that level so I wouldn't be surprised if I missed something.
[0]: https://www.engineeringtoolbox.com/air-specific-heat-capacit...
So at least part of the efficiency comes from not restarting the fan repeatedly.
So the blower uses less than 3% of the energy used by the burner. Some of the waste heat from the blower motor probably helps heat the home, so in reality it's probably less than that.
While startup current is significant (for a very short time), starting the fan up in the morning and again in the evening after work doesn't add that much to the total power drain, or even cycling it on/off a few times an hour to maintain temperature. Starting wattage on that motor is around 1800W, but only lasts for a few seconds.
The rate of heat flow across an insulator is proportional to the temperature difference across the insulator, if I recall correctly.
Suppose you want the inside of you house to be at temperature Tin when occupied and the temperature outside is Tout.
The rate of heat flow across the boundary between inside and outside will be k|Tin-Tout|.
Imagine two identical houses.
In house #1 we keep it at a constant Tin. In house #2 we keep it at Tin while occupied but when empty we let it move toward Tout, turning on heating or cooling when it is going to be occupied again to get it back to Tin when the people return.
For each house, if we plot |Tin-Tout| over time, the total energy used over a given interval will be proportional to the area under that |Tin-Tout| curve.
For house #1, that curve is simply a horizontal straight line.
For house #2, that curve matches the curve of #1 while the house is occupied, but during the unoccupied times falls toward zero until it is time to get it back to Tin in anticipation of people returning home. It then rises back up to Tin.
The area under house #2's curve will be less than the area under house #1's curve.
Thus, in terms of energy required to produce the desired temperature curve, house #2 uses less than house #1.
There are inefficiencies in producing and applying that energy, though, and so getting a given amount of energy applied to heating or cooling the house from your heating or cooling system will take more energy, and that extra energy might depend on Tin or Tout, so it is possible that this might be a big enough effect to counter the savings from letting the unoccupied temperature move toward Tout.