I recently read the following example in a textbook. Suppose your furnace is sized to heat your house to the target temperature (the temperature set on the thermostat) when the outside temperature is between 0 and 65 degrees F. Suppose it’s -10 degrees F outside, and your furnace is too small for the task. The textbook said the indoor temperature would equilibrate at 10 degrees below the target temperature.
But what if you only set your furnace to 55 degrees F? Will it still only reach 45 degrees F inside the house, as I inferred from the textbook? Would the furnace simply stay on longer to compensate for the lower outside temperature, so eventually the house would be at 55 degrees F?
Disregarding wind effects, heat loss from the house (measured in energy lost per unit time) is proportional to the difference between indoor and outdoor temperature. The furnace output (also measured in energy per unit time) is apparently powerful enough to just offset that loss and sustain the house at 65 degrees above the outside temperature, whatever that outside temperature may be.
So if your furnace (running continuously) is capable of keeping the house at 55 degrees when it’s -10F out, then when you set the thermostat at (or above) 55 degrees, the furnace will keep the house at 55 degrees by running continuously.
If the house is at some lower temperature to begin with, yes, it will take longer to reach 55 degrees than it would if it were warmer outside. In fact, if it’s -10F outside, the math predicts that the indoor temp will asymptotically approach 55F.
The explanation by Machine Elf leaves out a critical point with modern multistage furnaces. Set to 65 the high stage maximum output would run the whole time to reach 55. Setting the thermostat for 55 in you example would result in never reaching 55. The high output stage would cut out a couple degrees before 55 and only the low output stage would be active which would be incapable of raising it to 55. On an old one stage furnace Machine Elf had it right.