I'd say you have a precision error. The microcontroller reads the voltage from the potentiometer with a reference to something, usually Vcc but it can also be referencing ground, or some other internal voltage.
If your voltage changes, the value converted by the ADC will change as well. Voltage changes can come about from many things, including but not limited to ambient temperature changes, power supply load, component temperatures, etc.
If everything is fairly consistent, you may just have a rounding error. For example prior to reset you might get an arbitrary value of 10.4 (for example), but after reset or a second ADC read you get 10.5. The latter would be rounded up to 11 if using integer math.
You might want to modify your code to have fewer or broader "steps" to convert to time, so that input values do not necessarily have to be as precise.
Short answer: what you want can not be done this way.
As you assumed, a potentiometer itself forms a resistor divider. The two parts (let's call them upper and lower) divide the voltage proportionally to the two resistor values.
Now hen you connect something (in your case the heater) to the output of the potentiometer you in effect connect it in parallel to the lower resistor. Two resistors in parallel have a total resistance of 1/( 1/R1 + 1/R2). When one of the parallel resistors has a very low value this means that you can ignore the other one.
This is what happens in your setup: in effect you have a resistor divider consisting of
Your heater has a rather low resistance, so when you turn your potentiometer 'up' almost nothing happens, until the very end, when the resistance of the upper part of the potentiometer becomes comparable to that of the heater.
I don't know the resistance of the heater, but to create any heat it must be low. Combined with 24V this will result in a large current, for which your potentiometer is probably not designed. Don't be surprised when you see smoke coming out of your potentiometer.
Without switching, there is no way to do what you want effectively (that is, without wasting around 9 times the heat you get in your heater in the rest of the circuit). A simple solution using Pulse Width Modulation (PWM) could consist 555 chip circuit and a switching series transistor or FET.
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Based on new information:
You have a 24V DC source, a 10k potmeter, and you want it to produce 0 .. 2.5 V DC output? That's a very different ballgame, and an easy one.
The trick is not to reduce the output of the potmeter (because that suffers from the same load problem as I described above, although somewhat less because the resistors involved have a higher value), but the input. You can use the potentiometer itself as the lower resistor, you only need to add a top resistor (between the potentiometer and the 24V source).
The value is easy to calculate. You want 10k (your potentiometer) to get 2.5V. That leaves 24 - 2.5 = 21.5 for the top resistor. Each 1k drops 0.25V, hence 21.5V requires 21.5 / 0.25 = 86kOhm. 82k is a standard value.
Best Answer
simulate this circuit – Schematic created using CircuitLab
If the bottom part of the pot is 7.8K then the top should be 10K-7.8K = 2.2K.
For the feedback voltage to equal the reference (2.495V), the voltage at the top of the divider would have to be (10/7.8)*2.495V = 3.20V.