I don't think there's anything wrong with my math Richard.
You wrote:
Assuming that the softpot is 10K as advertised, putting 5V across it will result in 0.0005 amps (0.5 mA) which equals 0.0025 watts (2.5 mW)
I wrote:
With a supply voltage of 5V, power would be as low as 2.5mW for a 10k resistor
If you want to expose a 10k resistor to 0.5W you will need to feed it with just above 70V. Current is then about 7mA (70/10k) and power (70V*7mA) is 0.49W.
I'm merely suggesting that the power requirement stays fixed and that the sensor requires "heating" (0.5W) before it will reach its advertized characteristic. When "heating" the sensor however there is a big difference between 7mA and 100mA.