Now we're getting somewhere. If you're getting 50 at ambient (20 C?) then that's 50/1023*5 = 0.244 volts. The voltage divider you are forming has a gain of 1k/(1k+RT) where RT is your thermistor value. The output voltage from your assembly is then 5*1k/(1k+RT)=0.244 volts from which we can solve that RT=20k approximately.
Of course I could have probably suggested you measure that directly with a DMM
If that's true, then you don't have to do any interpolation. I'd replace the 1k resistor with a 10k-20k resistor (let's say 10k for now) then you have the equation:
analog voltage = 5*10k/(10k+RT)
from which you can solve for RT. Thermistors generally have well-known temperature-resistance properties that you can look up at various manufacturers. Given that you only need accuracy to 5 degrees though it would probably be easiest to just build a lookup table of analog voltage reading-->temperature. Use a commercial temperature gauge for the moment to build this lookup table (e.g., heat up your sensor and the temperature gauge at the same time and just take readings).
The Quick Shield
: breakout all 28 pins to quick-connect terminals