With a 24 volt supply, the current thru the limb of the bridge containing the thermistor is: -
- Nearly 11mA at 0C
- Nearly 19mA at 50C
At 0C this produces a power in the thermistor of 132mW and 63mW at 50C. To me this is a problem. The self-heating of the thermistor will create a significant measurement error at 0C compared to 50C. I'd use an instrumentation amplifier and keep the excitation levels much lower to prevent significant self-heating errors: -
The picture above shows the AD620 powered from +5V and 0V but a better power regime (for the OP) is +/- 15V with 0V connected to pin 5 (the reference input). The excitation voltage can be much lower (say 2.5 volts) and this means the self-heating effects are 1.5mW at 0C and 0.7mW at 50C i.e. trivial.
However, there is another potential problem of a single thermistor in a Wheatstone bridge and that is linearity. You may design a circuit that produces the correct voltages at the extremes of temperature you require but, between these two points, there is a non-linear mapping between voltage and temperature over and above the basic non-linearity of the thermistor's resistance change with temperature.
It might seem intuitive to place 2 thermistors at opposite sides of the bridge but, with voltage excitation, this won't improve linearity (but does double sensitivity). Here is a good document by ADI explaining all of this (called Bridge circuits by Walt Kester): -
However, if you adopted a current driven bridge configuration then you could use two thermistors and get perfect linearity: -
So, rounding up, try to avoid self-heating effects and, for greater linearity, use two elements in opposing faces of the bridge with the bridge excitation being a constant current like the one below: -
The important voltage to keep constant in the above diagram is that across R1 and, for high accuracy a precision shunt voltage reference would be used in place of R1.