I don't know how you're processing the digital data, but if you average over any integer number of cycles of the noise, you will essentially eliminate it.

Clarification: assuming the noise is fairly periodic at 200 Hz, then the period is 1/200s = 5ms. You're sampling at 10kHz, so there will be 50 samples in every 5ms cycle of noise. By averaging 50 samples (or 100, or 150, or 200, ...) you will average out the noise. Obviously, this lowers your effective sample rate by 50 (or 100, or 150, ...) but you still have not justified sampling temperature at 10kHz. :)

Better yet, find the source of the 200hZ noise and eliminate it.

I don't know how you're processing the digital data, but if you average over any integer number of cycles of the noise, you will essentially eliminate it.

Clarification: assuming the noise is fairly periodic at 200 Hz, then the period is 1/200s = 5ms. You're sampling at 10kHz, so there will be 50 samples in every 5ms cycle of noise. By averaging 50 samples (or 100, or 150, or 200, ...) you will average out the noise. Obviously, this lowers your effective sample rate by 50 (or 100, or 150, ...) but you still have not justified sampling temperature at 10kHz. :)

Better yet, find the source of the 200Hz noise and eliminate it.