Answer:
ΔTmin = 3.72 °C
Explanation:
With a 16-bit ADC, you get a resolution of [tex]2^{16}=65536[/tex] steps. This means that the ADC will divide the maximum 10V input into 65536 steps:
ΔVmin = 10V / 65536 = 152.59μV
Using the thermocouple sensitiviy we can calculate the smallest temperature change that 152.59μV represents on the ADC:
[tex]\Delta Tmin = \frac{\Delta Vmin}{41 \mu V/C}= 3.72 C[/tex]