The specific volume of mercury is given by the equation:
V(T) = V(0 degree C)x(1+1.82x10^-4T + 7.8x10^-9T), T in degree C
Suppose that you take an unlabeled mercury in glass thermometer, mark .01 degree C and 100 Degree C for the triple point and normal boiling point of water, and then place intermediate temperature points by linear interpolation. Because of the non-linear expansion of mercury, your temperature scale will differ a little from the true Celsius scale. What is the maximum error of the linearly marked thermometer? What other source of error do you need to consider if you are interested in constructing very accurate thermometers?