The specific volume of mercury is given by the equation. V=specific volume
V(T) = V(0 C)*(1+(1.82*10^-4)T + (7.8*10^-9)T^2) in degrees celsius
Suppose you take an unlabeled mecury-in-glass- thermometer, mark 0.01C and 100C for the triple point and normal boiling point of water, and then place intermediate temperature points by linear interpolation. Because 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?

Question

The specific volume of mercury is given by the equation. V=specific volume
V(T) = V(0 C)*(1+(1.82*10^-4)T + (7.8*10^-9)T^2) in degrees celsius 
Suppose you take an unlabeled mecury-in-glass- thermometer, mark 0.01C and 100C for the triple point and normal boiling point of water, and then place intermediate temperature points by linear interpolation. Because 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?

joannablackwelder · Answer

@abb0t @nincompoop @taramgrant0543664 @chmvijay

joannablackwelder · Answer

@Cuanchi

aaronq · Answer

Use excel to plot this, find the largest distance from the line to the curve 

It'll look something like this, now just use some math