Question

A faulty thermometer reads 2°C when dipped in ice at 0°C and 95°C when dipped in steam at 100°C. What would this thermometer read if placed in water at room temperature at 18°C?​

Answer 1

Answer:

The read will be 20.9[C]$x=\frac{(y-y_{1} )}{(y_{2} -y_{1} )}*(x_{2}-x_{1})+y_{1}\\ x=\frac{(18-0 )}{(95 -0 )}*(100-2)+2\\\\x= 20.9[C]$

Explanation:

This is a problem related to linear interpolation, Linear interpolation consists of tracing a line through two known points y = r (x) and calculating the intermediate values according to this line.

The equation of a known line two points (x1, y1)and (x2, y2) = (2,0) (100,95) is:

$\frac{(y-y_{1} )}{(y_{2} -y_{1} )}=\frac{(x-y_{1} )}{(x_{2}-x_{1} )}$

If we clear y from the equation we have:

[tex]y=\frac{(x-y_{1})}{(x_{2}-x_{1} )}*(y_{2} -y_{1} )+y_{1}  \\replacing