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

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

Step-by-step 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:

`((y-y_(1) ))/((y_(2) -y_(1) ))=((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