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?

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:

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

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

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?​

Respuesta :

Otras preguntas

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?