Answer to Question #130404 in Atomic and Nuclear Physics for vishnu

Question

consider a linear temperature scale x have -75.5º X of boiling point of water and -190ºX freezing point. what is the temperature of 2.00ºC on the X scale.

Answer 1

Let's consider X as a new temperature in linear scale and t - real temperature. Since the scale is linear the only possible connection between X and t is $X = at+b$, where a and b are constants. In is enough to know 2 points to completely define this equation. We know that freezing point for water is $t = 0^\circ C$ and $X = -190^\circ$. Putting this into the equation, we have

$-190 = a \cdot 0 + b \Rightarrow b= - 190$

Also, we know that for boiling point $t=100^\circ C, X = -75.5^\circ.$

$-75.5 = a \cdot 100 - 190 \Rightarrow a = 1.145$

So, $X(t) = 1.145t - 190$.

$X(2.0) = 1.145 \cdot 2 - 190 = -187.71^\circ$.

Answer: $-187.71^\circ$ .