Question

suppose that a new temperature scale has been devised on which the melting point of some compound (-102.8°c) and the boiling point of the same compound (82.2°c) are taken as 0° s and 100.° s, respectively, where s is the symbol for the new temperature scale. derive an equation relating a reading on this scale to a reading on the celsius scale. what would this thermometer read at 29.9°c? round your answer to 3 significant digits.

Explanation:

Step1: Find the relationship between the two scales

Let \(C\) be the Celsius temperature and \(S\) be the new - scale temperature.
We know two points: \((C_1=- 102.8,S_1 = 0)\) and \((C_2 = 82.2,S_2=100)\)
The general formula for a linear relationship is \(S=mC + b\)
First, find the slope \(m\): \(m=\frac{S_2 - S_1}{C_2 - C_1}=\frac{100 - 0}{82.2-(-102.8)}=\frac{100}{185}\)

Step2: Find the \(y\) - intercept \(b\)

Using the point \((C_1=-102.8,S_1 = 0)\) and \(S=mC + b\)
\(0=\frac{100}{185}\times(-102.8)+b\)
\(b=\frac{100\times102.8}{185}\)

The equation is \(S=\frac{100}{185}(C + 102.8)\)

Step3: Substitute \(C = 29.9\) into the equation

\(S=\frac{100}{185}(29.9+102.8)\)
\(S=\frac{100\times132.7}{185}\)
\(S=\frac{13270}{185}\approx71.7\)

Answer:

\(71.7\)