Suppose that a new temperature scale has been devised on which the melting point of ethanol \(\left(-117.3^{\circ} \mathrm{C}\right)\) and the boiling point of ethanol \(\left(78.3^{\circ} \mathrm{C}\right)\) are taken as \(0^{\circ} \mathrm{S}\) and \(100^{\circ} \mathrm{S}\), respectively, where \(\mathrm{S}\) 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 \(25^{\circ} \mathrm{C} ?\)

On the Celsius scale, the known points are -117.3°C and 78.3°C. Corresponding to these temperatures on the new scale are 0°\(S\) and 100°\(S\) respectively.

considering the points on the Celsius scale as \(x\), and on the \(S\) scale as \(y\), a linear equation can be established. The equation has the linear form \(y = mx + b\), where \(m\) is the slope, \(b\) is the y-intercept, \(x\) is the temperature in Celsius and \(y\) is the temperature in the \(S\) scale. The slope \(m\) is obtained by the formula \((y2 - y1) / (x2 - x1)\) where \(x1 = -117.3, y1 = 0, x2 = 78.3, y2 = 100\). Substituting these values yields \(m = (100 - 0) / (78.3 - (-117.3)) = 100 / 195.6 \approx 0.511\). Now to find \(b\) (which corresponds to the intercept where \(x = 0\)), we can use one of the known points and the slope obtained. Let's use \(x1, y1\), so \(b = y1 - m * x1\), which gives \(b = 0 - 0.511 * -117.3 \approx 60\). This allows us to describe the linear equation as \(y = 0.511x + 60\).

Use the derived formula to compute the temperature for 25°C in the \(S\) scale. It hurts, \(y = 0.511 * 25 + 60), resulting in an approximate value of 72.75°\(S\).