Answer to Question #119111 in Molecular Physics | Thermodynamics for Thandeka

Question

On the Rankine temperature scale, which is sometimes used in engineering applications, the freezing point of water is at 491.67 °R and the boiling point of water is at 671.67 °R.
(i) Determine a relationship (a mathematical equation) between the Rankine and Fahrenheit temperature scales. (ii) A healthy person has an oral temperature of 98.6F. Calculate what this reading would be on the Rankine temperature scale.

Answer 1

(i) We know that on the Fahrenheit scale the freezing point of water is 32 F and boiling point of water is 212 F (see https://en.wikipedia.org/wiki/Fahrenheit). The difference between temperatures is $212-32 = 180F.$ The difference between analogous temperatures on the Rankine scale is

$671.67-491.67 = 180 R.$ Therefore 1°R = 1°F. So we should take into account only the difference between the initial points of scales.

0°F = 491.67 -32 =459.67°R. Therefore, t(°F) = t(°R) - 459.67 and t(°R) = t(°F) + 459.67.

(ii) If a person has temperature 98.6°F, then on the Rankine scale his temperature will be

t(°R) = t(°F) + 459.67 = 98.6 + 459.67 = 558.27°R