Temperature Conversions: Jekyll and Hyde Scale Comparison

[Qube]

Homework Statement

On a new Jekyll temperature scale, water freezes at 17 degree(s) J and boils at 97 degree(s) J. On another new temperature scale, the Hyde scale, water freezes at 0 degree(s) oH and boils at 120 degree(s) oH. If methyl alcohol boils at 84 oH, what is its boiling point on the Jekyll scale?

Homework Equations

Jekyll has a FP 17 degrees higher than the Hyde scale.

The Attempt at a Solution

Alright. Freezing point on Jekyll is 17 degrees higher than it is on Hyde.

Therefore, any formula relating J and H must have the general form of:

J(H) = H + 17, where H = 0.

The general formula is:

J(H) = H[(97-17)/120] + 17

Checking my work:

H = 0, J = 17. Correct.

H = 120. J = 97. Correct.

What are some good ways of tackling this problem? Why does subtracting two points on the Jekyll scale divided by subtracting the two points on the Hyde scale multiplied by H work?

 

[DrClaude]

The best way is to tackle this mathematically.

You are looking for a formula of the form
$$
J(H) = a H + b
$$
(i.e., assuming that the two scales are linear).

You then have
$$
J(0) = 17 = a \times 0 +b
$$
and
$$
J(120) = 97 = a \times 120 + b
$$
or, in other words, two equations with two unknowns. The first is easily solved for $b = 17$, which allows you to solve the second for $a$.