Question

Chapter 18, Problem 006 On a linear X temperature scale, water freezes at -148.0ox and boils at 316.0oX. On a linear Y temperature scale, water freezes at -89.00°Y and boils at -26.00°Y. A temperature of 42.00°Y corresponds to what temperature on the X scale? Number Units the tolerance is +/-2%

Answer 1

The following is the general relation between 2 temperature scales X and Y

$\frac{y_B-y_F}{x_B-x_F}=\frac{y-y_F}{x-x_F}$

where $y_B\textup{ and }y_F$ are boiling and freezing points on Y scale and similarly $x_B\textup{ and }x_F$ are boiling and freezing points on X scale. This is obtained by equating the slope of FG and FB.

For our problem

$x_B=316,\ x_F=-148,\ y_B=-26,\ y_F=-89,\ y=42,\ x=?$

Insert the values

$\frac{63}{464}=\frac{131}{x+148}$

$x=131\times\frac{464}{63}-148=816.8$