# Comparing unknown temperature scales [closed]

Source: Principles of Physics by Resnick, Halliday, Walker. $9^{th}$ edition. Chapter 18. Problem 6.

On a linear $X$ temperature scale, water freezes at $-125.0^0X$ and boils at $360.0^0X$. On a linear $Y$ scale, water freezes at $-70.00^0Y$ and boils at $-30.00^0Y$. A temperature of $50.00^0Y$ corresponds to what temperature on the $X$ scale?

My approach:

I have a formula for such problems- $$\frac{X-(freezing \;point\; of\; water\;in\;X)}{(boiling\;point\;of\;water\;in\;X)-(freezing\;point\;of\;water\;in\;X)}=\frac{Y-(freezing \;point\; of\; water\;in\;Y)}{(boiling\;point\;of\;water\;in\;Y)-(freezing\;point\;of\;water\;in\;Y)}$$

So, $$\frac{X+125}{360+125}=\frac{Y+70}{-30+70}$$

Putting $Y=50.00^0$ in this equation, I get $X=1330.00^0$.

Approach of the solution manual:

Since the temperature scales are linear, we get $$Y=mX+c$$

From the given data, we can form two equtions and solving them simultaneously, we get $$m=0.08,\;\;c=-60$$

So, $$Y=0.08X-60$$

Putting $Y=50.00^0$, we get $X=1375.00^0$

Is there any error in my method?

• I solved the equations given in solution manual and got different results: m = 8/97 and c = -5790/97. With these values, the result is 1330. Seems like Kyle is right. – Wojciech Mar 14 '14 at 13:53
• @Tejas please keep in mind that posting some work and simply asking whether it's correct is not the sort of thing this site is for. – David Z Mar 15 '14 at 22:07

If you rearrange your equation to solve for $Y$, $$Y=(X+125)\frac{-30+70}{360+125}-70$$ This reduces to $$Y=\frac{8X-5790}{97}=0.08247X+59.6907\approx0.08X-60$$ which is about what the textbook obtains.