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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?

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  • $\begingroup$ 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. $\endgroup$ – Wojciech Mar 14 '14 at 13:53
  • $\begingroup$ @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. $\endgroup$ – David Z Mar 15 '14 at 22:07
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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.

The difference in values you are getting is completely due to the approximation that the solution manual uses. If you use the exact values, as you did, you will indeed get 1330. If you use the approximate values as the solution manual did, then you will indeed get 1375.

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