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For instance, I have the following problem:

convert:

$\frac{{m^2}^{\circ}C}{W}$ to $\frac{ ^{\circ} F * ft^{2} * hr}{Btu}$

Take $x \frac{{m^2}^{\circ}C}{W}$ for instance, how would I convert it to $y \frac{{m^2}^{\circ}F}{W}$? The reason that I am confused is that there is no conversion factor, to my knowledge, that I can multiply x by to get y.

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  • $\begingroup$ The same as any other time: if $T=x$ in Celsius then $T=1.8x+32$ in Fahrenheit. $\endgroup$ – Ian Oct 1 '15 at 21:21
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    $\begingroup$ @Ian, I don't think so, probably just the ratio of the size of a degree in each scale $\endgroup$ – MPW Oct 1 '15 at 21:31
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Use just the stretching factor I suppose.

$^\circ C$ and $^\circ F$ are used for a temperature range in this formulas.
So just use $5/9$ or $9/5$ without any shift.

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  • $\begingroup$ Makes sense, thank you. $\endgroup$ – FabulousGlobe Oct 1 '15 at 21:31
  • $\begingroup$ @FabulousGlobe Then you can mark it as an answer. After waiting maybe an hour, maybe someone has an even better explanation. =) $\endgroup$ – Antitheos Oct 1 '15 at 21:32
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    $\begingroup$ I was going to, but then it told me that I had to wait 15 minutes before marking an answer. Its marked now. $\endgroup$ – FabulousGlobe Oct 1 '15 at 21:36
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Notice, The temperature scales in $^\circ C$ & $^\circ F$ are co-related as follows $$\frac{^\circ F-32}{9}=\frac{^\circ C}{5}\implies ^\circ F=\frac{9({^\circ}C)+160}{5}$$

Hence, we can convert from $x\frac{m^2 {^\circ}C}{W}$ to $y\frac{m^2 {^\circ}F}{W}$ as follows $$\color{red}{y}=\color{red}{\left(\frac{9x+160}{5}\right)}\ \ \color{blue}{\frac{m^2 \ {^\circ}F}{W}}$$

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$\dfrac{{m^2}^{\circ}C}{W}$ to $\dfrac{ ^{\circ} F * ft^{2} * hr}{Btu}$

Btu/hr to W = 0.29307107 = X

Multiply existing value by $$ \dfrac{3.28 ^2 1.8}{X} $$

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