In a curious exercise I found, I know that the relationship between ºC and ºF can be given by: $$F=\frac95C+32$$

I also know that the relationship between ºC and ºK can be given by: $$C=K-273,15$$

I'm curious, how can I get the relationship between ºK and ºF?

I've tried replacing, but just got stuck in a endless loop.

The choices I have are:

A) $K=\frac{9F-288}{5}+273,15$

B) $\frac{K-273.15}{9}= \frac{F-32}{5}$

C) $\frac{F-32}{9} = \frac{K-273.15}{5}$

D) $\frac{5F-150}{9}-273.15$

  • 1
    $\begingroup$ Just replace $C$ with what it is: $F = \frac 9 5 (K - 273.15) + 32$ $\endgroup$ – user61527 May 9 '14 at 22:32
  • $\begingroup$ Added some details to the original question. $\endgroup$ – ddrjm May 9 '14 at 22:47

Solve each of the two equations for $C$, then set the resulting expressions equal to each other, eliminating $C$: $$ C=\frac59(F-32)$$ $$C=K-273.15$$ therefore $$\frac59(F-32) = K-273.15$$ From this you can see that answer (C) is correct.


\begin{align*} F &=\frac95 \underbrace{C}_{ K-273.15}+32 \\ &=\frac95 \left( K-273.15 \right) + 32 \end{align*}

Now which of $A, B, C, D$ is equivalent to this?

  • $\begingroup$ I can't point the finger exactly. But C seems to sand out, but I can't point why. or how did 9 and 5 got to where they got. $\endgroup$ – ddrjm May 9 '14 at 22:58
  • 2
    $\begingroup$ @ddrjm check! Try solving c) for $F$, and see if it looks the same. $\endgroup$ – Omnomnomnom May 9 '14 at 22:59
  • $\begingroup$ @Omnomnomnom Ok, I did try to solve it, but the closest I managed was the expression in the answer without the $\frac95$ which I'm not sure it is the right answer. $\endgroup$ – ddrjm May 9 '14 at 23:05

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