1
$\begingroup$

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

$\endgroup$
  • 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
1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

\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?

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.