Question

At what temperature, both the Celsius and Fahrenheit scales are equal?

Answer 1

since, $F^o=C^o\times \frac{9}{5}+\mathrm{32....}\left(1\right)$

and ${}^{o}C=F^o-32\times \frac{5}{9}.....\left(2\right)$

Since $C^o=F^o$ as given in the state

Hence, ${}^{o}C=(C\times \frac{9}{5})+32$

${}^{o}C-\frac{9}{5}{C}^o=32$

$\Rightarrow$ $-{\frac{4}{5}}^o=32$

$C^o=-{40}^o.....\left(A\right)$

where as, ${}^{o}F=(F\times \frac{9}{5})+{32}^o$

$\Rightarrow$ $F^o-{\frac{9}{5}}^{o}F=32$

$F=32\times \frac{5}{9}=-{40}^o$