Answer
$a.\quad-40^{\mathrm{o}}\mathrm{C}=-40^{\mathrm{o}}\mathrm{F}$.
$b.\quad 575 \mathrm{K}=575^{\mathrm{o}}\mathrm{F}$
Work Step by Step
Conversion equations$: \left[\begin{array}{ll}
T_{\mathrm{F}}=\frac{9}{5}T_{\mathrm{C}}+32^{\mathrm{o}} & 17-1\\
T_{\mathrm{C}}=\frac{5}{9}(T_{\mathrm{F}}-32^{\mathrm{o}}) & 17-2\\
T_{\mathrm{K}}=T_{\mathrm{C}}+273.15 & 17-3
\end{array}\right]$
$ a.\quad$
$T_{\mathrm{F}}=\displaystyle \frac{9}{5}T_{\mathrm{C}}+32^{\mathrm{o}}$,\yen qquad ...setting $T_{C}=T_{F}$
$T_{\mathrm{C}}=\displaystyle \frac{9}{5}T_{\mathrm{C}}+32^{\mathrm{o}}\qquad/\times 5$
$5T_{C}=9T_{\mathrm{C}}+160^{\mathrm{o}}$
$-4T_{C}=160^{\mathrm{o}}$
$T_{C}=-40^{\mathrm{o}}\mathrm{C}=-40^{\mathrm{o}}\mathrm{F}$.
$ b.\quad$
$ T_{\mathrm{K}}=T_{\mathrm{C}}+273.15\qquad$ ... use 17-2
$T_{\mathrm{K}}=\displaystyle \frac{5}{9}(T_{\mathrm{F}}-32^{\mathrm{o}})+273.15\qquad $...setting $T_{K}=T_{F}$
$\displaystyle \mathrm{T}=\frac{5}{9}\mathrm{T}-\frac{5}{9}\cdot 32+273.15$
$\displaystyle \frac{4}{9}\mathrm{T}=-\frac{5}{9}\cdot 32+273.15$
$\displaystyle \mathrm{T}=\frac{9}{4}(-\frac{5}{9}\cdot 32+273.15)=575$
$575 \mathrm{K}=575^{\mathrm{o}}\mathrm{F}$
