Answer
$a.\quad 127^{\mathrm{o}}\mathrm{C},\ 260^{\mathrm{o}}\mathrm{F}$
$b.\quad -178^{\mathrm{o}}\mathrm{C}$ , $ -289^{\mathrm{o}} \mathrm{F}$
$c.\quad 1.55\times 10^{7}\ {}^{\mathrm{o}}\mathrm{C},\ 2.79\times 10^{7} \ {}^{\mathrm{o}}\mathrm{F}$
Work Step by Step
Conversion formulas: $\left[\begin{array}{ll}
T_{\mathrm{F}}=\frac{9}{5}T_{\mathrm{C}}+32^{\mathrm{o}} & 17-1\\
T_{\mathrm{C}}=\frac{5}{9}(T_{\mathrm{F}}-32^{\mathrm{o}}) & 17-2\\
T_{\mathrm{K}}=T_{\mathrm{C}}+273.15 & 17-3
\end{array}\right]$
$ a.\quad$
Use eq.17-3
$T_{\mathrm{C}}=400-273.15=127^{\mathrm{o}}\mathrm{C}$
then eq. 17-1
$T_{\mathrm{F}}=\displaystyle \frac{5}{9}(126.85)+32=260^{\mathrm{o}}\mathrm{F}$
$b.\quad $
$T_{\mathrm{C}}=95-273.15=-178^{\mathrm{o}}\mathrm{C}$
$T_{\mathrm{F}}=\displaystyle \frac{5}{9}(-178.15)+32=-289^{\mathrm{o}}\mathrm{F} $
$ c.\quad$
$T_{\mathrm{C}}=1.55\times 10^{7}-273.15=1.55\times 10^{7}\ {}^{\mathrm{o}}\mathrm{C}$
$T_{\mathrm{F}}=\displaystyle \frac{9}{5}(1.55\times 10^{7})+32=2.79\times 10^{7}\ {}^{\mathrm{o}}\mathrm{F}$
