Answer
$a.\quad-18.0\mathrm{F}^{\mathrm{o}}$
$b.\quad-10.0\mathrm{C}^{\mathrm{o}}$
Work Step by Step
The temperature interval from freezing to boiling point of water is,
for the Celsius scale, $\Delta T_{\mathrm{C}}=100\mathrm{C}^{\mathrm{o}}$
and for the Fahrenheit scale$, \Delta T_{\mathrm{F}}=180\mathrm{F}^{\mathrm{o}}$.
$\displaystyle \frac{\Delta T_{\mathrm{C}}}{\Delta T_{\mathrm{F}}}=\frac{100}{180}=\frac{5}{9},$
given $\Delta T_{\mathrm{K}} =\Delta T_{\mathrm{C}}=-10\mathrm{C}^{\mathrm{o}}$
$ a.\quad$
$\displaystyle \frac{-10.0\mathrm{C}^{\mathrm{o}}}{\Delta T_{\mathrm{F}}}=\frac{5}{9},$
$\displaystyle \Delta T_{\mathrm{F}}=\frac{9}{5}(-10.0\mathrm{C}^{\mathrm{o}})=-18.0\mathrm{F}^{\mathrm{o}}$
$ b.\quad$
$\Delta T_{\mathrm{C}}=\Delta T_{\mathrm{K}}=-10.0\mathrm{C}^{\mathrm{o}}$
