Answer
$a.\quad\Delta T_{C}=15^{\mathrm{o}}\mathrm{C}$
$b.\quad\Delta T=15K$
Work Step by Step
$a.$
Given that, from freezing to boiling points of water at normal atmospheric pressure, there are 100 degrees on the Celsius scale, $\Delta T_{C}=100$
while there are 180 in the Fahrenheit scale, $\Delta T_{\mathrm{F}}=180$, we can write
$\displaystyle \frac{ \Delta T_{C}}{ \Delta T_{F}}=\frac{100}{180}=\frac{5}{9},$
$\displaystyle \Delta T_{C}=\frac{5}{9}\Delta T_{F}$
(a difference in temperature of one degree Fahrenheit equals $(\displaystyle \frac{5}{9})^{o}C$
$\displaystyle \Delta T_{C}=\frac{5}{9}(27)=15$
$b.$
The Kelvin and Celsius scales have the property that a difference in temperature of $1^{\mathrm{o}}C$ equals a difference in temperature of $1K$.
$\Delta T=\Delta T_{C}=15$
