Answer
For $ T=10^{\circ} \mathrm{C}$: $\mu=1.765 \times 10^{-5} \frac{\mathrm{N} \cdot 5}{\mathrm{~m}^2}$
For $ T=90^{\circ} \mathrm{C}$: $\mu=2.14 \times 10^{-5} \frac{\mathrm{N} \cdot 5}{\mathrm{~m}^2}$
Work Step by Step
$$
\begin{gathered}
\mu=\frac{C T^{\frac{3}{2}}}{T+S}=\frac{\left(1.458 \times 10^{-6}\right) T^{\frac{3}{2}}}{T+110.4 \mathrm{~K}} \\
\text { For } T=10^{\circ} \mathrm{C} \\
T=283.15 \mathrm{~K}, \\
\mu=\frac{\left(1.458 \times 10^{-6}\right)(283.15 \mathrm{k})^{3 / 2}}{283.15+110.4}=1.765 \times 10^{-5} \frac{\mathrm{N} \cdot 5}{\mathrm{~m}^2}
\end{gathered}
$$
From Table B.4, $\mu=1.76 \times 10^{-5} \frac{\mathrm{N} . \mathrm{s}}{\mathrm{m}^2}$
For $T=90^{\circ} \mathrm{C}$
$$
\mathrm{T}=363.15 \mathrm{~K} \text {, }
$$
$$
\mu=\frac{\left(1.458 \times 10^{-6}\right)(363.15)^{3 / 2}}{363.15+110.4}=2.13 \times 10^{-5} \frac{\mathrm{N} \cdot \mathrm{s}}{\mathrm{Nmm}^2}
$$
From Table B. $4, \mu=2.14 \times 10^{-5} \frac{\mathrm{N} \cdot \mathrm{S}}{\mathrm{m}^2}$
