Answer
$μ_{}=7.61^{\circ}\text{C/MPa}$
Work Step by Step
The enthalpy of steam at $2.5\ \mathrm{MPa}$ and $T=400^{\circ} \mathrm{C}$ is $h=3240.1\ \mathrm{~kJ} / \mathrm{kg}$. Now consider a throttling process from this state to $1.2\ \mathrm{MPa}$. The temperature of the steam at the end of this throttling process will be $$
\left.\begin{array}{l}
P=1.2\ \mathrm{MPa} \\
h=3240.1\ \mathrm{~kJ} / \mathrm{kg}
\end{array}\right\} T_2=390.10^{\circ} \mathrm{C}
$$ Thus the temperature drop during this throttling process is $$
\Delta T=T_2-T_1=390.10-400=-9.90 ^\circ\mathrm{C}
$$ The average Joule-Thomson coefficient for this process is determined from $$
\mu=\left(\frac{\partial T}{\partial P}\right)_h \equiv\left(\frac{\Delta T}{\Delta P}\right)_{h=3240\ . \mathrm{kJ}/ \mathrm{K}_{\mathrm{g}}}=\frac{(390.10-400) \mathrm{C}}{(1.2-2.5) \mathrm{MPa}}=7.61 ^\circ \mathrm{C} / \mathrm{MPa}
$$
