c
Let $\theta$ be the temperature of junction, $\mathrm{H}_{1}$ $\mathrm{H}_{2}$ and $\mathrm{H}_{3}$ the heat currents. Then
$\mathrm{H}_{1}=\mathrm{H}_{2}+\mathrm{H}_{3}$
or $\frac{30-\theta}{\left(\frac{30}{\mathrm{KA}}\right)}=\frac{\theta-20}{\left(\frac{20}{\mathrm{KA}}\right)}+\frac{\theta-10}{\left(\frac{10}{\mathrm{KA}}\right)}$
or $2\left(30^{\circ}-\theta\right)=3(\theta-20)+6(\theta-10)$
or $\theta=16.36^{\circ} \mathrm{C} \approx 16.4^{\circ} \mathrm{C}$
