Answer
$$\left(\frac{Q}{t}\right)_{\mathrm{Cu}}=134.3^{\circ} \mathrm{C} \approx 130^{\circ} \mathrm{C}$$
Work Step by Step
For temperature at the shared to stay constant the heat flow in the rods requisite be the same note that the cross-sectional areas and lengths are the same.
Use Eq. $14 5$ for heat conduction.
$$
\left(\frac{Q}{t}\right)_{\mathrm{Cu}}=\left(\frac{Q}{t}\right)_{\mathrm{Al}} \rightarrow k_{\mathrm{Cu}} A \frac{T_{\mathrm{hot}}}{\ell}=k_{\mathrm{Al}} A \frac{T_{\text {middle }}-T_{\mathrm{cool}}}{\ell} \rightarrow
$$
$$T_{\text {middle }}=\frac{k_{\mathrm{Cu}} T_{\mathrm{hot}}+k_{\mathrm{Al}} T_{\mathrm{col}}}{k_{\mathrm{Cu}}+k_{\mathrm{Al}}}$$$$=\frac{\left(380 \mathrm{J} / \mathrm{s} \cdot \mathrm{m} \cdot \mathrm{C}^{\circ}\right)\left(205^{\circ} \mathrm{C}\right)+\left(200 \mathrm{J} / \mathrm{s} \cdot \mathrm{m} \cdot \mathrm{C}^{\circ}\right)\left(0.0^{\circ} \mathrm{C}\right)}{380 \mathrm{J} / \mathrm{s} \cdot \mathrm{m} \cdot \mathrm{C}^{\circ}+200 \mathrm{J} / \mathrm{s} \cdot \mathrm{m} \cdot \mathrm{C}^{\circ}}$$$$=134.3^{\circ} \mathrm{C} \approx 130^{\circ} \mathrm{C}$$
