A copper rod and an aluminium rod, each 0.80 m long with a cross-section of 1.5 cm², lie side by…

Step 1. Copper. $385\ \text{W m}^{-1}\text{K}^{-1} \times 1.5 \times 10^{-4}\ \text{m}^2 \times \dfrac{100\ \text{K}}{0.80\ \text{m}} = 7.22\ \text{W}$.

Step 2. Aluminium. $205\ \text{W m}^{-1}\text{K}^{-1} \times 1.5 \times 10^{-4}\ \text{m}^2 \times \dfrac{100\ \text{K}}{0.80\ \text{m}} = 3.84\ \text{W}$.

Step 3. Total. $11.06\ \text{W}$.