Let's break down the problem: ### Data Given: - **Thermal conductivity, \( k \) = 16 W/m·°C** - **Rod length, \( L = 90 \) mm = 0.09 m** - **Equilateral triangle side, \( a = 10 \) mm = 0.01 m** - **Surface conductance, \( h = 90 \) W/m²·°C** - **Wall temperature, \( T_b = 450 \)°C** - **Surrounding environment temperature, \( T_\infty = 50 \)°C** --- ### Step 1: Area and Perimeter of Cross-section For an equilateral triangle of side \( a \): - **Area, \( A \):** \[ A = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (0.01)^2 = 4.33 \times 10^{-5} \text{ m}^2 \] - **Perimeter, \( P \):** \[ P = 3a = 3 \times 0.01 = 0.03 \text{ m} \] --- ### Step 2: Calculate the Fin Parameter \( m \): \[ m = \sqrt{\frac{hP}{kA}} \] Plug in the values: \[ m = \sqrt{\frac{90 \times 0.03}{16 \times 4.33 \times 10^{-5}}} \] \[ = \sqrt{\frac{2.7}{0.0006928}} \] \[ = \sqrt{3895.5} \approx 62.43 \text{ m}^{-1} \] --- ### Step 3: Heat Dissipated by the Rod (Fin) For a **long fin** (since the rod is long), the heat dissipated is: \[ Q = \sqrt{hPkA}(T_b - T_\infty) \] Calculate \( \sqrt{hPkA} \): \[ hPkA = 90 \times 0.03 \times 16 \times 4.33 \times 10^{-5} \] \[ = 90 \times 0.03 = 2.7 \] \[ 16 \times 4.33 \times 10^{-5} = 6.928 \times 10^{-4} \] \[ hPkA = 2.7 \times 6.928 \times 10^{-4} = 0.00187 \] \[ \sqrt{0.00187} = 0.0432 \] So, \[ Q = 0.0432 \times (450 - 50) \] \[ Q = 0.0432 \times 400 = 17.28 \text{ W} \] --- ### Step 4: Match to the Options The closest option is **17 W**. --- ## **Final Answer:** **The heat dissipated by the rod is:** \[ \boxed{17 \text{ W}} \]