Let's analyze the problem step by step: ### Given Data: - Length of the steel rod, \(L = 900\, \text{mm} = .9\, \text{m}\) - Cross-sectional area, \(A = 10\, \text{mm}^2 = 10 \times 10^{-6}\, \text{m}^2\) - Temperature change, \(\Delta T = 50^\circ C - ^\circ C = 50^\circ C\) - Thermal conductivity, \(\kappa = 50\, \text{W/m}^\circ C\) - Surface conductance, \(h = 10\, \text{W/m}^2\,^\circ C\) ### Objective: Calculate the heat dissipated \(Q\) the rod. ### Step 1: Calculate the heat to conduction: \[ Q_{cond} = \frac{\kappa \times A \times \Delta T}{L} \] \[ Q_{cond} = \frac{50 \times 10 \times 10^{-6} \times 50}{.9} \] \[ Q_{cond} = \frac{50 \times 10^{-5} \times 50}{.9} = \frac{50 \times 50 \times 10^{-5}}{.9} \] \[ Q_{cond} = \frac{250 \times 10^{-5}}{.9} = \frac{.025}{.9} \approx .0278\, \text{W} \] Step 2: Calculate the heat loss due to convection: \[ Q_{conv} = h \times A_{surface} \times \Delta T \] The surface area \(A_{surface} = \) circumference \(\times\) length: \[ A_{surface} = 2 \pi r \times L \] Where radius \(r = \sqrt{\frac{A}{\pi}}\): \[ r = \sqrt{\frac{10 \times 10^{-6}}{\pi}} \approx \sqrt{\frac{10 \times 10^{-6}}{3.1416}} \approx \sqrt{3.18 \times 10^{-6}} \approx 1.78 \times 10^{-3}\, \text{m} \] Now, \[ A_{surface} = 2 \pi r L = 2 \pi \times 1.78 \times 10^{-3} \times .9 \approx 2 \times 3.1416 \times 1.78 \times 10^{-3} \times .9 \] \[ A_{surface} \approx 6.2832 \times 1.78 \times 10^{-3} \times .9 \approx 6.2832 \times 1.602 \times 10^{-3} \approx 10.07 \times 10^{-3} \text{m}^2 \] Now, \[ Q_{conv} = h \times A_{surface} \times \Delta T = 10 \times 10.07 \times 10^{-3} \times 50 \] \[ Q_{conv} \approx 10 \times .01007 \times 50 = .5035\, \text{W} \] ### Step 3: Total heat dissipation: The heat dissipated by the rod is primarily due to convection, so the power \(Q \approx .5035\, \text{W}\). ### Closest option: **Option 1.7 W** is the closest to our calculation. --- ## **Answer: 1.7 W**

Let's recheck the calculations step by step. ## Given Data: - **Length of the steel rod,** \(L = 900\, \text{mm} = 0.9\, \text{m}\) - **Cross-sectional area,** \(A = 10\, \text{mm}^2 = 10 \times 10^{-6}\, \text{m}^2\) - **Temperature change,** \(\Delta T = 50^\circ C - 0^\circ C = 50^\circ C\) - **Thermal conductivity,** \(\kappa = 50\, \text{W/m}^\circ C\) - **Surface conductance,** \(h = 10\, \text{W/m}^2\,^\circ C\) ## Objective: Calculate the heat dissipated \(Q\) by the rod. ### Step 1: Heat Transfer by Conduction Using the formula: \[ Q_{cond} = \frac{\kappa \times A \times \Delta T}{L} \] Substituting the values: \[ Q_{cond} = \frac{50 \times 10 \times 10^{-6} \times 50}{0.9} \] Calculating: \[ Q_{cond} = \frac{2500 \times 10^{-6}}{0.9} = \frac{2.5 \times 10^{-3}}{0.9} \approx 0.00278\, \text{W} \] ### Step 2: Heat Loss due to Convection First, calculate the surface area \(A_{surface}\): 1. Calculate radius \(r\) from the cross-sectional area: \[ r = \sqrt{\frac{A}{\pi}} = \sqrt{\frac{10 \times 10^{-6}}{\pi}} \approx 1.78 \times 10^{-3}\, \text{m} \] 2. Calculate surface area: \[ A_{surface} = 2 \pi r L = 2 \pi \times 1.78 \times 10^{-3} \times 0.9 \] \[ A_{surface} \approx 2 \times 3.1416 \times 1.78 \times 10^{-3} \times 0.9 \approx 0.01007\, \text{m}^2 \] Calculating \(Q_{conv}\): \[ Q_{conv} = h \times A_{surface} \times \Delta T = 10 \times 0.01007 \times 50 \] \[ Q_{conv} \approx 10 \times 0.01007 \times 50 \approx 5.035\, \text{W} \] ### Step 3: Total Heat Dissipation Since convection is the dominant form of heat loss: \[ Q \approx Q_{conv} \approx 5.035\, \text{W} \] ### Conclusion The total heat dissipated by the rod is approximately **5.035 W**. ## **Final Answer: 5.035 W** This value does not match the earlier suggestion of **1.7 W**. The correct calculation shows that the heat dissipated is higher due to considerable convection effects.