Question

One end of a rod of length 20 cm is inserted in a furnace at 800 K. The sides of the rod are covered with an insulating material and the other end emits radiation like a blackbody. The temperature of this end is 750 K in the steady state. The temperature of the surrounding air is 300 K. Assuming radiation to be the only important mode of energy transfer between the surrounding and the open end of the rod, find the thermal conductivity of the rod. Stefan constant σ = 6.0 × 10⁻⁸ W m⁻² K⁻⁴.

Answer 1

Stefan's constant, σ = 6 × 10⁻⁸ W m⁻² K⁻⁴
l = 0.2 m
T₁ = 300 K
T₂ = 750 K
$$\begin{gathered} \frac{\Delta Q}{\mathrm{\Delta }t}=\epsilon A\sigma \left(T_2^4-T_1^4\right) \\ \frac{\Delta Q}{\Delta t\mathit{}A}\mathit{=}\mathit{}\epsilon \sigma \mathit{}\left(T_{\mathit{2}}^{\mathit{4}}\mathit{-}{T}_{\mathit{1}}^{\mathit{4}}\right)\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{.}\mathit{.}\mathit{.}\mathit{}\mathit{}\mathit{}\left(1\right) \\ \mathrm{Also},\mathit{}\frac{\Delta Q}{\Delta t}\mathit{}\mathit{=}\mathit{}\frac{K\mathit{}A\mathit{}\left({\theta }_{\mathit{1}}\mathit{-}{\theta }_{\mathit{2}}\right)}{l}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{} \\ \frac{\mathrm{\Delta Q}}{\mathrm{\Delta }t·\mathrm{A}}=\frac{\mathrm{K}\left(800-750\right)}{0.2}...\left(2\right) \\ \mathrm{From}\left(1\right)\mathrm{and}\left(2\right), \\ \mathrm{\epsilon \sigma }\left(T_2^4-T_1^4\right)=\frac{K\times 50}{0.2} \\ 1\times 6\times {10}^{-8}\left({750}^4-{300}^4\right)=\frac{K\times 50}{0.2} \\ K=74\mathrm{w}/\mathrm{m}°\mathrm{C} \end{gathered}$$