Let's break down the problem step by step: **Given:** - **Emissivity (\(\epsilon\)) of tungsten filament:** - \(\epsilon = 0.5\) for \(\lambda < 1\,\mu m\) - \(\epsilon = 0.15\) for \(\lambda > 1\,\mu m\) - **Temperatures:** \(T_1 = 1500\,K\), \(T_2 = 2500\,K\) We are to find the **average emissivity** at each temperature, and also the **absorptivity** and **reflectivity**. --- ## 1. **Average Emissivity Calculation** The **average emissivity** is the ratio of the total emitted energy (weighted by the spectral emissivity) to the total blackbody emissive power: \[ \overline{\epsilon} = \frac{\int_0^\infty \epsilon_\lambda E_{b,\lambda}(T) d\lambda}{\int_0^\infty E_{b,\lambda}(T) d\lambda} \] Where: - \(\epsilon_\lambda\) is the spectral emissivity - \(E_{b,\lambda}(T)\) is the blackbody spectral emissive power Given the stepwise emissivity function: \[ \epsilon_\lambda = \begin{cases} 0.5 & \lambda < 1\,\mu m \\ 0.15 & \lambda > 1\,\mu m \end{cases} \] So: \[ \overline{\epsilon} = 0.5 \cdot f_{<1} + 0.15 \cdot f_{>1} \] where \(f_{<1}\) is the fraction of blackbody energy at \(\lambda < 1\,\mu m\), and \(f_{>1}\) is at \(\lambda > 1\,\mu m\). But \(f_{>1} = 1 - f_{<1}\), so: \[ \overline{\epsilon} = 0.5 \cdot f_{<1} + 0.15 \cdot (1 - f_{<1}) = 0.15 + 0.35 \cdot f_{<1} \] So all we need is \(f_{<1}\) at each temperature. --- ### **a) At 1500 K** Find \(f_{<1}\) (fraction of blackbody radiation below \(1\,\mu m\)). We use the normalized blackbody function: \[ x = \frac{c_2}{\lambda T} \] where \(c_2 = 1.4388 \times 10^{-2}\) m·K. Calculate for \(\lambda = 1\,\mu m = 1 \times 10^{-6}\) m: \[ x = \frac{1.4388 \times 10^{-2}}{1 \times 10^{-6} \times T} = \frac{1.4388 \times 10^{-2}}{T \times 10^{-6}} \] At \(T = 1500\,K\): \[ x_{1500} = \frac{1.4388 \times 10^{-2}}{1.5 \times 10^3 \times 10^{-6}} = \frac{1.4388 \times 10^{-2}}{1.5 \times 10^{-3}} \approx 9.592 \] Now, \(f_{<1}\) is the fraction of blackbody energy below \(\lambda = 1\,\mu m\) at \(T = 1500\,K\), which we can obtain from blackbody radiation tables or Planck’s law calculators. **From standard blackbody tables:** - At \(x = 9.59\), the fraction \(F_{0 \rightarrow \lambda}\) is approximately **0.01** (1%). So: \[ \overline{\epsilon}_{1500} = 0.15 + 0.35 \times 0.01 = 0.15 + 0.0035 = \boxed{0.1535} \] --- ### **b) At 2500 K** \[ x_{2500} = \frac{1.4388 \times 10^{-2}}{2.5 \times 10^3 \times 10^{-6}} = \frac{1.4388 \times 10^{-2}}{2.5 \times 10^{-3}} \approx 5.755 \] From blackbody tables: - At \(x = 5.76\), \(F_{0 \rightarrow \lambda}\) is approximately **0.055** (5.5%). So: \[ \overline{\epsilon}_{2500} = 0.15 + 0.35 \times 0.055 = 0.15 + 0.01925 = \boxed{0.169} \] --- ## 2. **Absorptivity and Reflectivity** - For an opaque surface: \(\epsilon + \rho = 1\) (no transmission) - **At thermal equilibrium**, by **Kirchhoff’s Law**: **\(\alpha = \epsilon\)** So, at each temperature: ### **At 1500 K:** \[ \alpha = \epsilon = 0.1535 \] \[ \rho = 1 - \epsilon = 1 - 0.1535 = 0.8465 \] ### **At 2500 K:** \[ \alpha = \epsilon = 0.169 \] \[ \rho = 1 - 0.169 = 0.831 \] --- ## **Summary Table** | Temperature (K) | Avg Emissivity (\(\overline{\epsilon}\)) | Absorptivity (\(\alpha\)) | Reflectivity (\(\rho\)) | |-----------------|---------------------------|--------------------------|-------------------------| | 1500 | 0.1535 | 0.1535 | 0.8465 | | 2500 | 0.169 | 0.169 | 0.831 | --- ### **Summary of Procedure** 1. **Calculate \(x = c_2 / (\lambda T)\) for \(\lambda = 1\,\mu m\) and each \(T\)** 2. **Find fraction of blackbody energy below 1 µm (\(f_{<1}\)) from tables** 3. **Plug into \(\overline{\epsilon} = 0.15 + 0.35 f_{<1}\)** 4. **Absorptivity equals emissivity (\(\alpha = \overline{\epsilon}\))** 5. **Reflectivity \(\rho = 1 - \overline{\epsilon}\)** --- ### **Final Answers** #### **(a) At 1500 K:** - **Average Emissivity:** \(\boxed{0.1535}\) - **Absorptivity:** \(\boxed{0.1535}\) - **Reflectivity:** \(\boxed{0.8465}\) #### **(b) At 2500 K:** - **Average Emissivity:** \(\boxed{0.169}\) - **Absorptivity:** \(\boxed{0.169}\) - **Reflectivity:** \(\boxed{0.831}\) --- **If you need the detailed calculation for \(f_{<1}\) using Planck's Law, let me know!**

Certainly! Here's what each notation in the problem and solution represents: - **\(\epsilon\)**: Emissivity of the tungsten filament at a specific wavelength. It indicates how effectively the filament emits thermal radiation compared to a perfect blackbody at the same temperature. - **\(\overline{\epsilon}\)**: The *average emissivity* over the entire spectrum at a given temperature, accounting for the spectral variation of emissivity and the distribution of emitted energy. - **\(\alpha\)**: Absorptivity of the filament, representing the fraction of incident radiation absorbed. For opaque surfaces in thermal equilibrium, \(\alpha \approx \epsilon\). - **\(\rho\)**: Reflectivity of the filament, indicating the fraction of incident radiation reflected. For opaque surfaces, \(\rho = 1 - \alpha\). - **\(T\)**: Temperature of the filament, given in Kelvin (K). - **\(\lambda\)**: Wavelength of radiation, in micrometers (\(\mu m\)) or meters (m). - **\(f_{<1}\)**: Fraction of the total blackbody radiation energy emitted at wavelengths less than 1 µm. - **\(F_{0 \rightarrow \lambda}\)**: The cumulative fraction of blackbody energy emitted at wavelengths from 0 to \(\lambda\). - **\(x\)**: A dimensionless variable related to wavelength and temperature, defined as \(x = \frac{c_2}{\lambda T}\), where: - **\(c_2\)**: Second radiation constant, approximately \(1.4388 \times 10^{-2}\) m·K. - **\(E_{b,\lambda}(T)\)**: Spectral blackbody emissive power at wavelength \(\lambda\) and temperature \(T\). - **\(E_b(T)\)**: Total blackbody emissive power at temperature \(T\), integrated over all wavelengths. - **\(c_2\)**: The second radiation constant, used in Planck’s law calculations, approximately \(1.4388 \times 10^{-2}\) m·K. --- **In summary:** - The spectral emissivity \(\epsilon_\lambda\) varies with wavelength. - The average emissivity \(\overline{\epsilon}\) is computed by weighting the spectral emissivity with the blackbody spectral energy distribution. - Absorptivity \(\alpha\) is equal to the average emissivity \(\overline{\epsilon}\) under thermal equilibrium. - Reflectivity \(\rho\) is simply \(1 - \alpha\) for opaque surfaces. Let me know if you'd like clarification on any specific part!