91Ó°ÊÓ

We are given the spectral distribution of the emissivity as a function of wavelength: \(\varepsilon_{\lambda} = \begin{cases} 0.2, & 0 \leq \lambda \leq 2\;\mu m \\ 0.4, & 2 \leq \lambda \leq 8\;\mu m \\ 0.85, & \lambda > 8\;\mu m \end{cases}\) To determine the total hemispherical emissivity, we must calculate the average value over all wavelengths. For this, we can use the following formula: \(\varepsilon = \frac{\int_{0}^{\infty} \varepsilon_{\lambda} d\lambda}{\int_{0}^{\infty} d\lambda}\) Given the spectral distribution, we can rewrite the formula as: \(\varepsilon = \frac{\int_{0}^{2}0.2\:d\lambda + \int_{2}^{8}0.4\:d\lambda + \int_{8}^{\infty}0.85\:d\lambda}{\int_{0}^{\infty} d\lambda}\) The denominator is infinite, so it doesn't make sense in this context. Instead, we can calculate the area under the curve of the emissivity function: \(\varepsilon = \int_{0}^{2}0.2\:d\lambda + \int_{2}^{8}0.4\:d\lambda + \int_{8}^{\infty}0.85\:d\lambda\) Now, we can find the total hemispherical emissivity: \(\varepsilon_t = 0.2(\lambda) \big|_{0}^{2} + 0.4(\lambda) \big|_{2}^{8} + 0.85(\infty)\) \(\varepsilon_t = (0.4 - 0) + (3.2 - 0.8) + 0.85(\infty)\) The third term is infinite, which is not physically reasonable. However, in practice, the wavelength range is limited by the characteristics of the material and the detector. It is common to truncate the integral at a maximum wavelength, such as 15 or 20 µm, considering the emissivity does not vary significantly for higher wavelengths in most cases. Here, let's assume a maximum wavelength of 20 µm, so we can calculate the total hemispherical emissivity: \(\varepsilon_t = 0.2(2) + 0.4(6) + 0.85(12)\) \(\varepsilon_t = 0.4 + 2.4 + 10.2 = 13\)

To calculate the total emissive power E, we can use the following formula, which relates emissive power to total hemispherical emissivity and the blackbody emissive power: \(E = \varepsilon \sigma T^4\) Where: \(\sigma\) is the Stefan-Boltzmann constant (\(\sigma = 5.6703 \times 10^{-8} Wm^{-2}K^{-4}\)) \(T\) is the temperature Given the problem, we want the total emissive power at a temperature of 2000 K: \(E = 13 \times 5.6703 \times 10^{-8} \times (2000)^4\) \(E = 13 \times 5.6703 \times 10^{-8} \times 1.6 \times 10^{13}\) \(E = 1.4755 \times 10^6 Wm^{-2}\)

To plot the emissivity as a function of temperature for \(500 \leq T \leq 3000 K\), we will calculate values for each temperature in the range, assuming the temperature dependence for emissivity is negligible (emissivity doesn't change much over this temperature range). However, the proper analysis and interpretation of such plots require accurate information about the material and/or experimental setups. Additionally, the variation could be explained by considering factors like changes in surface-to-volume ratio, lattice vibrations, chemical composition, etc., which are beyond the scope of this basic exercise.