A horizontal opaque flat plate is well insulated on the edges and the lower surface. The top surface has an area of \(5 \mathrm{~m}^{2}\), and it experiences uniform irradiation at a rate of \(5000 \mathrm{~W}\). The plate absorbs \(4000 \mathrm{~W}\) of the irradiation, and the surface is losing heat at a rate of \(500 \mathrm{~W}\) by convection. If the plate maintains a uniform temperature of \(350 \mathrm{~K}\), determine the absorptivity, reflectivity, and emissivity of the plate.

First, let's calculate the total energy balance of the plate. The plate receives irradiation, absorbs some of that energy, and loses energy by convection. The remaining energy should be equal to the energy emitted by the plate as radiation. Given that the plate absorbs \(4000 \mathrm{~W}\) of the irradiation and loses \(500 \mathrm{~W}\) by convection, we can calculate the energy emitted by the plate as: Energy emitted = \(4000 \mathrm{~W}\) (absorbed) - \(500 \mathrm{~W}\) (lost by convection) = \(3500 \mathrm{~W}\)

Absorptivity is defined as the fraction of incident radiation that is absorbed by a surface. We know that the plate absorbs \(4000 \mathrm{~W}\) of the irradiation and the total irradiation is \(5000 \mathrm{~W}\). Therefore, the absorptivity \((\alpha)\) can be calculated as: Absorptivity \((\alpha)\) = \(\frac{\text{absorbed energy}}{\text{incident energy}}\) \(\alpha\) = \(\frac{4000 \mathrm{~W}}{5000 \mathrm{~W}}\) = \(0.8\)

Reflectivity \((\rho)\) is defined as the fraction of incident radiation that is reflected by a surface. Since an opaque surface either absorbs or reflects all the incident radiation, and we have already calculated absorptivity, we can simply subtract absorptivity from 1 to find the reflectivity: Reflectivity \((\rho)\) = \(1 - \alpha\) \(\rho\) = \(1 - 0.8\) = \(0.2\)

Emissivity \((\varepsilon)\) is the effectiveness of a surface in emitting energy as thermal radiation and can be found using the Stefan-Boltzmann law: Emitted energy = \(\varepsilon \sigma A T^4\) Where \(\sigma\) is the Stefan-Boltzmann constant, \(A\) is the surface area, and \(T\) is the temperature in Kelvin. We know the emitted energy is \(3500 \mathrm{~W}\), the surface area is \(5 \mathrm{~m}^{2}\), and the temperature is \(350 \mathrm{~K}\). The Stefan-Boltzmann constant is \(5.67 \times 10^{-8} \mathrm{Wm}^{-2}\mathrm{K}^{-4}\). Rearrange the Stefan-Boltzmann law to solve for emissivity: \(\varepsilon\) = \(\frac{\text{emitted energy}}{\sigma A T^4}\) \(\varepsilon\) = \(\frac{3500 \mathrm{~W}}{(5.67 \times 10^{-8} \mathrm{Wm}^{-2}\mathrm{K}^{-4}) (5 \mathrm{~m}^2) (350 \mathrm{~K})^4}\) \(\varepsilon \approx 0.89\) Thus, the absorptivity of the plate is \(0.8\), its reflectivity is \(0.2\), and its emissivity is approximately \(0.89\).