A circular ceramic plate that can be modeled as a blackbody is being heated by an electrical heater. The plate is \(30 \mathrm{~cm}\) in diameter and is situated in a surrounding ambient temperature of \(15^{\circ} \mathrm{C}\) where the natural convection heat transfer coefficient is \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the efficiency of the electrical heater to transfer heat to the plate is \(80 \%\), determine the electric power that the heater needs to keep the surface temperature of the plate at \(200^{\circ} \mathrm{C}\).

We are given the diameter of the plate (\(30\,\text{cm}\)), so we can calculate the area of the plate using the formula for the area of a circle: \[A = \pi r^2\] where \(A\) is the area of the circle, \(r\) is the radius of the circle, and \(\pi\) is a constant (\(\approx 3.14159\)). The radius is half of the diameter, so \(r = \frac{30}{2} = 15\,\text{cm}\). We need to convert this to meters in order to match the unit of the heat transfer coefficient. Therefore, \(r = 0.15\,\text{m}\). Now we can calculate the area: \[A = \pi (0.15\,\text{m})^2 = 0.070685\,\text{m}^2\]

For a blackbody, we can use the following formula to calculate the heat loss by radiation: \[q_\text{rad} = \sigma A(T_\text{plate}^4 - T_\text{ambient}^4)\] where \(q_\text{rad}\) is the heat loss by radiation, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67\times 10^{-8}\,\text{W/m}^2 \text{K}^4\)), \(A\) is the area of the plate, \(T_\text{plate}\) is the surface temperature of the plate in Kelvin, and \(T_\text{ambient}\) is the surrounding ambient temperature in Kelvin. First, we need to convert the temperatures to Kelvin: \[T_\text{plate} = 200^{\circ} \mathrm{C} + 273.15 = 473.15\,\text{K}\] \[T_\text{ambient} = 15^{\circ} \mathrm{C} + 273.15 = 288.15\,\text{K}\] Now, we can calculate the heat loss by radiation: \[q_\text{rad} = (5.67\times 10^{-8}\,\text{W/m}^2 \text{K}^4) (0.070685\,\text{m}^2)(473.15^4 - 288.15^4) = 109.87\,\text{W}\]

The formula for heat loss by natural convection is: \[q_\text{conv} = h_\text{conv} A(T_\text{plate} - T_\text{ambient})\] where \(q_\text{conv}\) is the heat loss by natural convection, \(h_\text{conv}\) is the natural convection heat transfer coefficient, \(A\) is the area of the plate, \(T_\text{plate}\) is the surface temperature of the plate in Kelvin, and \(T_\text{ambient}\) is the surrounding ambient temperature in Kelvin. We are given the heat transfer coefficient, \(h_\text{conv} = 12\,\text{W/m}^2\,\text{K}\). We can now calculate the heat loss by natural convection: \[q_\text{conv} = (12\,\text{W/m}^2\,\text{K})(0.070685\,\text{m}^2)(473.15 - 288.15) = 157.08\,\text{W}\]

We can find the total heat loss by adding the heat loss by radiation and the heat loss by natural convection: \[q_\text{total} = q_\text{rad} + q_\text{conv} = 109.87\,\text{W} + 157.08\,\text{W} = 266.95\,\text{W}\]

The efficiency of the electrical heater is given as \(80\%\). In order to find the electric power required to maintain the surface temperature of the plate, we can use the following formula: \[P_\text{electric} = \frac{q_\text{total}}{\text{efficiency}}\] where \(P_\text{electric}\) is the required electric power and efficiency is given as a decimal (i.e., \(80\% = 0.8\)). Now we can calculate the required electric power: \[P_\text{electric} = \frac{266.95\,\text{W}}{0.8} = 333.69\,\text{W}\] So, the electric power needed to keep the surface temperature of the plate at \(200^{\circ}\,\text{C}\) is approximately \(333.69\,\text{W}\).