A furnace is of cylindrical shape with a diameter of \(1.2 \mathrm{~m}\) and a length of \(1.2 \mathrm{~m}\). The top surface has an emissivity of \(0.70\) and is maintained at \(500 \mathrm{~K}\). The bottom surface has an emissivity of \(0.50\) and is maintained at \(650 \mathrm{~K}\). The side surface has an emissivity of \(0.40\). Heat is supplied from the base surface at a net rate of \(1400 \mathrm{~W}\). Determine the temperature of the side surface and the net rates of heat transfer between the top and the bottom surfaces, and between the bottom and side surfaces.

To calculate the surface areas of the top, bottom, and side surfaces, we need to use the formulas for the area of a circle and the lateral surface area of a cylinder. The area of a circle is given by \(A = \pi r^2\), and the lateral surface area of a cylinder is given by \(A = 2 \pi r h\), where \(r\) is the radius, and \(h\) is the height. Given that the diameter of the furnace is \(1.2 \mathrm{~m}\), the radius is \(r = \frac{1.2 \mathrm{~m}}{2} = 0.6 \mathrm{~m}\). Top and bottom surface areas are the same, so: $$ A_{\text{top}} = A_{\text{bottom}} = \pi (0.6)^2 \mathrm{~m^2} = 1.131 \mathrm{~m^2} $$ Now, let's calculate the side surface area: $$ A_{\text{side}} = 2 \pi (0.6 \mathrm{~m})(1.2 \mathrm{~m}) = 4.523 \mathrm{~m^2} $$

The Stefan-Boltzmann Law states that the net heat transfer rate between two surfaces is: $$ q = \epsilon \sigma A (T_{\text{hot}}^4 - T_{\text{cold}}^4) $$ where \(\epsilon\) is the emissivity of the surface, \(\sigma\) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} \mathrm{W/m^2 K^4})\), \(A\) is the area, and \(T\) is the temperature in Kelvin. For the net heat transfer rate between the top surface and bottom surface: $$ q_{\text{top-bottom}} = \frac{\epsilon_{\text{top}} \sigma A_{\text{top}} T_{\text{top}}^4 - \epsilon_{\text{bottom}} \sigma A_{\text{bottom}} T_{\text{bottom}}^4}{1 - \epsilon_{\text{top}} \epsilon_{\text{bottom}}} $$ Plugging in the given values: $$ q_{\text{top-bottom}} = \frac{0.70 (5.67 \times 10^{-8} \mathrm{W/m^2 K^4}) (1.131 \mathrm{~m^2}) (500 \mathrm{~K})^4 - 0.50 (5.67 \times 10^{-8} \mathrm{W/m^2 K^4}) (1.131 \mathrm{~m^2}) (650 \mathrm{~K})^4}{1 - 0.70 \times 0.50} $$ After calculating, we get: $$ q_{\text{top-bottom}} = 637.12 \mathrm{~W} $$ Now since the net heat rate supplied at the bottom is \(1400 \mathrm{~W}\), the heat transfer rate between the bottom and side surfaces can be calculated as: $$ q_{\text{bottom-side}} = 1400 \mathrm{~W} - q_{\text{top-bottom}} = 1400 \mathrm{~W} - 637.12 \mathrm{~W} = 762.88 \mathrm{~W} $$

To find the temperature of the side surface, we will use the Stefan-Boltzmann Law to find the ratio \(T_{\text{side}}^4 - T_{\text{bottom}}^4\). From the given values, we can write the formula as: $$ q_{\text{bottom-side}} = \epsilon_{\text{side}} \sigma A_{\text{side}} (T_{\text{side}}^4 - T_{\text{bottom}}^4) $$ Now solving for \(T_{\text{side}}\): $$ T_{\text{side}}^4 - T_{\text{bottom}}^4 = \frac{q_{\text{bottom-side}}}{\epsilon_{\text{side}} \sigma A_{\text{side}}} $$ Plugging in the values, $$ T_{\text{side}}^4 - (650 \mathrm{~K})^4 = \frac{762.88 \mathrm{~W}}{0.40 (5.67 \times 10^{-8} \mathrm{W/m^2 K^4}) (4.523 \mathrm{~m^2})} $$ After calculating, we get: $$ T_{\text{side}}^4 \approx 1.82 \times 10^{11} \mathrm{K^4} $$ Finally, taking the fourth root: $$ T_{\text{side}} \approx 565 \mathrm{~K} $$ In conclusion, the temperature of the side surface is approximately \(565 \mathrm{~K}\). The net heat transfer rates are \(637.12 \mathrm{~W}\) between the top and the bottom surfaces, and \(762.88 \mathrm{~W}\) between the bottom and side surfaces.