A 3-m-internal-diameter spherical tank made of 1 -cm-thick stainless steel is used to store iced water at \(0^{\circ} \mathrm{C}\). The tank is located outdoors at \(25^{\circ} \mathrm{C}\). Assuming the entire steel tank to be at \(0^{\circ} \mathrm{C}\) and thus the thermal resistance of the tank to be negligible, determine \((a)\) the rate of heat transfer to the iced water in the tank and \((b)\) the amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24 -hour period. The heat of fusion of water at atmospheric pressure is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\). The emissivity of the outer surface of the tank is \(0.75\), and the convection heat transfer coefficient on the outer surface can be taken to be \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assume the average surrounding surface temperature for radiation exchange to be \(15^{\circ} \mathrm{C}\).

The surface area of a sphere is given by: \(A = 4\pi r^{2}\) Where: \(A\) = surface area \(r\) = radius of the sphere Given the internal diameter of the tank is 3m: \(r = \frac{3}{2} \mathrm{m}\) Plugging the value of \(r\) into the formula, we get: \(A = 4\pi (\frac{3}{2})^2\) \(A = 9\pi \mathrm{m}^2\)

The convection heat transfer (\(Q_{conv}\)) is given by: \(Q_{conv} = h \cdot A \cdot (T_{\infty} - T_{s})\) Where: \(h\) = convection heat transfer coefficient (\(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)) \(A\) = surface area of the tank (from step 1) \(T_{\infty}\) = ambient temperature (\(25^{\circ} \mathrm{C}\)) \(T_s\) = surface temperature of the tank (assumed to be \(0^{\circ} \mathrm{C}\)) Plugging the values into the formula, we get: \(Q_{conv} = 30 \cdot 9\pi \cdot (25 - 0)\) \(Q_{conv} = 6750\pi \mathrm{~W}\) The radiative heat transfer (\(Q_{rad}\)) is given by: \(Q_{rad} = \epsilon \cdot \sigma \cdot A(T_{surr}^4 - T_{s}^4)\) Where: \(\epsilon\) = emissivity of the outer surface of the tank (0.75) \(\sigma\) = Stefan-Boltzmann constant (\(5.6703 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^4\)) \(A\) = surface area of the tank (from step 1) \(T_{surr}\) = average surrounding surface temperature for radiation exchange (\(15^{\circ} \mathrm{C}\) + 273.15) \(T_s\) = surface temperature of the tank (assumed to be \(0^{\circ} \mathrm{C}\) + 273.15) Plugging the values into the formula, we get: \(Q_{rad} = 0.75 \cdot 5.6703 \times 10^{-8} \cdot 9\pi \cdot ((15 + 273.15)^4 - (0 + 273.15)^4)\) \(Q_{rad} \approx 657.38 \mathrm{~W}\) Now we sum the convective and radiative heat transfer to find the total rate of heat transfer (\(Q_{total}\)): \(Q_{total} = Q_{conv} + Q_{rad}\) \(Q_{total} \approx 6750\pi + 657.38 \mathrm{~W}\) \((a)\) The rate of heat transfer to the iced water in the tank is \(\approx 6750\pi + 657.38 \mathrm{~W}\).

In order to calculate the amount of ice melted during a 24-hour period, we need to find the energy received by the ice during the time period. Energy received (\(E_{total}\)) is equal to the rate of heat transfer multiplied by the time period. Given the time period of 24-hours, we convert it to seconds: \(24 \mathrm{~hours} = 24 \times 60 \times 60 \mathrm{~s} = 86400 \mathrm{~s}\) Therefore: \(E_{total} = Q_{total} \cdot t\) \(E_{total} \approx (6750\pi + 657.38) \cdot 86400 \mathrm{~J}\) Now, we can find the mass of ice that melts by using the heat of fusion of water (\(h_{if} = 333.7 \mathrm{~kJ/kg}\)). We convert heat of fusion to J/kg: \(h_{if} = 333.7 \times 10^3 \mathrm{~J/kg}\) The mass of ice melted (\(m_{melted}\)) is given by: \(m_{melted} = \frac{E_{total}}{h_{if}}\) \(m_{melted} \approx \frac{(6750\pi + 657.38) \cdot 86400}{333.7 \times 10^3} \mathrm{~kg}\) \((b)\) The amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24-hour period is \(\approx \frac{(6750\pi + 657.38) \cdot 86400}{333.7 \times 10^3} \mathrm{~kg}\).