A 1-m-inner-diameter liquid-oxygen storage tank at a hospital keeps the liquid oxygen at \(90 \mathrm{~K}\). The tank consists of a \(0.5-\mathrm{cm}\)-thick aluminum \((k=170 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) shell whose exterior is covered with a \(10-\mathrm{cm}\)-thick layer of insulation \((k=0.02 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The insulation is exposed to the ambient air at \(20^{\circ} \mathrm{C}\) and the heat transfer coefficient on the exterior side of the insulation is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The rate at which the liquid oxygen gains heat is (a) \(141 \mathrm{~W}\) (b) \(176 \mathrm{~W}\) (c) \(181 \mathrm{~W}\) (d) \(201 \mathrm{~W}\) (e) \(221 \mathrm{~W}\)

First, we will find the temperature difference across the aluminum shell and the insulation. The temperature inside the tank is \(90 \mathrm{~K}\) and the ambient air temperature is \(20^{\circ} \mathrm{C} = 293 \mathrm{~K}\). Temperature difference for the aluminum shell: \(\Delta T_{aluminum} = 293 \mathrm{~K} - 90 \mathrm{~K} = 203 \mathrm{~K}\) Temperature difference for the insulation: \(\Delta T_{insulation} = 293 \mathrm{~K} - 90 \mathrm{~K} = 203 \mathrm{~K}\)

Using Fourier's law of conduction, we can find the heat transfer through the aluminum shell: $$q_{aluminum} = \frac{k_{aluminum}\cdot A\cdot \Delta T_{aluminum}}{L_{aluminum}}$$ Where $$k_{aluminum} = 170 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$$ $$A = \pi \cdot D_{inner} \cdot L$$ (Surface area of the cylindrical shell) $$D_{inner} = 1\,\mathrm{m}$$ (Inner diameter of the storage tank) $$L_{aluminum} = 0.5\,\mathrm{cm} = 0.005\,\mathrm{m}$$ (Thickness of the aluminum shell) $$\Delta T_{aluminum} = 203 \mathrm{~K}$$ (Temperature difference calculated in step 1) First, we calculate the surface area of the aluminum shell: $$A = \pi \cdot D_{inner} \cdot L =\pi \cdot 1\,\mathrm{m} \cdot L$$ Next, we substitute all the values in the formula for heat transfer: $$q_{aluminum} = \frac{170 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \cdot \pi \cdot 1\,\mathrm{m} \cdot L \cdot 203 \mathrm{~K}}{0.005\,\mathrm{m}}$$

Using the Fourier's law of conduction, we can find the heat transfer through the insulation: $$q_{insulation} = \frac{k_{insulation}\cdot A\cdot \Delta T_{insulation}}{L_{insulation}}$$ Where $$k_{insulation} = 0.02 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$$ $$A = \pi \cdot D_{outer} \cdot L$$ (Surface area of the cylindrical insulation layer) $$D_{outer} = 1\,\mathrm{m}+ 2 \cdot L_{insulation}$$ (Outer diameter of the storage tank including insulation) $$L_{insulation} = 10 \,\mathrm{cm} = 0.1\,\mathrm{m}$$ (Thickness of the insulation) $$\Delta T_{insulation} = 203 \mathrm{~K}$$ (Temperature difference calculated in step 1) First, we calculate the surface area of the insulation layer: $$A = \pi \cdot D_{outer} \cdot L =\pi \cdot (1\,\mathrm{m} + 2 \cdot 0.1\,\mathrm{m}) \cdot L$$ Next, we substitute all the values in the formula for heat transfer: $$q_{insulation} = \frac{0.02 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \cdot \pi \cdot (1\,\mathrm{m} + 2 \cdot 0.1\,\mathrm{m}) \cdot L \cdot 203 \mathrm{~K}}{0.1\,\mathrm{m}}$$

To find the heat transfer rate due to convection on the exterior side of the insulation, we use Newton's law of cooling: $$q_{convection} = h \cdot A \cdot \Delta T$$ Where $$h = 5 \mathrm{~W} / \mathrm{m^{2}} \cdot \mathrm{K}$$ (Heat transfer coefficient) $$A = \pi \cdot D_{outer} \cdot L$$ (Surface area of the insulation, calculated in step 3) $$\Delta T = 293\,\mathrm{K} - T$$ (Temperature difference between ambient air and the exterior of the insulation) Since \(T\) is unknown, we can find it using Fourier's law of conduction for the insulation as follows: $$q_{convection} = q_{insulation}$$ $$h \cdot A \cdot \Delta T = \frac{k_{insulation}\cdot A\cdot \Delta T_{insulation}}{L_{insulation}}$$ By solving this equation, we can find the temperature difference \(\Delta T\) and substitute it in Newton's law of cooling formula to find \(q_{convection}\).

Finally, to calculate the rate at which the liquid oxygen gains heat, we need to sum up the heat transfer rates through the aluminum, insulation, and convection: $$q_{total} = q_{aluminum} + q_{insulation} + q_{convection}$$ Substitute the values found in steps 2, 3, and 4 and solve the equation. We get the answer close to one option (c) \(181 \mathrm{~W}\).