A large cast iron container \((k=52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=\) \(1.70 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\) ) with 5 -cm- thick walls is initially at a uniform temperature of \(0^{\circ} \mathrm{C}\) and is filled with ice at \(0^{\circ} \mathrm{C}\). Now the outer surfaces of the container are exposed to hot water at \(60^{\circ} \mathrm{C}\) with a very large heat transfer coefficient. Determine how long it will be before the ice inside the container starts melting. Also, taking the heat transfer coefficient on the inner surface of the container to be \(250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer to the ice through a \(1.2-\mathrm{m}\)-wide and \(2-\mathrm{m}\)-high section of the wall when steady operating conditions are reached. Assume the ice starts melting when its inner surface temperature rises to \(0.1^{\circ} \mathrm{C}\).

To determine if the lumped capacitance analysis is valid for our problem, we need to calculate the Biot number (Bi). The Biot number represents the ratio of internal conductive resistance to the external convective resistance. The lumped capacitance is valid when Bi is less than or equal to 0.1. Bi is given by: \(\mathrm{Bi} = \cfrac{hL_{c}}{k}\) where \(h\) is the heat transfer coefficient, \(L_c\) is the characteristic length, and \(k\) is the thermal conductivity of the material. Since we have a container with 5-cm-thick walls and considering the inner surface of the container, the characteristic length will be half of the thickness, which is \(2.5\,\mathrm{cm} = 0.025\,\mathrm{m}\). Let's calculate Bi using the given values: \(h = 250\,\cfrac{\mathrm{W}} {\mathrm{m}^2\cdot \mathrm{K}}\) \(k = 52\,\cfrac{\mathrm{W}} {\mathrm{m}\cdot \mathrm{K}}\) \(L_c = 0.025\,\mathrm{m}\) \(\mathrm{Bi}= \cfrac{(250\,\cfrac{\mathrm{W}}{\mathrm{m}^2\cdot\mathrm{K}})(0.025\,\mathrm{m})}{(52\,\cfrac{\mathrm{W}}{\mathrm{m}\cdot \mathrm{K}})} = 0.1202\) Since the Biot number is greater than 0.1, lumped capacitance analysis is not valid. We have to use the unsteady-state conduction analysis instead.

Since the lumped capacitance analysis is not valid, we need to use the unsteady-state conduction method. The Fourier number (Fo) is used to find the time it takes to reach a certain temperature change. The Fourier number is given by: \(\mathrm{Fo} = \cfrac{\alpha t}{L_c^2}\) We are given the thermal diffusivity of the material: \(\alpha = 1.70 \times 10^{-5} \cfrac{\mathrm{m}^2}{\mathrm{s}}\) Since the inner surface of the cast iron can be approximated as following the semi-infinite solid solution, we will use this equation to find the temperature change as a function of time and position: \(\theta = \cfrac{T - T_i}{T_{\infty} - T_i} = \operatorname{erfc}(\cfrac{x}{2(L_c \alpha t)^{1/2}})\) where \(T_i\) is the initial temperature, \(T_\infty\) is the outer surface temperature, \(x\) is the distance into the solid (thickness of the container), \(\operatorname{erfc}\) denotes the complementary error function, and \(\theta\) is the non-dimensional temperature change function. We want to find the time it takes for the inner surface of the container (x = 0.05 m) to reach \(0.1^{\circ} \mathrm{C}\). Plugging in the known values: \(0.1 = \operatorname{erfc}(\cfrac{0.05}{2(0.025(1.70\times10^{-5}t)^{1/2}}})\) Solve for \(t\) numerically: \(t = 5275\,\mathrm{s}\)

Under steady-state conditions, the rate of heat transfer to the ice at the inner surface can be determined using the heat transfer coefficient provided and the temperature difference between the inner surface and the ice. To find the rate of heat transfer (Q) over the given area, we will use the equation: \(Q = hA(T_{s,i} - T_{ice})\) where A is the area, \(T_{s,i}\) is the inner surface temperature, and \(T_{ice}\) is the ice temperature. Plugging in the known values: \(h = 250\,\cfrac{\mathrm{W}}{\mathrm{m}^2\cdot\mathrm{K}}\) \(A = 1.2\,\mathrm{m}\times 2\,\mathrm{m} = 2.4\,\mathrm{m}^2\) \(T_{s,i} = 0.1^{\circ}\,\mathrm{C}\) \(T_{ice} = 0^{\circ}\,\mathrm{C}\) \(Q = (250\,\cfrac{\mathrm{W}}{\mathrm{m}^2\cdot\mathrm{K}})(2.4\,\mathrm{m}^2)(0.1\,\mathrm{K}) = 60\,\mathrm{W}\) So, the rate of heat transfer to the ice is 60 W when steady-state operating conditions are reached. In conclusion, it will take approximately 5275 seconds (1.47 hours) for the ice inside the container to start melting, and the rate of heat transfer to the ice through a \(1.2\,\mathrm{m}\)-wide and \(2\,\mathrm{m}\)-high section of the wall is 60 W under steady-state operating conditions.