Consider a stainless steel spoon \((k=15.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\varepsilon=0.6\) ) that is partially immersed in boiling water at \(95^{\circ} \mathrm{C}\) in a kitchen at \(25^{\circ} \mathrm{C}\). The handle of the spoon has a cross section of about \(0.2 \mathrm{~cm} \times 1 \mathrm{~cm}\) and extends \(18 \mathrm{~cm}\) in the air from the free surface of the water. The spoon loses heat by convection to the ambient air with an average heat transfer coefficient of \(h=13 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) as well as by radiation to the surrounding surfaces at an average temperature of \(T_{\text {surr }}=\) \(295 \mathrm{~K}\). Assuming steady one- dimensional heat transfer along the spoon and taking the nodal spacing to be \(3 \mathrm{~cm}\), (a) obtain the finite difference formulation for all nodes, (b) determine the temperature of the tip of the spoon by solving those equations, and \((c)\) determine the rate of heat transfer from the exposed surfaces of the spoon.

Define the variables and parameters for the problem, such as thermal conductivity (k), emissivity (ε), dimensions of the spoon, heat transfer coefficient (h), and the surrounding temperature (T_surr). - Thermal conductivity, k = 15.1 W/mK - Emissivity, ε = 0.6 - Cross-sectional area of the spoon handle, A = 0.2 cm * 1 cm = 0.2e-4 m² - Length of the spoon handle exposed to air, L = 18 cm = 0.18 m - Heat transfer coefficient, h = 13 W/m²K - Surrounding temperature, T_surr = 295 K

The heat transfer along the spoon handle can be solved using the finite difference method. Divide the handle into nodes with equal spacing of 3 cm. - Nodal spacing, Δx = 3 cm = 0.03 m For each node, we will write an energy balance, which gives us the finite difference equation. A general energy balance on the i-th node would look like: (Q_in)_i + (Q_out)_i = 0 The conductive heat transfer brings energy into the node and convective and radiative heat transfer removes energy. Write the energy balance on each node (from node 1 to node 6): - For node 1: `k * A * (T2-T1)/Δx - h * P * Δx * (T1-T_ambient) - ε * σ * A * (T1^4 - T_surr^4) = 0` - For nodes 2 to 5: `(Q_in)_i = k * A * (Ti-1 - Ti)/Δx` `(Q_out)_i = k * A * (Ti - Ti+1)/Δx + h * P * Δx * (Ti-T_ambient) + ε * σ * A * (Ti^4 - T_surr^4)` `(Q_in)_i + (Q_out)_i = 0` - For node 6 (in contact with boiling water): `(Q_in)_i = 0` `(Q_out)_i = k * A * (T6 - T5)/Δx + h * P * Δx * (T6-T_ambient) + ε * σ * A * (T6^4 - T_surr^4) + Q_bw` `(Q_in)_i + (Q_out)_i = 0` Where Q_bw is the heat transfer rate from the boiling water to node 6. Solve these equations to get the temperatures of each node.

We need to solve the system of equations from step 2 to find the temperatures at each node. This can be achieved using a variety of numerical methods, such as the Gauss-Seidel method or the Newton-Raphson method. The numerical method used should be able to deal with the non-linear nature of the radiative term (Ti^4 - T_surr^4). Once we have calculated the temperatures at all the nodes, we can determine the temperature at the tip of the spoon, which is the temperature at node 1 (T1).

As we have the temperature of all nodes, we can use these values to compute the heat transfer rate from the exposed surfaces of the spoon. We will calculate conductive, convective, and radiative heat transfer components separately and then sum them. - Conductive heat transfer rate: `Q_cond = k * A * (T2 - T1)/Δx` - Convective heat transfer rate: `Q_conv = h * P * Δx * (T1 - T_ambient)` - Radiative heat transfer rate: `Q_rad = ε * σ * A * (T1^4 - T_surr^4)` Total Heat Transfer Rate (Q_total) = Q_cond + Q_conv + Q_rad We now have the temperature of the tip of the spoon (T1) and the rate of heat transfer from the exposed surfaces of the spoon (Q_total).