A stainless steel saucepan, with a base that is made of \(0.350-\mathrm{cm}-\) thick steel \([\kappa=46.0 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})]\) fused to a \(0.150-\mathrm{cm}\) thickness of copper $[\kappa=401 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})],\( sits on a ceramic heating element at \)104.00^{\circ} \mathrm{C} .\( The diameter of the pan is \)18.0 \mathrm{cm}$ and it contains boiling water at \(100.00^{\circ} \mathrm{C} .\) (a) If the copper-clad bottom is touching the heat source, what is the temperature at the copper-steel interface? (b) At what rate will the water evaporate from the pan?

Copper layer: - Thickness, \(d_1 = 0.150\) cm - Thermal conductivity, \(\kappa_1 = 401 \,\mathrm{W} /(\mathrm{m} \cdot \mathrm{K})\) Steel layer: - Thickness, \(d_2 = 0.350\) cm - Thermal conductivity, \(\kappa_2 = 46.0 \,\mathrm{W} /(\mathrm{m} \cdot \mathrm{K})\) Diameter of the pan, \(D = 18.0\) cm Temperature of the heat source, \(T_1 = 104.00^{\circ} \mathrm{C}\) Temperature of boiling water, \(T_W = 100.00^{\circ} \mathrm{C}\)

To calculate the area of the pan, we'll use the formula for the area of a circle: \(A = \pi r^2\) The radius is half of the diameter: \(r = \frac{D}{2} = \frac{18.0 \, \mathrm{cm}}{2} = 9.0 \, \mathrm{cm} = 0.09 \, \mathrm{m}\) Now we can find the area: \(A = \pi(0.09 \, \mathrm{m})^2 = 0.0254 \, \mathrm{m^2}\)

We know that the heat conduction is the same through both layers since there is no other loss. Using the formula for heat conduction rate through a layer, we can find Q for copper. \(Q = \kappa_1 A\frac{T_1 - T_2}{d_1}\) Now we need to express the thickness in meters: \(d_1 = 0.150 \, \mathrm{cm} = 0.00150 \, \mathrm{m}\) \(Q = 401 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}} \cdot 0.0254 \, \mathrm{m^2} \cdot \frac{104.00^{\circ} \mathrm{C} - T_2}{0.00150 \, \mathrm{m}}\) Now, we will do the same for the steel layer: \(Q = \kappa_2 A\frac{T_2 - T_W}{d_2}\) \(d_2 = 0.350 \, \mathrm{cm} = 0.00350 \, \mathrm{m}\) \(Q = 46.0 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}} \cdot 0.0254 \, \mathrm{m^2} \cdot \frac{T_2 - 100.00^{\circ} \mathrm{C}}{0.00350 \, \mathrm{m}}\)

Since the heat conduction rate (Q) is the same for both layers, we can set the two equations equal to each other and solve for T2. \(46.0 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}} \cdot 0.0254 \, \mathrm{m^2} \cdot \frac{T_2 - 100.00^{\circ} \mathrm{C}}{0.00350 \, \mathrm{m}} = 401 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}} \cdot 0.0254 \, \mathrm{m^2} \cdot \frac{104.00^{\circ} \mathrm{C} - T_2}{0.00150 \, \mathrm{m}}\) Solve for \(T_2\): \(T_2 = 101.54^{\circ} \mathrm{C}\) So, the temperature at the copper-steel interface is \(101.54^{\circ} \mathrm{C}\).

We will use the same heat conduction rate (Q) for steel to find the heat transfer: \(Q = 46.0 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}} \cdot 0.0254 \, \mathrm{m^2} \cdot \frac{101.54^{\circ} \mathrm{C} - 100.00^{\circ} \mathrm{C}}{0.00350 \, \mathrm{m}}\) \(Q \approx 1246 \, \mathrm{W}\) Now, we can calculate the rate of water evaporation using the power formula: \(P = mc\Delta T\) Here, m is the mass of water evaporating per second, c is the specific heat of water (\(\approx 4.18 \, \mathrm{J/g\cdot K}\)), and \(\Delta T\) is the change in temperature (\(101.54^{\circ} \mathrm{C} - 100.00^{\circ} \mathrm{C}\)). We have Q=P, so: \(1246 \, \mathrm{W} = m \cdot 4.18 \, \mathrm{J/g\cdot K} \cdot 1.54 \, \mathrm{K}\) Solve for m: \(m \approx 194 \, \mathrm{g/s}\) So, the rate of water evaporation is approximately \(194 \, \mathrm{g/s}\).