Question

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?

Short answer

Answer: The temperature at the copper-steel interface is approximately \(101.54^{\circ} \mathrm{C}\), and the rate of water evaporation is approximately \(194 \, \mathrm{g/s}\).

Step-by-step solution

List given values

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, \(T_W = 100.00^{\circ} \mathrm{C}\)

Calculate the area of the pan

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: \(A = \pi(0.09 \, \mathrm{m})^2 = 0.0254 \, \mathrm{m^2}\)

Calculate the heat conduction rate (Q)

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: \(Q = \kappa_2 A\frac{T_2 - T_W}{d_2}\) \(d_2 = 0.350 \, \mathrm{cm} = 0.00350 \, \mathrm{m}\) \(Q = 46.0 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}} \cdot 0.0254 \, \mathrm{m^2} \cdot \frac{T_2 - 100.00^{\circ} \mathrm{C}}{0.00350 \, \mathrm{m}}\)

Find the temperature at the copper-steel interface (T2)

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 \(101.54^{\circ} \mathrm{C}\).

Calculate the heat transfer and rate of water evaporation

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 \(194 \, \mathrm{g/s}\).