Question

A \(9-\mathrm{cm}\)-diameter potato $\left(\rho=1100 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=3900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.$, \(k=0.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and $\alpha=1.4 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}$ ) that is initially at a uniform temperature of \(25^{\circ} \mathrm{C}\) is baked in an oven at $170^{\circ} \mathrm{C}$ until a temperature sensor inserted into the center of the potato indicates a reading of \(70^{\circ} \mathrm{C}\). The potato is then taken out of the oven and wrapped in thick towels so that almost no heat is lost from the baked potato. Assuming the heat transfer coefficient in the oven to be $40 \mathrm{~W} / \mathrm{m}^{2}\(. \)\mathrm{K}\(, determine \)(a)$ how long the potato is baked in the oven and \((b)\) the final equilibrium temperature of the potato after it is wrapped. Solve this problem using the analytical one-term approximation method.

Short answer

\(t = -\frac{(0.045)^2}{1.4 \times 10^{-7} \pi^2} \ln \left[\frac{\pi(70-25)}{4(170-25)}\right] = 4724.64\,\text{s}\) The time it takes to bake the potato is approximately 4724.64 seconds, or around 78.74 minutes. #tag_title#Step 4: Determine the final equilibrium temperature of the potato after being wrapped in thick towels#tag_content#After removing the potato from the oven, it is wrapped in thick towels, which act as an insulator. Since the towels will minimize heat loss, we can assume that the final equilibrium temperature of the potato will be close to the desired center temperature, which is 70°C. In conclusion, it takes approximately 78.74 minutes to bake the potato at 170°C, and the final equilibrium temperature after being wrapped in thick towels is approximately 70°C.

Step-by-step solution

Analyze the given data and derive necessary formulas

Given: Potato diameter: \(D = 9\,\text{cm}\) Density of potato: \(\rho = 1100\,\text{kg/m}^3\) Specific heat of potato: \(c_p = 3900\,\text{J/kg}\cdot\text{K}\) Potato thermal conductivity: \(k = 0.6\,\text{W/m}\cdot\text{K}\) Potato thermal diffusivity: \(\alpha = 1.4 \times 10^{-7}\,\text{m}^2\,\text{s}\) Initial temperature of potato: \(T_i = 25^\circ\text{C}\) Oven temperature: \(T_{\infty} = 170^\circ\text{C}\) Desired center temperature of potato: \(T_c = 70^\circ\text{C}\) Heat transfer coefficient in the oven: \(h = 40\,\text{W/m}^2\,\text{K}\) We need to derive the following formulas: 1. Biot number: \(\operatorname{Bi}=\frac{h D}{2 k}\) 2. Fourier number: \(Fo= \frac{\alpha t}{\left(\frac{D}{2}\right)^2}\) 3. One-term approximation equation: \(T_c-T_i=\frac{4\left(T_{\infty}-T_i\right)}{\pi}\sum_{n=0}^{\infty} \frac{(-1)^n}{(2 n+1)} e^{\frac{-\alpha t(2 n+1)^2 \pi^2}{\left(\frac{D}{2}\right)^2}}\)

Determine the Biot number (Bi) and Fourier number (Fo)

Now, we will calculate the Biot number (Bi) and Fourier number (Fo). Biot number: \(\operatorname{Bi}=\frac{h D}{2 k} = \frac{40\cdot0.09}{2\cdot0.6} = 3\) The Biot number, Bi > 0.1, which indicates that the temperature variation inside the potato is important and can't be neglected.

Use the one-term approximation method to find the time to bake the potato

We will use the one-term approximation equation and the given data to find the time it takes to bake the potato. \(T_c-T_i= \frac{4\left(T_{\infty}-T_i\right)}{\pi}\sum_{n=0}^{\infty} \frac{(-1)^n}{(2 n+1)} e^{\frac{-\alpha t(2 n+1)^2 \pi^2}{\left(\frac{D}{2}\right)^2}}\) Considering only the first term (\(n=0\)) of the series, we have: \(T_c-T_i= \frac{4\left(T_{\infty}-T_i\right)}{\pi} e^{\frac{-\alpha t\pi^2}{\left(\frac{D}{2}\right)^2}}\) Now, we can solve the equation for time (\(t\)): \(t = -\frac{\left(\frac{D}{2}\right)^2}{\alpha \pi^2} \ln \left[\frac{\pi(T_c-T_i)}{4(T_{\infty}-T_i)}\right]\)