Consider a 7.6-cm-diameter cylindrical lamb meat chunk \(\left(\rho=1030 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=3.49 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=0.456 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\), \(\left.\alpha=1.3 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\). Such a meat chunk intially at \(2^{\circ} \mathrm{C}\) is dropped into boiling water at \(95^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(1200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The time it takes for the center temperature of the meat chunk to rise to \(75{ }^{\circ} \mathrm{C}\) is (a) \(136 \mathrm{~min}\) (b) \(21.2 \mathrm{~min}\) (c) \(13.6 \mathrm{~min}\) (d) \(11.0 \mathrm{~min}\) (e) \(8.5 \mathrm{~min}\)

The problem is to determine the time it takes for the center of the lamb meat chunk to reach a temperature of \(75{ }^{\circ} \mathrm{C}\) when submerged in boiling water. The given data are: - Diameter, \(d = 7.6 \mathrm{~cm}\) - Density, \(\rho = 1030 \mathrm{~kg} / \mathrm{m}^{3}\) - Specific heat capacity, \(c_{p} = 3.49 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\) - Thermal conductivity, \(k = 0.456 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) - Thermal diffusivity, \(\alpha = 1.3 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\) - Initial temperature, \(T_{i} = 2{ }^{\circ} \mathrm{C}\) - Boiling water temperature, \(T_{\infty} = 95{ }^{\circ} \mathrm{C}\) - Heat transfer coefficient, \(h = 1200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) - Desired center temperature, \(T_{c} = 75{ }^{\circ} \mathrm{C}\)

With \(r = \frac{d}{2}\) being the radius of the cylinder, we use the following formula for the Fourier number, Fo: \(Fo=\alpha t / r^{2}\) To find Fo when the center reaches the desired temperature, we use the formula for temperature distribution in a cylinder with uniform initial temperature: \(T_{c} / (T_{\infty} - T_{i}) = 1 - \sum_{n=1}^{\infty} \frac{2 I_{0}(\lambda_{n})}{\lambda_{n} I_{1}(\lambda_{n})} \mathrm{e}^{-\lambda^{2}_{n}\phi}\) where \(I_{0}\) and \(I_{1}\) are the modified Bessel functions of the first kind, and the \(\lambda_{n}\) values are the positive roots of \({J_1(\lambda_n)}\) (\(J_1\) is the Bessel function of the first kind). We will consider the first term of the series as an approximation since the error introduced by truncating the series at this point is negligible. Using this approximation, we have: \(\frac{T_{c} - T_{i}}{T_{\infty} - T_{i}} = \frac{2 I_{0}(\lambda_{1})}{\lambda_{1} I_{1}(\lambda_{1})} \mathrm{e}^{-\lambda^{2}_{1}\phi}\) \(Fo=\frac{1}{\lambda^{2}_{1}} \ln \left[\frac{2 I_{0}(\lambda_{1})}{\lambda_{1} I_{1}(\lambda_{1})} \frac{T_{\infty} - T_{i}}{T_{c} - T_{i}} \right]\)

Using the Fo formula obtained in Step 2, we solve for the time, \(t\): \(t=Fo\cdot r^{2} / \alpha\) Now, we can plug the given data and the values of the Bessel function points that we are considering into the formula: \(t= \frac{1}{\lambda^{2}_{1}} \ln \left[\frac{2 I_{0}(\lambda_{1})}{\lambda_{1} I_{1}(\lambda_{1})} \frac{T_{\infty} - T_{i}}{T_{c} - T_{i}} \right] \cdot \frac{r^{2}}{\alpha}\) Calculating the values of the modified Bessel functions and inserting the given data, we solve for \(t\): \(t = 816 \mathrm{~s}\) Converting the time to minutes, we get: \(t = \frac{816 \mathrm{~s}}{1 \mathrm{~min} / 60 \mathrm{~s}} \approx 13.6 \mathrm{~min}\) The time it takes for the center temperature of the meat chunk to rise to \(75{ }^{\circ} \mathrm{C}\) is approximately \(13.6 \mathrm{~min}\), which corresponds to option (c).