An experiment is to be conducted to determine heat transfer coefficient on the surfaces of tomatoes that are placed in cold water at \(7^{\circ} \mathrm{C}\). The tomatoes \((k=0.59 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=\) \(\left.0.141 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}, \rho=999 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=3.99 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) with an initial uniform temperature of \(30^{\circ} \mathrm{C}\) are spherical in shape with a diameter of \(8 \mathrm{~cm}\). After a period of 2 hours, the temperatures at the center and the surface of the tomatoes are measured to be \(10.0^{\circ} \mathrm{C}\) and \(7.1^{\circ} \mathrm{C}\), respectively. Using analytical one-term approximation method (not the Heisler charts), determine the heat transfer coefficient and the amount of heat transfer during this period if there are eight such tomatoes in water.

First, let's calculate the Biot number (Bi) and Fourier number (Fo). We have the following formulas for Bi and Fo: Bi = \(\frac{hD_{s}}{k}\) Fo = \( \frac{\alpha t}{D_{s}^2}\) where \(h\) is the heat transfer coefficient, \(D_{s}\) is the radius of the tomato (which is half of the diameter), \(k\) is the thermal conductivity, \(\alpha\) is the thermal diffusivity, and \(t\) is the time. Let's plug in the known values: \(D_s =8 \mathrm{~cm} / 2=4 \mathrm{~cm}\) (half of the diameter) Now, we can compute Fo: Fo = \(\frac{0.141\times10^{-6}\mathrm{~m}^{2} / \mathrm{s} \times 2 \mathrm{~hours}}{(0.04\mathrm{~m})^2} = 0.02547\) The next step is to calculate the Biot number, but we need the heat transfer coefficient that we don't have yet. However, we have the one-term approximation method formula to find it.

We are given the temperature at the surface of the tomato(\(T_s\)) and at its center (\(T_c\)). Let's calculate the temperature ratio \(T_r\) for one-term method: \(T_{r}=\frac{T_{s}-T_c}{\mathrm{T}(0)-T_c} = \frac{7.1^{\circ} \mathrm{C}-10.0^{\circ} \mathrm{C}}{30.0^{\circ} \mathrm{C}-10.0 }=-0.2917\)

As we have Bi and temperature ratio T_r related using the formula for one-term method: \(T_{r}=\frac{2}{Bi}\left[\text{exp}\left(-\frac{Bi^2}{4Fo}\right)-\text{exp}\left(-\frac{Bi^2}{Fo}\right)\right]\) We can rearrange the formula above and solve for the heat transfer coefficient h: \(h=\frac{2k[\text{exp}\left(-\frac{Bi^2}{4Fo}\right)-\text{exp}\left(-\frac{Bi^2}{Fo}\right)]}{T_{r}D_s}\) We can approximate Bi as 3 due to negative reciprocal of T_r and substitute the values: \(h = \frac{2 \cdot 0.59 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} [\text{exp}\left(-\frac{3^2}{4 \cdot 0.02547}\right)-\text{exp}\left(-\frac{3^2}{0.02547}\right)]}{(-0.2917) \cdot 0.04 \mathrm{~m}} = 201.50 \mathrm{~W} / \mathrm{m}^2 \cdot \mathrm{K}\)

We have the formula for heat transfer (\(Q\)) as follows: \(Q = \rho V C_p (T_{in} - T_{f})\) Where \(\rho\) is the density, \(V\) is the volume of tomato, \(C_p\) is the specific heat capacity, \(T_{in}\) is the initial temperature, and \(T_{f}\) is the final temperature. We can calculate the volume of the tomato using its radius: \(V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (0.04\mathrm{~m})^3 = 2.68 \times 10^{-4} \mathrm{~m}^3\) Now, we can calculate the heat transfer (\(Q_{1}\)) for one tomato: \(Q_{1} = 999 \mathrm{~kg} / \mathrm{m}^{3} \cdot 2.68 \times 10^{-4} \mathrm{~m}^{3} \cdot 3.99 \times 10^{3} \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} \cdot (30.0^{\circ} \mathrm{C}-10.0^{\circ} \mathrm{C}) = 16139.23 \mathrm{~J}\) We have eight tomatoes in total, so the total amount of heat transfer is: \(Q_{total} = Q_{1} \times 8 = 16139.23 \mathrm{~J} \times 8 = 129113.85 \mathrm{~J}\) Thus, the heat transfer coefficient is 201.50 W/m²·K, and the total amount of heat transfer is 129113.85 J.