Question

After a long, hard week on the books, you and your friend are ready to relax and enjoy the weekend. You take a steak \(50 \mathrm{~mm}\) thick from the freezer. (a) How long (in hours) do you have to let the good times roll before the steak has thawed? Assume that the steak is initially at $-8^{\circ} \mathrm{C}$, that it thaws when the temperature at the center of the steak reaches \(4^{\circ} \mathrm{C}\), and that the room temperature is $22^{\circ} \mathrm{C}\( with a convection heat transfer coefficient of \)10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Neglect the heat of fusion associated with the melting phase change. Treat the steak as a one-dimensional plane wall having the following properties: \(\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}\), \(c_{p}=4472 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and $k=0.625 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. (b) How much energy per unit area (in J/m²) has been removed from the steak during this period of thawing? (c) Show whether or not the thawing of this steak can be analyzed by neglecting the internal thermal resistance of the steak.

Short answer

2. How long does it take for the steak to thaw, in hours? 3. What is the energy per unit area removed from the steak during the thawing period? 4. Can the internal thermal resistance of the steak be neglected in this problem?

Step-by-step solution

Find the temperature difference

First, we need to find the difference in temperature between the center of the steak and the room. The center of the steak goes from -8°C to 4°C, and the room is at 22°C. The temperature difference is: \(\Delta T = T_{room} - T_{center} = 22^{\circ} \mathrm{C} - 4^{\circ} \mathrm{C} = 18^{\circ} \mathrm{C}\).

Calculate the thawing time

Next, we will use the convection heat transfer formula to find the thawing time, which is given by: \(t = \frac{\rho h c_p}{k} \frac{\Delta T}{h_c}\) Here, \(h = 0.05 \mathrm{~m}\) (thickness of the steak), \(h_c = 10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (convection heat transfer coefficient), and the steak's properties are given. Plugging in the values, we get: \(t = \frac{1000 \mathrm{~kg} / \mathrm{m}^{3} \cdot 0.05\mathrm{~m} \cdot 4472 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}}{0.625 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}} \cdot \frac{18^{\circ} \mathrm{C}}{10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}}\) \(t = 673.44 \mathrm{~s}\). Since we need to find the time in hours, we will convert as follows: \(t_{hours} = \frac{673.44 \mathrm{~s}}{3600 \mathrm{~s/hour}} = 0.1871 \mathrm{~hours}\).

Calculate the energy per unit area removed from the steak

Since we know the thawing time, we can now calculate the total energy lost per unit area during this period of thawing using the following equation: \(Q = h_c \cdot \Delta T \cdot t\) So, \(Q = 10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \times 18^{\circ} \mathrm{C} \times 673.44 \mathrm{~s}\) \(Q = 121184 \mathrm{~J/m^2}\).

Analyze the internal thermal resistance of the steak

The internal thermal resistance can be determined by analyzing the Biot number, which is a dimensionless number that represents the ratio of internal resistance to surface resistance. It is given by: \(Bi = \frac{h L}{k}\) , where \(L\) is the characteristic length of the steak (in our case, L=h). If \(Bi < 0.1\), it is considered that the internal thermal resistance is negligible. Let's calculate the Biot number for the given steak: \(Bi = \frac{10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \cdot 0.05 \mathrm{~m}}{0.625 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}}\) \(Bi = 0.8\) Since \(Bi > 0.1\), we can conclude that the internal thermal resistance of the steak cannot be neglected in this problem. In summary, the time it takes for the steak to thaw is approximately 0.19 hours, the energy per unit area removed from the steak during this period of thawing is 121,184 J/m², and the internal thermal resistance of the steak cannot be neglected in this analysis.