Question

Consider a 7.6-cm-diameter cylindrical lamb meat chunk $\left(\rho=1030 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=3.49 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=0.456 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.$, \(\alpha=1.3 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\) ). 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) \(8.5 \mathrm{~min}\)

Short answer

Answer: _____ minutes

Step-by-step solution

Recall the relationship between temperature and time using the lumped capacitance method

In this case, we'll be using the lumped capacitance method to determine the relationship between temperature and time (assuming it is valid for this problem). The equation for temperature versus time is: \(T(t) - T_\infty = (T_0 - T_\infty)e^{-\frac{hA}{\rho V c_p}t}\) Where: - \(T(t)\): Temperature of the meat chunk at time \(t\) - \(T_\infty\): Temperature of the water (95°C) - \(T_0\): Initial temperature of the meat chunk (2°C) - \(h\): Heat transfer coefficient (1200 W/m²·K) - \(A\): Surface area of the cylindrical meat chunk - \(\rho\): Density of the meat (1030 kg/m³) - \(V\): Volume of the meat chunk - \(c_p\): Specific heat capacity of meat (3.49 kJ/kg·K) - \(t\): Time (s)

Check validity of lumped capacitance method

In order for the lumped capacitance method to be valid, the Biot number (Bi) must be less than 0.1. Calculate Bi as follows: \(Bi = \frac{hL_c}{k}\) Where: - \(L_c\): Characteristic length (in this case, \(L_c = \frac{V}{A}\)) - \(k\): Thermal conductivity (0.456 W/m·K)

Calculate the surface area and volume

The surface area \(A\) and volume \(V\) of a cylinder are given by the following formulas: \(A = 2 \pi r^2 + 2 \pi rh\) \(V = \pi r^2h\) Where: - \(r\): Radius of the cylinder (half of the diameter; 3.8 cm) - \(h\): Height of the cylinder (not given, assume it is equal to the diameter; 7.6 cm)

Solve the temperature-time equation for time

Now, we will express all lengths in meters (1 cm = 0.01 m) and replace the given values into the temperature-time equation. We need to find the time \(t\) when the center temperature of the meat chunk is 75°C. We will rearrange the temperature-time equation to isolate \(t\) on one side: \(t = -\frac{\rho V c_p}{hA} \ln \left(\frac{T(t) - T_\infty}{T_0 - T_\infty}\right)\) Plug in the given values and calculate the time \(t\).

Convert the time from seconds to minutes

Convert the calculated time \(t\) from seconds to minutes by dividing the result by 60. Compare the calculated time with the given options and choose the best match.