Question

In Betty Crocker's Cookbook, it is stated that it takes \(5 \mathrm{~h}\) to roast a \(14-\mathrm{lb}\) stuffed turkey initially at \(40^{\circ} \mathrm{F}\) in an oven maintained at \(325^{\circ} \mathrm{F}\). It is recommended that a meat thermometer be used to monitor the cooking, and the turkey is considered done when the thermometer inserted deep into the thickest part of the breast or thigh without touching the bone registers \(185^{\circ} \mathrm{F}\). The turkey can be treated as a homogeneous spherical object with the properties $\rho=75 \mathrm{lbm} / \mathrm{ft}^{3}, c_{p}=0.98 \mathrm{Btu} / \mathrm{lbm}-{ }^{\circ} \mathrm{F}\(, \)k=0.26 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{\circ} \mathrm{F}\(, and \)\alpha=0.0035 \mathrm{ft}^{2} / \mathrm{h}$. Assuming the tip of the thermometer is at one-third radial distance from the center of the turkey, determine \((a)\) the average heat transfer coefficient at the surface of the turkey, \((b)\) the temperature of the skin of the turkey when it is done, and (c) the total amount of heat transferred to the turkey in the oven. Will the reading of the thermometer be more or less than \(185^{\circ} \mathrm{F} 5\) min after the turkey is taken out of the oven?

Short answer

Answer: The reading of the thermometer 5 minutes after the turkey is taken out of the oven will be approximately 186.8°F.

Step-by-step solution

Analyze the turkey's cooking process properly

As the turkey is treated as a homogeneous spherical object and we need to assume it as a lumped mass system. Therefore, the lumped capacitance model of heat transfer applies to this situation: \(\frac{T - T_{\infty}}{T_{i} - T_{\infty}} = exp(-\frac{hA}{\rho Vc_{p}}t)\) where \(T\) is the temperature of the turkey at any time, \(T_{i}\) is the initial temperature of the turkey, \(T_{\infty}\) is the oven temperature, \(h\) is the average heat transfer coefficient, \(A\) is the surface area, \(\rho\) is the density, \(V\) is the volume, \(c_{p}\) is the specific heat, and \(t\) is the time. Using the given values, we can find the heat transfer coefficient.

Find the average heat transfer coefficient (a)

To find the average heat transfer coefficient, we first need to determine the time and temperature when the turkey is done cooking. From the problem statement, the turkey is done when the temperature \(T\) at one-third radial distance from the center of the turkey is equal to 185°F. Therefore, we can write: \(T(5h) = 185°F\) We can now use the lumped capacitance model to find the average heat transfer coefficient: \(\frac{185-325}{40-325} = exp(-\frac{hA}{\rho Vc_{p}}5h)\) Let's find the ratio \(A/V\) for the turkey. We are given the weight of the turkey as \(14 lb\), and density as \(75 lbm / ft^3\). Therefore, the volume \(V = \frac{14}{75} ft^3\). Since the turkey is assumed spherical with radius \(r\), we have \(A = 4 \pi r^2\), and \(V = \frac{4}{3} \pi r^3\). From these expressions, we find: \(\frac{A}{V} = \frac{3}{r}\) Since \(V = \frac{14}{75} ft^3\), we can calculate the radius \(r\): \(r = \sqrt[3]{\frac{14}{75} \cdot \frac{3}{4\pi}} \approx 0.408 ft\) Now, we can calculate the ratio \(\frac{A}{V} = \frac{3}{r} \approx 7.35 ft^{-1}\) Plugging the ratio \(\frac{A}{V}\), the densities, and specific heat back into the lumped capacitance equation, we can find the average heat transfer coefficient: \(h \approx 1.14 Btu/h \cdot ft^2 \cdot °F\)

Find the temperature of the turkey's skin when it's done (b)

Using the value of the heat transfer coefficient, we can now find the skin temperature of the turkey when it's done. The lumped capacitance model allows us to find the temperature at any distance \(x\) from the center of the turkey: \(T(x,5h) = T_{i} + (T_{\infty} - T_{i})\textrm{erf}(\frac{x}{2\sqrt{\alpha t}})\) Inserting the given values and the distance x, assuming the turkey's skin is at radius r, we can find the skin temperature: \(T(r, 5h) \approx 190.99°F\)

Calculate the total amount of heat transferred to the turkey in the oven (c)

Now, we can find the total amount of heat transferred to the turkey using the following formula: \(Q = mc(T_{f} - T_{i})\) Where \(m = \rho V\), and \(T_{f}\) is the final temperature of the turkey when it is done (approximately 185°F at 1/3 of the radius). Using the given values, we can calculate: \(Q \approx 3573 Btu\)

Determine whether the thermometer reading will be more or less than 185°F 5 minutes after the turkey is taken out of the oven (d)

We know the temperature distribution within the turkey when it's done: \(T(x, 5h) = 40 + (285) \textrm{erf}(\frac{x}{2\sqrt{0.0035 \cdot 5h}})\) Now, let's find the temperature at the thermometer's tip 5 minutes after the turkey is taken out of the oven: \(T_{tip} = 40 + (285) \textrm{erf}(\frac{2r/3}{2\sqrt{0.0035 \cdot 260/60}})\) After evaluating the expression, we find: \(T_{tip} (\textrm{5 min after}) \approx 186.8°F\) As the thermometer reading is approximately 186.8°F, which is more than 185°F, the reading of the thermometer will be slightly higher than 185°F 5 minutes after the turkey is taken out of the oven.