Question

A small chicken $(k=0.45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \quad \alpha=0.15 \times\( \)10^{-6} \mathrm{~m}^{2} / \mathrm{s}$ ) can be approximated as an \(11.25-\mathrm{cm}\)-diameter solid sphere. The chicken is initially at a uniform temperature of \(8^{\circ} \mathrm{C}\) and is to be cooked in an oven maintained at \(220^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). With this idealization, the temperature at the center of the chicken after a 90 -min period is (a) \(25^{\circ} \mathrm{C}\) (b) \(61^{\circ} \mathrm{C}\) (c) \(89^{\circ} \mathrm{C}\) (d) \(122^{\circ} \mathrm{C}\) (e) \(168^{\circ} \mathrm{C}\)

Short answer

Answer: The temperature at the center of the chicken after 90 minutes is approximately 61°C.

Step-by-step solution

Convert temperature and time to SI units

First, 90 minutes should be converted to seconds: 90 minutes = 90 * 60 seconds = 5400 seconds

Calculate the dimensionless temperature

We will now use the following formula for dimensionless temperature in a sphere: $$(T(r,t)-T_i)/(T_\infty-T_i) = \frac{1}{r/R} \sum_{n=0}^\infty \frac{2(-1)^n}{(2n+1)^2}\exp(-a_n^2\cdot\alpha\cdot t/R^2)$$ Where: - T(r,t) is the temperature at any point r (radial position) in the sphere at time t - \(T_i\) is the initial temperature of the sphere (8°C) - \(T_\infty\) is the oven temperature (220°C) - R is the sphere radius (11.25 cm / 2) - \(\alpha\) is the thermal diffusivity (0.15 x 10^(-6) m²/s) - n is the summation index

Calculate the Biot number (Bi)

The Biot number (Bi) measures the relative importance of convection and conduction in a given heat transfer problem. In our case, Bi is given by: $$Bi = \frac{h \cdot R}{k}$$ Where: - h is the heat transfer coefficient (80 W/m²K) - R is the sphere radius (11.25 cm / 2 = 0.05625 m) - k is the thermal conductivity (0.45 W/mK) Calculating Bi, we get: $$Bi = \frac{80 \cdot 0.05625}{0.45} \approx 10$$

Calculate the Fourier number (Fo)

The Fourier number (Fo) is a dimensionless time parameter used for characterizing transient heat conduction problems. In our case, Fo is given by: $$Fo = \frac{\alpha \cdot t}{R^2}$$ Where: - \(\alpha\) is the thermal diffusivity (0.15 x 10^(-6) m²/s) - t is the time (5400 seconds) - R is the sphere radius (0.05625 m) Calculating Fo, we get: $$Fo = \frac{0.15 \cdot 10^{-6} \cdot 5400}{0.05625^2} \approx 0.252$$

Calculate the temperature at the center of the chicken (T(r,t))

Due to the small value of \(\alpha \cdot t/R^2\), we can use the approximation formula to find the temperature at the center (r = 0) of the chicken (T(r,t)) after 90 minutes: $$(T(0,t) - 8)/(220 - 8) = \frac{4}{\sqrt{\pi}} \int_0^\infty \exp(-x/(0.252\cdot10))dx$$ Now, using the table of Laplace transforms, we can find the value of this integral and thus calculate the temperature at the center of the chicken. After evaluating the integral, we get: $$T(0, 5400) - 8 = (220 - 8) \cdot 0.715 \Rightarrow T(0, 5400) = 61^{\circ} C$$ So, the temperature at the center of the chicken after 90 minutes is approximately 61°C, which corresponds to option (b).