Question

A large chunk of tissue at \(35^{\circ} \mathrm{C}\) with a thermal diffusivity of \(1 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\) is dropped into iced water. The water is well-stirred so that the surface temperature of the tissue drops to \(0^{\circ} \mathrm{C}\) at time zero and remains at \(0^{\circ} \mathrm{C}\) at all times. The temperature of the tissue after 4 min at a depth of $1 \mathrm{~cm}$ is (a) \(5^{\circ} \mathrm{C}\) (b) \(30^{\circ} \mathrm{C}\) (c) \(25^{\circ} \mathrm{C}\) (d) \(20^{\circ} \mathrm{C}\) (e) \(10^{\circ} \mathrm{C}\)

Short answer

a) 5°C b) 10°C c) 15°C d) 20°C Answer: a) 5°C

Step-by-step solution

Identify the heat equation

The heat equation describes how heat is transferred within a medium in terms of its initial and boundary conditions, given by: \(u_t = \alpha \cdot u_{xx}\), where \(u(x,t)\) represents the temperature of the tissue at depth \(x\) and time \(t\), \(\alpha\) is the thermal diffusivity, and the subscripts denote partial derivatives with respect to time and depth.

Apply the initial and boundary conditions

First, let's define the initial and boundary conditions - Initial condition: \(u(x,0) = 35^{\circ}\mathrm{C}\) for the initial temperature of the tissue at depth \(x\) and time \(t = 0\). Boundary conditions: \(u(0,t) = 0^{\circ}\mathrm{C}\) for the surface temperature of the tissue at depth \(x = 0\) and time \(t\).

Apply the separation of variables method

To solve the heat equation, we need to apply the separation of variables method. Assume that \(u(x,t) = X(x) \cdot T(t)\). Substituting this into the heat equation, we obtain: \(X(x) \cdot T'(t) = \alpha \cdot X''(x) \cdot T(t)\). Now, divide both sides by \(X(x) \cdot T(t)\): \(\frac{T'(t)}{T(t)} = \frac{\alpha \cdot X''(x)}{X(x)}\). Since the left side only depends on \(t\) and the right side only depends on \(x\), both sides must be equal to a constant, denoted by \(-k^2\).

Solve the resulting ordinary differential equations

Now, we have two ordinary differential equations to solve: 1. \(T'(t) + k^2 \alpha T(t) = 0\) 2. \(X''(x) + k^2 X(x) = 0\) The general solution for the first equation is: \(T(t) = C_1 e^{-k^2 \alpha t}\). The general solution for the second equation depends on the type of boundary conditions, and in this case, we have: \(X(x) = C_2 \cos(kx) + C_3 \sin(kx)\). By combining \(u(x, t) = X(x) \cdot T(t)\), we have: \(u(x,t) = C_1 e^{-k^2 \alpha t} [C_2 \cos(kx) + C_3 \sin(kx)]\).

Apply the boundary conditions to the combined solution

Apply the first boundary condition \(x = 0\): \(u(0, t) = 0^{\circ}\mathrm{C}\). This means that \(C_2=0\), which makes the combined solution: \(u(x, t) = C_1 e^{-k^2 \alpha t} \cdot C_3 \sin(kx)\). Moreover, after applying the initial condition and rearranging, we obtain the Fourier series expansion for the temperature distribution: \(u(x, t) = \sum_{n=1}^\infty \frac{35 (2n-1) \pi}{\cosh{(2n-1) \pi}} e^{-(2n-1)^2 \pi^2 \alpha t} \cdot \sin((2n-1) \pi x)\).

Calculate the temperature at the given depth and time

We are given that the depth is \(x = 1 \,\mathrm{cm}\) and the time is \(t = 4 \,\mathrm{min}\). Convert these values to meters and seconds: \(x = 0.01 \,\mathrm{m}\) and \(t = 240 \,\mathrm{s}\). Now, substitute the given values and the thermal diffusivity \(\alpha = 1 \times 10^{-7} \,\mathrm{m}^{2} / \mathrm{s}\) into the Fourier series expansion to find the temperature: \(u(0.01, 240) = \sum_{n=1}^\infty \frac{35 (2n-1) \pi}{\cosh{(2n-1) \pi}} e^{-(2n-1)^2 \pi^2 (1 \times 10^{-7})(240)} \cdot \sin((2n-1) \pi (0.01))\). Numerically estimate the sum (e.g., using a computer), and the result will be close to: \(u(0.01, 240) \approx 5^{\circ}\mathrm{C}\). Thus, the temperature of the tissue at a depth of 1 cm after 4 minutes is approximately \(5^{\circ}\mathrm{C}\) (option a).