This problem involves heat conduction along a rod with insulated sides, which can be modeled using the heat equation. In summary:

The heat equation for a one-dimensional rod is:

$\frac{\partial U(x,t)}{\partial t} = \alpha \frac{\partial^2 U(x,t)}{\partial x^2}$

where:

• $U(x,t)$ is the temperature at position $x$ and time $t$.
• $\alpha$ is the thermal diffusivity constant of the rod.
• $0 \leq x \leq L$, where $L = 30 \, \text{cm}$.

We need to account for the initial conditions and the change in boundary conditions at the ends of the rod.

---

Step 1: Initial Conditions

The temperature distribution at $t = 0$ (when steady-state prevails) is a linear function between 20°C at $x = 0$ and 80°C at $x = 30$ cm:

$U(x, 0) = 20 + \frac{80 - 20}{30} \cdot x = 20 + 2x$

Thus, the initial temperature distribution is:

$U(x, 0) = 20 + 2x$

---

Step 2: Boundary Conditions

After the sudden change, the ends of the rod are maintained at 0°C. This gives us new boundary conditions:

$U(0, t) = 0, \quad U(30, t) = 0 \quad \text{for all} \quad t > 0$

---

Step 3: Solving the Heat Equation

The general solution to the heat equation with boundary conditions $U(0, t) = U(L, t) = 0$ is given by the Fourier sine series:

$U(x, t) = \sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi x}{L}\right) e^{-\alpha \left(\frac{n\pi}{L}\right)^2 t}$

The coefficients $b_n$ are determined by the initial temperature distribution $U(x, 0)$.

---

Step 4: Finding Fourier Coefficients $b_n$

We use the formula for the Fourier sine series coefficients:

$b_n = \frac{2}{L} \int_0^L \left(20 + 2x\right) \sin\left(\frac{n\pi x}{L}\right) \, dx$

Here, $L = 30 \, \text{cm}$. We need to solve the integral:

$b_n = \frac{2}{30} \int_0^{30} \left(20 + 2x\right) \sin\left(\frac{n\pi x}{30}\right) \, dx$

This integral can be evaluated using integration by parts.

---

Step 5: Writing the Temperature Function $U(x, t)$

Once the coefficients $b_n$ are found, the final solution will be:

$U(x, t) = \sum_{n=1}^{\infty} b_n \sin\left(\frac{n\pi x}{30}\right) e^{-\alpha \left(\frac{n\pi}{30}\right)^2 t}$

This function gives the temperature distribution along the rod at any position $x$ and time $t$.

---

Would you like me to continue with the detailed evaluation of the coefficients $b_n$?

---

Related Questions:

1. How does the Fourier sine series method work in solving heat equations?
2. What assumptions are involved in applying the heat equation to a rod?
3. How does the thermal diffusivity $\alpha$ affect the temperature distribution over time?
4. What other boundary conditions can arise in heat conduction problems?
5. How do numerical methods compare with analytical methods for heat conduction problems?

---

Tip: Fourier series solutions work well for heat equations, but for more complex geometries or variable thermal properties, numerical methods like the finite difference method are often required.