1. **Problem Statement:** We want to model heat flow in a copper bar of length 10 cm, insulated on the sides, with ends held at 0°C, and initial temperature distribution given by $f(x) = 4 - 0.8 |x - 5|$. 2. **Mathematical Model of Heat Flow:** Heat flow in a solid is modeled by the heat equation: $$\frac{\partial u}{\partial t} = \alpha \frac{\partial^2 u}{\partial x^2}$$ where $u(x,t)$ is the temperature at position $x$ and time $t$, and $\alpha$ is the thermal diffusivity. 3. **Thermal Diffusivity Calculation:** Thermal diffusivity $\alpha$ is given by: $$\alpha = \frac{k}{\rho c}$$ where - $k$ = thermal conductivity = 0.94 cal/(cm s °C) - $\rho$ = density = 8.96 g/cm³ - $c$ = specific heat = 0.093 cal/(g °C) Calculate $\alpha$: $$\alpha = \frac{0.94}{8.96 \times 0.093} = \frac{0.94}{0.83328} \approx 1.128$$ 4. **Boundary Conditions:** The bar is insulated on the sides and ends are held at 0°C: $$u(0,t) = 0, \quad u(10,t) = 0$$ 5. **Initial Condition:** $$u(x,0) = f(x) = 4 - 0.8 |x - 5|$$ 6. **Solution Method:** Use separation of variables and Fourier series expansion. The eigenfunctions for the boundary conditions are: $$\phi_n(x) = \sin\left(\frac{n \pi x}{10}\right), \quad n=1,2,3,...$$ The solution is: $$u(x,t) = \sum_{n=1}^\infty b_n e^{-\alpha \left(\frac{n \pi}{10}\right)^2 t} \sin\left(\frac{n \pi x}{10}\right)$$ 7. **Fourier Coefficients $b_n$:** $$b_n = \frac{2}{10} \int_0^{10} f(x) \sin\left(\frac{n \pi x}{10}\right) dx = \frac{1}{5} \int_0^{10} (4 - 0.8 |x - 5|) \sin\left(\frac{n \pi x}{10}\right) dx$$ Split integral at $x=5$ due to absolute value: $$b_n = \frac{1}{5} \left[ \int_0^5 (4 - 0.8 (5 - x)) \sin\left(\frac{n \pi x}{10}\right) dx + \int_5^{10} (4 - 0.8 (x - 5)) \sin\left(\frac{n \pi x}{10}\right) dx \right]$$ Simplify inside integrals: - For $0 \leq x \leq 5$: $f(x) = 4 - 0.8(5 - x) = 4 - 4 + 0.8x = 0.8x$ - For $5 < x \leq 10$: $f(x) = 4 - 0.8(x - 5) = 4 - 0.8x + 4 = 8 - 0.8x$ So: $$b_n = \frac{1}{5} \left[ \int_0^5 0.8 x \sin\left(\frac{n \pi x}{10}\right) dx + \int_5^{10} (8 - 0.8 x) \sin\left(\frac{n \pi x}{10}\right) dx \right]$$ 8. **Final Temperature Distribution:** $$u(x,t) = \sum_{n=1}^\infty b_n e^{-1.128 \left(\frac{n \pi}{10}\right)^2 t} \sin\left(\frac{n \pi x}{10}\right)$$ This series converges to the temperature at any position $x$ and time $t$. --- **Summary:** - Heat equation with $\alpha \approx 1.128$ - Boundary conditions $u(0,t)=u(10,t)=0$ - Initial condition $f(x) = 4 - 0.8 |x-5|$ - Solution via Fourier sine series with coefficients $b_n$ as above This fully models the heat flow and temperature distribution in the copper bar over time.