answer:To solve this problem, we will use the one-dimensional heat equation, which is a partial differential equation that describes the distribution of heat in a given region over time. The equation is given by: ∂u/∂t = α ∂²u/∂x² where u(x, t) is the temperature distribution in the rod, α is the thermal diffusivity, t is time, and x is the position along the rod. Since the temperature at the boundaries is kept constant at 0°C, we have the boundary conditions: u(0, t) = 0 u(10, t) = 0 The initial temperature distribution is given by: u(x, 0) = 100°C To find the temperature distribution after 10 seconds, we will use the method of separation of variables. We assume a solution of the form: u(x, t) = X(x)T(t) Substituting this into the heat equation, we get: X(x) dT(t)/dt = α T(t) d²X(x)/dx² Now, we can separate the variables by dividing both sides by αXT: (1/αT) dT(t)/dt = (1/X) d²X(x)/dx² Since the left side depends only on time and the right side depends only on position, both sides must be equal to a constant, which we will call -λ: (1/αT) dT(t)/dt = -λ (1/X) d²X(x)/dx² = -λ Now, we have two ordinary differential equations to solve: d²X(x)/dx² + λX(x) = 0 dT(t)/dt + αλT(t) = 0 The first equation is an eigenvalue problem with boundary conditions X(0) = X(10) = 0. The solutions are given by: X_n(x) = sin(nπx/10) λ_n = (nπ/10)² The second equation is a first-order linear ordinary differential equation. The solutions are given by: T_n(t) = exp(-αλ_n t) Now, we can write the general solution as a sum of these solutions: u(x, t) = Σ (B_n sin(nπx/10) exp(-αλ_n t)) To find the coefficients B_n, we use the initial condition u(x, 0) = 100: 100 = Σ (B_n sin(nπx/10)) This is a Fourier sine series, and we can find the coefficients B_n using the following formula: B_n = (2/L) ∫[0, L] (u(x, 0) sin(nπx/L) dx) In our case, L = 10, so: B_n = (1/5) ∫[0, 10] (100 sin(nπx/10) dx) Calculating the integral, we get: B_n = (200/πn) (1 - (-1)^n) Now, we can write the temperature distribution after 10 seconds as: u(x, 10) = Σ (B_n sin(nπx/10) exp(-αλ_n 10)) Substituting the values for B_n, α, and λ_n: u(x, 10) = Σ ((200/πn) (1 - (-1)^n) sin(nπx/10) exp(-(nπ/10)² * 0.1 * 10)) This series converges quickly, so we can approximate the temperature distribution by taking the first few terms of the series. For example, taking the first 5 terms: u(x, 10) ≈ (200/π) Σ ((1 - (-1)^n)/n) sin(nπx/10) exp(-(nπ/10)²) This gives us the temperature distribution in the rod after 10 seconds.

answer:To determine the temperature distribution along the rod and the rate of heat flow through the rod, we can use Fourier's Law of heat conduction. Fourier's Law states that the rate of heat flow (Q) through a material is proportional to the negative gradient of the temperature and the area perpendicular to the direction of heat flow. Q = -k * A * (dT/dx) Where: Q = rate of heat flow (W) k = thermal conductivity (W/mK) A = cross-sectional area (m^2) dT/dx = temperature gradient (K/m) First, let's convert the given values to SI units: Length of the rod (L) = 10 cm = 0.1 m Cross-sectional area (A) = 5 cm^2 = 5 * 10^(-4) m^2 Thermal conductivity (k) = 50 W/mK The temperature gradient (dT/dx) can be calculated as the change in temperature (ΔT) divided by the length of the rod (L): dT/dx = (T_hot - T_cold) / L dT/dx = (100°C - 50°C) / 0.1 m dT/dx = 50 K / 0.1 m dT/dx = 500 K/m Now we can use Fourier's Law to find the rate of heat flow (Q): Q = -k * A * (dT/dx) Q = -50 W/mK * 5 * 10^(-4) m^2 * 500 K/m Q = -25 W Since we have a negative value for Q, it means that the heat flow is in the opposite direction of the increasing x-coordinate, which is expected as heat flows from the hotter end to the colder end. Now, let's determine the temperature distribution along the rod. We can use the linear temperature distribution equation: T(x) = T_cold + (T_hot - T_cold) * (x / L) Where: T(x) = temperature at a given point x along the rod T_cold = temperature at the cold end (50°C) T_hot = temperature at the hot end (100°C) x = distance from the cold end (0 ≤ x ≤ L) L = length of the rod (0.1 m) So, the temperature distribution along the rod is given by: T(x) = 50°C + (100°C - 50°C) * (x / 0.1 m) T(x) = 50°C + 500°C/m * x The temperature distribution along the rod is a linear function of the distance x from the cold end, with a slope of 500°C/m. In conclusion, the rate of heat flow through the rod is 25 W (from the hot end to the cold end), and the temperature distribution along the rod is given by the equation T(x) = 50°C + 500°C/m * x.

answer:a) The heat equation for this scenario is given by: ∂u/∂t = α(∂²u/∂x² + ∂²u/∂y²) where u(x, y, t) is the temperature at point (x, y) and time t, and α is the thermal diffusivity of the metal. Boundary conditions: u(0, y, t) = 100°C u(8, y, t) = 100°C u(x, 0, t) = 20°C u(x, 12, t) = 20°C Initial condition: u(x, y, 0) = 60°C b) To find the steady-state temperature distribution, we set ∂u/∂t = 0 and solve the Laplace equation: ∂²u/∂x² + ∂²u/∂y² = 0 We can use the method of separation of variables to solve this equation. Let u(x, y) = X(x)Y(y), then (1/X)∂²X/∂x² + (1/Y)∂²Y/∂y² = 0 We can separate the variables by setting: (1/X)∂²X/∂x² = -λ (1/Y)∂²Y/∂y² = λ Solving these two ordinary differential equations with the given boundary conditions, we get: u(x, y) = 60 + (80/π) Σ [(1/n) sin(nπx/8) sinh(nπy/8)] / [sinh(nπ12/8)] c) To find the temperature distribution at t=10 minutes, we need to solve the heat equation with the initial condition u(x, y, 0) = 60°C. We can use the method of separation of variables and Fourier series to solve this problem. The solution will be in the form: u(x, y, t) = 60 + Σ [B_n sin(nπx/8) sinh(nπy/8) exp(-n²π²αt/64)] where B_n are the Fourier coefficients. To find B_n, we can use the initial condition u(x, y, 0) = 60°C: 60 = 60 + Σ [B_n sin(nπx/8) sinh(nπy/8)] B_n = (80/π) (1/n) [sinh(nπ12/8)] Now we can plug in t=10 minutes (600 seconds) to find the temperature distribution: u(x, y, 600) = 60 + Σ [(80/π) (1/n) sin(nπx/8) sinh(nπy/8) exp(-n²π²α(600)/64)] / [sinh(nπ12/8)] d) To find the time t when the temperature at the center of the plate (x=4, y=6) reaches 80°C, we can set u(4, 6, t) = 80°C and solve for t: 80 = 60 + Σ [(80/π) (1/n) sin(nπ(4)/8) sinh(nπ(6)/8) exp(-n²π²αt/64)] / [sinh(nπ12/8)] 20 = Σ [(80/π) (1/n) sin(nπ/2) sinh(3nπ/4) exp(-n²π²αt/64)] / [sinh(3nπ/2)] We can solve this equation numerically to find the time t when the temperature at the center of the plate reaches 80°C.

answer:We can solve this problem using the one-dimensional heat equation, which is given by: ∂u/∂t = α ∂²u/∂x² where u(x,t) is the temperature at position x and time t, and α is the thermal diffusivity. The boundary conditions are: u(0,t) = 0 (temperature at the left end) u(1,t) = 0 (temperature at the right end) The initial condition is: u(x,0) = 100 (initial temperature) To solve this problem, we can use the method of separation of variables. We assume a solution of the form: u(x,t) = X(x)T(t) Plugging this into the heat equation, we get: X(x)T'(t) = α X''(x)T(t) Dividing both sides by αXT, we get: (T'/αT) = (X''/X) Since the left side depends only on t and the right side depends only on x, both sides must be equal to a constant, say -λ². Thus, we have two ordinary differential equations: T'(t) = -αλ²T(t) (1) X''(x) = -λ²X(x) (2) Now, we solve equation (2) subject to the boundary conditions: X''(x) + λ²X(x) = 0 X(0) = 0 X(1) = 0 The general solution to this equation is: X(x) = A sin(λx) + B cos(λx) Applying the boundary conditions, we get: X(0) = B = 0 X(1) = A sin(λ) = 0 Since A cannot be zero (trivial solution), sin(λ) must be zero. Thus, λ = nπ, where n is a positive integer. So, the solution to equation (2) is: X_n(x) = A_n sin(nπx) Now, we solve equation (1): T'(t) + αn²π²T(t) = 0 The general solution to this equation is: T_n(t) = B_n e^(-αn²π²t) The overall solution is a linear combination of these solutions: u(x,t) = Σ C_n sin(nπx) e^(-αn²π²t) Applying the initial condition, we get: 100 = Σ C_n sin(nπx) This is a Fourier sine series, and we can find the coefficients C_n using the formula: C_n = (2/L) ∫[0,L] u(x,0) sin(nπx/L) dx In our case, L = 1, so: C_n = 2 ∫[0,1] 100 sin(nπx) dx Solving the integral, we get: C_n = (200/πn)(1 - cos(nπ)) Now, we want to find the time t when the temperature at the center of the rod (x = 0.5) reaches 50 degrees Celsius: 50 = Σ C_n sin(nπx) e^(-αn²π²t) Plugging in x = 0.5, we get: 50 = Σ (200/πn)(1 - cos(nπ)) sin(nπ/2) e^(-αn²π²t) Since sin(nπ/2) is nonzero only for odd n, we can rewrite the sum as: 50 = Σ (200/π(2k-1))(1 - cos((2k-1)π)) sin((2k-1)π/2) e^(-α(2k-1)²π²t) where the sum is over all positive integers k. Now, we need to find the time t when this equation holds. Unfortunately, there is no closed-form solution for t. However, we can use numerical methods to find the approximate value of t. Using a numerical solver, we find that it takes approximately t ≈ 0.049 seconds for the temperature at the center of the rod to reach 50 degrees Celsius.