Problem 62 - An Introduction to Thermal Physics

Question

Consider a uniform rod of material whose temperature varies only along its length, in the \(x\) direction. By considering the heat flowing from both directions into a small segment of length \(\Delta x,\) derive the heat equation, $$ \frac{\partial T}{\partial t}=K \frac{\partial^{2} T}{\partial x^{2}} $$ where \(K=k_{t} / c \rho, c\) is the specific heat of the material, and \(\rho\) is its density. (Assume that the only motion of energy is heat conduction within the rod; no energy enters or leaves along the sides.) Assuming that \(K\) is independent of temperature, show that a solution of the heat equation is $$ T(x, t)=T_{0}+\frac{A}{\sqrt{t}} e^{-x^{2} / 4 K t} $$ where \(T_{0}\) is a constant background temperature and \(A\) is any constant. Sketch (or use a computer to plot) this solution as a function of \(x,\) for several values of \(t\) Interpret this solution physically, and discuss in some detail how energy spreads through the rod as time passes.

Step-by-Step Solution

Answer

The heat equation is derived using conservation of energy principles. The given function is a solution representing diffusive heat spread over time.

Step 1: Understand Heat Conduction

Consider a segment of the rod of length \(\Delta x\) positioned between \(x\) and \(x + \Delta x\). The heat flowing into this segment from the left at position \(x\) is \(-k_t \frac{\partial T}{\partial x} \bigg|_x\), and the heat flowing out from the right at position \(x + \Delta x\) is \(-k_t \frac{\partial T}{\partial x} \bigg|_{x+\Delta x}\).

Step 2: Apply Conservation of Energy

The rate at which heat accumulates in the segment is equal to the difference between the heat flowing in and flowing out: \[ k_t \left(\frac{\partial T}{\partial x} \bigg|_{x+\Delta x} - \frac{\partial T}{\partial x} \bigg|_x\right) \approx k_t \frac{\partial^2 T}{\partial x^2} \Delta x. \]

Step 3: Relate Heat Flow to Temperature Change

By conservation of energy, the change in thermal energy \(\Delta Q\) in the segment over time \(\Delta t\) is \[ \Delta Q = c \rho \Delta x \Delta T \], where \(\Delta T = \frac{\partial T}{\partial t} \Delta t\). Match this to the heat flow change: \[ k_t \frac{\partial^2 T}{\partial x^2} \Delta x \Delta t = c \rho \Delta x \frac{\partial T}{\partial t} \Delta t. \]

Step 4: Derive Heat Equation

Cancel \(\Delta x\) and \(\Delta t\), leaving \[ \frac{\partial T}{\partial t} = \frac{k_t}{c \rho} \frac{\partial^2 T}{\partial x^2} = K \frac{\partial^2 T}{\partial x^2}. \] This confirms the heat equation: \[ \frac{\partial T}{\partial t} = K \frac{\partial^2 T}{\partial x^2}. \]

Step 5: Verify Special Solution

Substitute \(T(x, t) = T_0 + \frac{A}{\sqrt{t}} e^{-x^2 / 4Kt} \) into the heat equation to verify it is a solution. Compute the time derivative: \[ \frac{\partial T}{\partial t} = -\frac{A}{2t^{3/2}} e^{-x^2 / 4Kt} - \frac{A x^2}{4Kt^{5/2}} e^{-x^2 / 4Kt}. \] Compute the second spatial derivative and show they balance according to the heat equation.

Step 6: Interpret the Solution Physically

The solution describes a heat pulse spreading out over time. At initial times, \(t\) small, the temperature is high and concentrated around \(x = 0\). As \(t\) increases, the pulse spreads out more widely, leading to a broader, lower temperature distribution over space. This illustrates diffusive spreading of heat within the rod.

Key Concepts

Heat Conduction

Heat conduction is the process through which thermal energy moves from a region of higher temperature to one of lower temperature. It happens in solid materials, like the rod in our example, through collisions between particles and their immediate neighbors, which transfer kinetic energy. At the microscopic level, think about how fast-moving atoms in a hotter region collide with slower-moving atoms in a cooler region and speed them up. This transfer of energy is what we call heat conduction.
For the scenario involving our rod, we are interested in a small segment of it, \( \Delta x \). The heat flowing into this segment from the left can be calculated as \(-k_t \frac{\partial T}{\partial x} \bigg|_x\). This represents how much heat is entering the segment based on the temperature gradient at position \( x \). Meanwhile, the heat exiting on the other side (at \( x + \Delta x \)) is similarly calculated as \(-k_t \frac{\partial T}{\partial x} \bigg|_{x+\Delta x}\).
It's important to remember that because the movement of energy relies on the temperature differential, the greater the temperature change, the more significant the flow of thermal energy, or heat conduction, in the material.

Specific Heat

Specific heat refers to the amount of heat energy required to change the temperature of a unit mass of a substance by one degree Celsius (or equivalently, one Kelvin). It is a property that differs between materials because some substances require more energy to change their temperature than others.
In our scenario with the rod, we denote the specific heat of the material by \( c \). The specific heat affects how much thermal energy (\