Problem 8 - A Modern Course in Statistical Physics

Question

Consider an isotropic fluid contained in a rectangular box with sides of length \(L_{x}=L, L_{y}\) \(=2 L\), and \(L_{z}=3 L\). Assume that the temperature of the fluid at time \(t=0\) has a distribution \(T(r, 0)\), but the fluid is initially at rest. Assume that the thermal expansivity of the fluid is very small so that coupling to pressure variations can be neglected. (a) Show that under these conditions the temperature variations satisfy the heat equation, \(\partial T(r, t) / \partial t=-\kappa \nabla^{2} T(r, t)\). What is \(\kappa\) ? (b) If the walls of the box conduct heat and are maintained at temperature, \(T_{0}\), approximately how long does it take for the system to reach equilibrium. (c) If the walls of the box are insulators, approximately how long does it take for the system to reach equilibrium? (Hint: No heat currents flow through the walls of insulators.)

Step-by-Step Solution

Answer

a) The temperature variations satisfy the heat equation with \( \kappa = \alpha \). b) Time to reach equilibrium with conducting walls: \( \frac{9L^2}{\kappa} \). c) Time to reach equilibrium with insulating walls: \( \frac{9L^2}{\kappa} \).

Step 1a: Understanding the Heat Equation

The heat equation is a partial differential equation representing the distribution of temperature in a given region over time. It is given by \[ \frac{\partial T(r, t)}{\partial t} = -\kappa abla^2 T(r, t) \] where \(\kappa\) is the thermal diffusivity constant.

Step 2a: Deriving the Heat Equation for an Isotropic Fluid

For an isotropic fluid at rest with negligible pressure variations, temperature variations affect the system as: \[ \frac{\partial T(r, t)}{\partial t} = abla \cdot (\alpha abla T(r, t)) \] Simplifying for an isotropic medium (constant \(\alpha\)): \[ \frac{\partial T(r, t)}{\partial t} = \alpha abla^2 T(r, t) \] Thus, comparing with the heat equation, we identify \(\kappa\) as thermal diffusivity (\(\kappa = \alpha\)).

Step 1b: Heat Conduction with Conducting Walls

Walls conducting heat held at temperature \( T_0 \). Over time, system approaches equilibrium where temperature stabilizes at \( T_0 \). Characteristic time scale defined by length, \( L \), and diffusivity, \( \kappa \): \[ t_{eq} \sim \frac{L^2}{\kappa} \] Given dimensions affect diffusivity calculation. Assuming rectangle side dimensions of 3L (longest): \[ t_{eq} \sim \frac{(3L)^2}{\kappa} \approx \frac{9L^2}{\kappa} \]

Step 2b: Computing Time to Equilibrium with Conducting Walls

Substituting assumed default \( \kappa \) value from material specifics: \[ t_{eq} \approx \frac{9L^2}{\kappa} \] Thus approximated time for system to reach equilibrium.

Step 1c: Heat Conduction with Insulating Walls

Insulating walls imply no heat flows through boundary; system in isolated thermal state. Heat transfer occurs solely within fluid. Longest dimension determines thermal conduction time across fluid: \[ t_{eq} \approx \frac{L^2}{\kappa} \] Relate longest dimension again: \(3L\). So: \[ t_{eq} \approx \frac{(3L)^2}{\kappa} \approx \frac{9L^2}{\kappa} \]

Key Concepts

Thermal Diffusivity

Thermal diffusivity, denoted as \( \kappa \), is a measure of how quickly heat spreads through a material. It is defined as the thermal conductivity divided by the product of density and specific heat capacity at constant pressure. This means that materials with high thermal diffusivity can conduct heat quickly throughout their mass. Mathematically, it is given by:
\[ \kappa = \frac{k}{\rho c_p} \]
where:

• \( k \) is thermal conductivity,
• \( \rho \) is density, and
• \( c_p \) is specific heat capacity at constant pressure.

In the context of the heat equation, \( \kappa \) appears as a constant that influences the rate of temperature change over time. It essentially determines how fast an object will reach thermal equilibrium.

Isotropic Fluid

An isotropic fluid has the same properties in all directions. This means that its thermal conductivity and other related physical properties are uniform regardless of the direction in which they are measured. For the given exercise, this simplification allows us to use a uniform thermal diffusivity, \( \kappa \), simplifying the heat equation to:

\[ \frac{\partial T(r, t)}{\partial t} = -\kappa abla^2 T(r, t) \]
In other words, the rate of change of temperature at any point within the fluid only depends on how the temperature varies spatially around that point. Having an isotropic property means we do not have to worry about directional biases when solving the equation, since the thermal characteristics are the same in every direction.

Heat Conduction

Heat conduction in a material signifies the transfer of thermal energy through the material due to temperature differences. This is described by Fourier's Law, which states that the heat flux (rate of heat transfer per unit area) is proportional to the negative gradient of temperature and the material’s thermal conductivity. For the heat equation provided:
\[ \frac{\partial T(r, t)}{\text{partial t}} = -\kappa abla^2 T(r, t) \]
This equation essentially models how temperature at a point evolves over time due to heat conduction.
When the walls of the box are conducting, heat is transferred from the fluid to the walls, causing the system to eventually reach the walls' temperature. The time taken to reach equilibrium can be estimated by the formula \( t_{eq} \approx \frac{L^2}{\kappa} \), considering the largest dimension of the box. If the walls are insulating, the system remains isolated, and the heat only spreads within the fluid. The time to equilibrium remains dependent on the thermal diffusivity and the longest dimension but assumes that no heat escapes through the boundaries.