Question

A \(0.3\)-cm-thick, 12-cm-high, and 18-cm-long circuit board houses 80 closely spaced logic chips on one side, each dissipating \(0.06 \mathrm{~W}\). The board is impregnated with copper fillings and has an effective thermal conductivity of \(16 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). All the heat generated in the chips is conducted across the circuit board and is dissipated from the back side of the board to the ambient air. Determine the temperature difference between the two sides of the circuit board. Answer: \(0.042^{\circ} \mathrm{C}\)

Short answer

The temperature difference between the two sides of the circuit board is 0.042°C.

Step-by-step solution

Find the total heat generated by the chips

First, we need to find the total amount of heat generated by the 80 chips. Given that each chip dissipates 0.06 W, we can find the total heat generated as follows: Total heat generated(Q) = Number of chips × Heat generated by each chip Q = 80 × 0.06 W

Calculate the area of the board

Now, we need to find the area of the board where the heat is being dissipated. The board is 12 cm high and 18 cm long, so the area (\(A\)) can be calculated as follows: A = height × length = 12 cm × 18 cm = 216 cm^2 Convert the area to m^2 A = 216 cm^2 × (0.01 m / 1 cm)^2 = 0.0216 m^2

Calculate the heat transfer rate

To calculate the heat transfer rate, we will use Fourier's Law of heat conduction. The law states that the heat transfer through a material is proportional to the temperature gradient and the thermal conductivity of the material: q = -k × (dT/dx) Where q is the heat transfer rate (W/m^2), k is the thermal conductivity (16 W/m⋅K), dT is the temperature difference (T2 - T1), and dx is the thickness of the material (0.3 cm).

Solve for the temperature difference

Now, we will rearrange the equation and solve for dT: q = -k × (dT/dx) -> dT = - (q × dx) / k We also need to convert the thickness of the board to meters and multiply q by the area, A: dx = 0.3 cm × (0.01 m / 1 cm) = 0.003 m Total heat transfer rate (Q) = q × A dT = - (Q × dx) / (k × A) Now, substitute the values we found in Steps 1-3 and solve for dT: Total heat generated Q = 80 × 0.06 W = 4.8 W dT = - (4.8 W × 0.003 m) / (16 W/m⋅K × 0.0216 m^2) = 0.042°C The temperature difference between the two sides of the circuit board is 0.042°C.

Key concepts

Fourier's Law of heat conduction

When dealing with heat conduction in solids, one of the most fundamental principles we encounter is Fourier's Law of heat conduction. This law offers a simple yet powerful relationship that explains how heat energy moves through a material. Imagine you're holding one end of a metal rod, and the other end is heated. The heat doesn't teleport to your hand, right? Instead, it travels gradually, moving from atom to atom in the metal. Fourier's Law quantifies this process, stating that the heat transfer rate (\( q \) ) is directly proportional to the negative gradient of temperatures and the area through which heat is being transferred.

Mathematically, it is represented as \( q = -k \frac{dT}{dx} \), where \( k \) denotes the thermal conductivity of the material, and \( \frac{dT}{dx} \) is the temperature gradient or how quickly the temperature changes with respect to distance. The negative sign indicates that heat flows from higher to lower temperatures. Applying this law not only allows us to understand how heat moves through solids but also to calculate the exact temperature difference when combined with the known values of thermal conductivity and the physical dimensions of the material.

Thermal Conductivity

Thermal conductivity, symbolized as \( k \) in equations, is a measure of a material's ability to conduct heat. It can be thought of as the material's thermal efficiency: materials with high thermal conductivity, like most metals, are really good at transferring heat, whereas materials with low thermal conductivity, such as wood or foam insulation, are much less efficient and therefore good insulators.

When we say a circuit board is impregnated with copper fillings and has an effective thermal conductivity of \( 16 \mathrm{W}/\mathrm{m}\cdot\mathrm{K} \), we're describing how well the board can move heat energy across it. This property is critically important in electronics, as excessive heat can damage components. Materials with high thermal conductivity help spread and dissipate that heat, protecting the delicate parts inside devices. Understanding thermal conductivity is not just academic; it's also practical, informing decisions from kitchen cookware to aerospace engineering.

Temperature Gradient

Temperature gradient is the rate at which temperature changes with distance in a particular direction, usually denoted as \( \frac{dT}{dx} \) in the context of heat conduction. Think of a mountain’s gradient: if it's steep, the altitude changes dramatically over a short distance. Temperature gradient works similarly; a steep temperature gradient means a large temperature change over a small space.

The temperature gradient is essential in calculating heat transfer through a material. It's the driving force described in Fourier's Law; without a difference in temperature—without a 'thermal slope' to 'slide down'—heat would not move. In our circuit board example, the temperature gradient is what causes the heat generated by the chips to flow across the board, resulting in a temperature difference that can be calculated with the other parameters in Fourier's Law.