Question

A \(0.3\)-cm-thick, \(12-\mathrm{cm}\)-high, and \(18-\mathrm{cm}\)-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

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

Step-by-step solution

Calculate the total heat generated by the logic chips

First, we need to find the total heat generated by all 80 logic chips. Each chip dissipates 0.06 W, so we can multiply this value by the total number of chips to find the total heat generation. \(Q_{total} = n \times Q_{chip}\) Where \(n\) is the number of logic chips and \(Q_{chip}\) is the heat dissipated by each chip.

Define the heat conduction equation

Next, we define the heat conduction equation and insert all the given values: \(Q_{total} = k \frac{A \Delta T}{d}\) Where \(Q_{total}\) is the total heat generation, \(k\) is the thermal conductivity of the circuit board (\(16 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\)), \(A\) is the area of the circuit board, \(\Delta T\) is the temperature difference we are trying to find, and \(d\) is the thickness of the circuit board (\(0.3\) cm).

Calculate the area of the circuit board

Now, we need to calculate the area of the circuit board. The height and length are given as \(12 \text{ cm}\) and \(18 \text{ cm}\), respectively. \(A = height \times length = 12 \text{ cm} \times 18 \text{ cm} = 216 \text{ cm}^2\) To match the units with thermal conductivity, we should convert the area to \(\text{m}^2\): \(A = 216 \text{ cm}^2 \times \frac{1 \text{ m}^2}{10000 \text{ cm}^2} = 0.0216 \text{ m}^2\)

Substitute the values and find the temperature difference

Finally, we can substitute all the given values and the calculated area into the heat conduction equation: \(Q_{total} = k \frac{A \Delta T}{d}\) First, convert the thickness of the circuit board from cm to m: \(d = 0.3 \text{ cm} \times \frac{1 \text{ m}}{100\text{ cm}} = 0.003 \text{ m}\) Now, substitute the values: \(80 \times 0.06 \text{ W} = 16 \text{~W / m} \cdot \text{ K} \times \frac{0.0216 \text{ m}^2 \Delta T}{0.003 \text{ m}}\) Solve for the temperature difference, \(\Delta T\): \(\Delta T = \frac{80 \times 0.06 \text{~W} \times 0.003 \text{ m}}{16 \text{~W} / \text{m} \cdot \text{K} \times 0.0216 \text{ m}^2} = 0.042^{\circ} \text{C}\) So, the temperature difference between the two sides of the circuit board is \(0.042^{\circ} \text{C}\).