A cylindrical resistor element on a circuit board dissipates \(1.2 \mathrm{~W}\) of power. The resistor is \(2 \mathrm{~cm}\) long, and has a diameter of \(0.4 \mathrm{~cm}\). Assuming heat to be transferred uniformly from all surfaces, determine \((a)\) the amount of heat this resistor dissipates during a 24-hour period, \((b)\) the heat flux, and \((c)\) the fraction of heat dissipated from the top and bottom surfaces.

Understanding how to calculate heat energy is essential for analyzing how electrical components, like resistors, manage energy conversion. In the given exercise, the resistor dissipates energy in the form of heat, which is a byproduct of its resistance to the electrical current. To calculate the heat energy (\(Q\)) that the resistor emits over time, we apply the formula:

\(Q = P \times t\), where \(P\) is the power dissipated by the resistor, and \(t\) is the time period over which the power is dissipated. The power, measured in watts (\(W\)), is a rate of energy transfer. When you multiply this rate by time, you get energy, which is measured in joules (\(J\)). For instance, our resistor dissipates \(1.2 W\) over a 24-hour period, which translates into \(103.68 \times 10^3 J\) of heat energy using the calculation from the provided solution.

Moving on to the concept of heat flux, which is essentially a measure of how quickly heat is transferred per area. It tells us the intensity of the heat being transferred. For our resistor example, the heat flux is determined by dividing the power dissipation by the surface area through which the heat is being conducted. Mathematically, it's expressed as:

\(\phi = \frac{P}{A}\), where \(\phi\) is the heat flux (\(W/m^2\)), \(P\) is the power dissipation, and \(A\) is the surface area. When you have a high heat flux, it means that a lot of heat is passing through a small area very quickly, which can be critical for the design and reliability of electrical components. In the exercise, the calculated heat flux is an impressive \(1496.26 W/m^2\), indicating that the small resistor emits heat quite densely across its surface.

The surface area of a cylinder is a pivotal factor in heat dissipation calculations, as it determines how much area is available for heat transfer. To find the total surface area of a cylindrical object—like our resistor—you need to calculate the areas of both the curved surface and the circular end caps. The formula for a cylinder's total surface area is:

\(A = 2\pi rh + 2\pi r^2\), where \(r\) is the radius and \(h\) is the height of the cylinder. The first term \(2\pi rh\) accounts for the curved surface area, while the second term \(2\pi r^2\) covers the area of the two end caps. By summing these areas, you get the total surface area available for heat to dissipate into the surrounding environment. For our exercise example, with a height of \(0.02 m\) and radius of \(0.002 m\), the calculated total surface area is \(8.02 \times 10^{-4} m^2\).

Finally, let's discuss the heat transfer rate, which tells us how fast heat is moving away from a source over time. It can be thought of as a speed measurement for thermal energy moving from one place to another. This rate is crucial in the context of electronic components because it affects their performance and longevity. The rate at which heat is transferred is directly proportional to the temperature difference between the component and its environment. If not managed properly, excessive heat can lead to reduced efficiency and potential failure. In our exercise, the power dissipated by the resistor (\(1.2 W\)) serves as an indication of the heat transfer rate, as it quantifies the thermal energy being released by the resistor every second. This is the steady state heat transfer rate and it highlights the importance of designing resistors to effectively disperse heat over time.