A transistor with a height of \(0.4 \mathrm{~cm}\) and a diameter of \(0.6 \mathrm{~cm}\) is mounted on a circuit board. The transistor is cooled by air flowing over it with an average heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the air temperature is \(55^{\circ} \mathrm{C}\) and the transistor case temperature is not to exceed \(70^{\circ} \mathrm{C}\), determine the amount of power this transistor can dissipate safely. Disregard any heat transfer from the transistor base.

First, we need to find the surface area of the transistor. We are given its height and diameter, so we can assume it has a cylindrical shape. We can, therefore, use the formula for the surface area of a cylinder without its bottom surface (base) to calculate the transistor's surface area, since we are disregarding heat transfer from the transistor base: Surface area (A) = Lateral area of the cylinder = \(2 \pi r h\), where r is the radius, and h is the height. Since the diameter is given, we can find the radius by dividing the diameter by 2: \(r = \frac{0.6 \, \mathrm{cm}}{2} = 0.3 \, \mathrm{cm}\) Now, we can calculate the surface area (A): \(A = 2 \pi (0.3 \, \mathrm{cm})(0.4 \, \mathrm{cm}) = 0.75398 \, \mathrm{cm}^{2}\) Convert the surface area to square meters: \(A = 0.75398 \, \mathrm{cm}^{2} \cdot \frac{1 \, \mathrm{m^2}}{10^4 \, \mathrm{cm}^{2}} = 7.5398 \times 10^{-5} \, \mathrm{m}^2\)

Next, we will apply Newton's Law of Cooling to relate the heat transfer coefficient (h), the surface area of the transistor (A), the temperature difference between the transistor and air (ΔT), and the power that can be dissipated safely (Q): \(Q = h \cdot A \cdot \Delta T\) We have the heat transfer coefficient (h) and the surface area (A), and we can find the temperature difference (ΔT) by subtracting the air temperature from the maximum transistor case temperature: \(\Delta T = 70^{\circ} \mathrm{C} - 55^{\circ} \mathrm{C} = 15^{\circ} \mathrm{C}\) or \(15 \, \mathrm{K}\) Now, we can find the power (Q) that can be dissipated safely: \(Q = (30 \, \frac{\mathrm{W}}{\mathrm{m}^2 \cdot \mathrm{K}}) \cdot (7.5398 \times 10^{-5} \, \mathrm{m}^2) \cdot (15 \, \mathrm{K})\) \(Q = 0.33929 \, \mathrm{W}\)

The amount of power that can be safely dissipated by this transistor is approximately 0.339 W.