Question

An electronic device dissipating \(20 \mathrm{~W}\) has a mass of \(20 \mathrm{~g}\), a specific heat of \(850 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and a surface area of \(4 \mathrm{~cm}^{2}\). The device is lightly used, and it is on for \(5 \mathrm{~min}\) and then off for several hours, during which it cools to the ambient temperature of \(25^{\circ} \mathrm{C}\). Taking the heat transfer coefficient to be \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the temperature of the device at the end of the 5 -min operating period. What would your answer be if the device were attached to an aluminum heat sink having a mass of \(200 \mathrm{~g}\) and a surface area of \(80 \mathrm{~cm}^{2}\) ? Assume the device and the heat sink to be nearly isothermal.

Short answer

Answer: The final temperature of the device is 377.94°C without the heat sink, and 56.8°C with the heat sink attached.

Step-by-step solution

Convert units

First, let's convert the given values to SI units: Mass of the device: \(m_{1} = 20\,\text{g} \times \frac{1\,\text{kg}}{1000\,\text{g}} = 0.02\,\text{kg}\) Surface area of the device: \(A_{1} = 4\,\text{cm}^2 \times \frac{1\,\text{m}^2}{10000\,\text{cm}^2} = 4\times10^{-4}\,\text{m}^2\) Operating time: \(t = 5\,\text{min} \times \frac{60\,\text{s}}{1\,\text{min}} = 300\,\text{s}\)

Calculate the heat gained and temperature change without heat sink

Now, let's calculate the heat gained by the device during the operating time, and then calculate the temperature change. Heat lost by the device: \(Q_1 = P \times t = 20\,\text{W} \times 300\,\text{s} = 6000\,\text{J}\) Temperature change of the device: \(\Delta T_1 = \frac{Q_1}{m_1 \times c_p} = \frac{6000\,\text{J}}{0.02\,\text{kg} \times 850\,\text{J/kg·K}} \approx 352.94\,\text{K}\) Final temperature of the device without the heat sink is: \(T_{1f} = T_{1i} + \Delta T_1 = 25^{\circ}\text{C} + 352.94\,\text{K} \approx 377.94^{\circ}\text{C}\)

Calculate the mass and surface area of the aluminum heat sink

Now, let's consider the aluminum heat sink: Mass of the heat sink: \(m_{2} = 200\,\text{g} \times \frac{1\,\text{kg}}{1000\,\text{g}} = 0.2\,\text{kg}\) Surface area of the heat sink: \(A_{2} = 80\,\text{cm}^2 \times \frac{1\,\text{m}^2}{10000\,\text{cm}^2} = 8\times10^{-3}\,\text{m}^2\)

Calculate the heat gained and temperature change with the heat sink

Now, let's calculate the heat gained and temperature change of the device and heat sink together: Total mass and specific heat: \(m_{total} = (m_1 + m_2) = 0.02\,\text{kg} + 0.2\,\text{kg} = 0.22\,\text{kg}\) Total surface area: \(A_{total} = A_1 + A_2 = 4\times10^{-4}\,\text{m}^2 + 8\times10^{-3}\,\text{m}^2 = 8.4\times10^{-3}\,\text{m}^2\) Heat lost by the device and heat sink: \(Q_2 = Q_1 = 6000\,\text{J}\) (assume nearly isothermal) Temperature change of the device and heat sink: \(\Delta T_2 = \frac{Q_2}{m_{total}\times c_p} = \frac{6000\,\text{J}}{0.22\,\text{kg} \times 850\,\text{J/kg·K}} \approx 31.8\,\text{K}\) Final temperature of the device with the heat sink is: \(T_{2f} = T_{2i} + \Delta T_2 = 25^{\circ}\text{C} + 31.8\,\text{K} \approx 56.8^{\circ}\text{C}\) So, the final temperature of the device is \(377.94^{\circ}\text{C}\) without the heat sink, and \(56.8^{\circ}\text{C}\) with the heat sink attached.

Key concepts

Thermal Management of Electronic Devices

When it comes to electronic devices, one of the major concerns is heat generation and its proper management. Without effective thermal management strategies, electronics can overheat, leading to reduced efficiency, performance, and even failures.

Heat production in electronics is a consequence of electrical resistance and inefficiency of components. To ensure the smooth operation of these devices, it's important to control their temperature. This is done by dissipating the generated heat into the surrounding environment or finding ways to minimize its production.

One common solution, illustrated by the exercise, is the use of a heat sink. A heat sink is designed to maximize the surface area in contact with the cooling medium, such as air. In our example, we saw a drastic difference in temperature when the device used a heat sink. The material properties of a heat sink, along with its size and shape, are crucial for effective heat dissipation.

Specific Heat Capacity

Specific heat capacity is a property that measures the amount of heat required to change the temperature of a substance by a degree Celsius (or Kelvin). It's denoted by the symbol 'c' and commonly expressed in units of joules per kilogram per Kelvin (J/kg·K).

The higher the specific heat capacity, the more heat is required to increase a material's temperature. This is a critical property for thermal management in electronics. Materials with high specific heat capacities are able to absorb more heat before their temperature rises significantly. In the exercise, the specific heat capacity of the device plays a key role in determining how much the temperature will increase during operation. It's also an important factor when a heat sink is attached, as the heat sink increases the overall capacity to store heat energy without significant temperature increases.

Heat Transfer Coefficient

The heat transfer coefficient is a measure of how well heat is transferred between a surface and a fluid flowing past it. It is symbolized by 'h' and usually expressed in units of watts per square meter per Kelvin (W/m²·K). This coefficient is influenced by a number of factors including the properties of the fluid, the velocity of the fluid, and the characteristics of the surface.

In practical terms, a higher heat transfer coefficient suggests that a surface, such as the heat sink in our exercise, is more effective at transferring heat away from itself and into the surrounding medium. It plays a fundamental role in the thermal management of electronic devices, as it helps in predicting how efficiently a device can be cooled under different operational conditions. For efficient cooling, engineers seek out materials and designs that maximize this coefficient.