Question

An electronic device dissipating \(18 \mathrm{~W}\) 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: Without an aluminum heat sink, the final temperature of the electronic device is \(342.65^{\circ}C\). With an aluminum heat sink, the final temperature is \(53.88^{\circ}C\).

Step-by-step solution

Convert quantities to SI units

We first need to convert all given quantities to SI units. The mass, surface area, and time should be converted to kg, m², and seconds, respectively. Mass: \(20 g = 0.02 kg\) Surface area: \(4 cm^2 = 4 \times 10^{-4} m^2\) Time: \(5 min = 300 s\) For the heat sink: Mass: \(200 g = 0.2 kg\) Surface area: \(80 cm^2 = 8 \times 10^{-3} m^2\) Now that we have the correct units, we can proceed with the calculation.

Calculate the heat absorbed by the device

To find the heat absorbed by the device, we can use the equation: \(q = P \times t\) where \(q\) is the heat absorbed, \(P\) is the power dissipated, and \(t\) is the time duration. For this exercise, we have: \(q = 18 W \times 300 s\) \(q = 5400 J\)

Calculate the temperature rise of the device

We can use the specific heat formula to find the temperature rise of the device: \(\Delta T = \frac{q}{mc}\) where \(\Delta T\) is the temperature rise, \(m\) is the mass of the device, and \(c\) is the specific heat. For this exercise, we have: \(\Delta T = \frac{5400 J}{(0.02 kg)(850 J/kg\cdot K)}\) \(\Delta T = 317.65 K\)

Calculate the final temperature

To find the final temperature, we can add the temperature rise to the ambient temperature: \(T_f = T_i + \Delta T\) where \(T_f\) is the final temperature and \(T_i\) is the initial temperature. For this exercise, we have: \(T_f = 25 + 317.65 = 342.65^{\circ}C\)

Calculate the temperature rise of the device with a heat sink

We will now account for the heat sink by adjusting the mass and surface area that absorb heat from the device. The total mass with heat sink: \(m_{total} = 0.02 kg + 0.2 kg = 0.22 kg\) The total surface area with heat sink: \(A_{total} = 4 \times 10^{-4} m^2 + 8 \times 10^{-3} m^2 = 8.4 \times 10^{-3} m^2\) Since we have assumed the device and the heat sink to be isothermal, we can use the specific heat of the aluminum heat sink. Now, we can calculate the temperature rise \(\Delta T\) using the same equation as in step 3: \(\Delta T = \frac{5400 J}{(0.22 kg)(850 J/kg\cdot K)}\) \(\Delta T = 28.88 K\)

Calculate the final temperature with a heat sink

Adding the temperature rise to the ambient temperature to find the final temperature: \(T_f = 25 + 28.88 = 53.88^{\circ}C\) The final temperature of the electronic device at the end of the 5-minute operating period without a heat sink is \(342.65^{\circ}C\), and with an aluminum heat sink, it is \(53.88^{\circ}C\).