Question

An electronic device dissipating \(25 \mathrm{W}\) has a mass of \(20 \mathrm{g}\) and a specific heat of \(850 \mathrm{J} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\). The device is lightly used, and it is on for 5 min and then off for several hours, during which it cools to the ambient temperature of \(25^{\circ} \mathrm{C}\) Determine the highest possible temperature of the device at the end of the 5 -min operating period. What would your answer be if the device were attached to a 0.5 -kg aluminum heat sink? Assume the device and the heat sink to be nearly isothermal.

Short answer

Answer: When the electronic device is attached to a 0.5-kg aluminum heat sink, the highest possible temperature at the end of the 5-minute operating period is 41.15°C.

Step-by-step solution

Calculate the total heat generated

To find out the total heat generated, we will use the equation Q = P * t, where Q is the heat generated, P is the power (25 W), and t is the time in seconds. We are given that the electronic device dissipates 25 W of power for 5 minutes, so we first convert the time into seconds: 5 minutes * 60 seconds/minute = 300 seconds Now, we will calculate the total heat generated: Q = P * t = 25 W * 300 seconds = 7500 J

Calculate the final temperature of the electronic device

To find the final temperature of the electronic device, we will use the formula: Q = mcΔT where Q is the heat generated (7500 J), m is the mass of the electronic device (20 g), c is the specific heat capacity of the electronic device (850 J/kg·°C), and ΔT is the change in temperature. We will first convert the mass to kg to match the unit in specific heat: m = 20 g * (1 kg / 1000 g) = 0.02 kg Now we will rearrange the formula to find ΔT: ΔT = Q / (mc) = 7500 J / (0.02 kg * 850 J/kg·°C) = 441.18°C To find the final temperature of the electronic device, we will add the initial temperature: Final Temperature = Initial Temperature + ΔT = 25°C + 441.18°C = 466.18°C So, the highest possible temperature of the device at the end of the 5-minute operating period is 466.18°C.

Calculate the final temperature when the device is attached to a heat sink

The device is now attached to a 0.5-kg aluminum heat sink. The total mass will now be the sum of both device and heat sink masses: m_total = m_device + m_heatsink = 0.02 kg + 0.5 kg = 0.52 kg The specific heat of aluminum is 900 J/kg·°C. Since we are assuming the device and the heat sink are nearly isothermal, we can use the weighted average of their specific heat capacities to find the combined specific heat: c_total = (m_device * c_device + m_heatsink * c_heatsink) / m_total = (0.02 kg * 850 J/kg·°C + 0.5 kg * 900 J/kg·°C) / 0.52 kg = 894.62 J/kg·°C Now, we will use the same formula as in step 2 to find the change in temperature, but using the total mass and combined specific heat: ΔT = Q / (m_total * c_total) = 7500 J / (0.52 kg * 894.62 J/kg·°C) = 16.15°C Final Temperature = Initial Temperature + ΔT = 25°C + 16.15°C = 41.15°C So, when the device is attached to the aluminum heat sink, the highest possible temperature at the end of the 5-minute operating period is 41.15°C.

Key concepts

Specific Heat Capacity

Imagine you are heating up two different pans on a stove, one made of aluminum and the other made of cast iron. You've probably noticed that the aluminum pan heats up much more quickly than the cast iron one. This difference is largely due to a concept in thermodynamics called specific heat capacity.

Specific heat capacity, often symbolized as c, is the amount of heat energy required to raise the temperature of one kilogram of a substance by one degree Celsius. It's a property inherent to the material and plays a critical role in understanding how that material interacts with heat. In our textbook problem, the specific heat capacity of the electronic device material is given as 850 J/kg·°C. This value indicates how much energy it takes to change the device's temperature, which is crucial for determining how long the device can operate before it becomes too hot.

Using the formula Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature, we can calculate the heat energy that results in a specific temperature increase of an object. The practical implication of understanding specific heat capacity includes designing systems that are efficient in energy consumption and effective in managing temperature changes, like our electronic device.

Thermal Equilibrium

When you place a hot cup of coffee in a cooler room, over time, the coffee cools down while the room gets slightly warmer, until they both reach the same temperature. This process is a classic example of thermal equilibrium. It is the point at which two objects that are in contact with each other no longer exchange heat energy because they have reached the same temperature.

In thermodynamic terms, when a system reaches thermal equilibrium, there's no net heat flow between its components or with its surroundings. In our textbook problem, the device reaches thermal equilibrium with the ambient surroundings at 25°C after being turned off for several hours. This concept is a fundamental principle of the zeroth law of thermodynamics, which states that if two systems are each in thermal equilibrium with a third one, then they are in thermal equilibrium with each other. A deeper understanding of thermal equilibrium is essential when analyzing heat transfer in various engineering applications, including electronics cooling, climate control in buildings, and even meteorology.

Heat Sink Application

Overheating is the nemesis of electronic devices, and that's where a heat sink comes in as a superhero. A heat sink is designed to dissipate heat from a hot object—typically an electronic component—to prevent it from overheating.

In our textbook exercise, we examined the scenario of an electronic device with and without a heat sink. The addition of an aluminum heat sink significantly lowers the highest possible temperature of the device after operating for 5 minutes. This occurs because the heat generated by the device is spread over a larger mass with a high specific heat capacity—aluminum has a specific heat of 900 J/kg·°C. Aluminum is commonly used as a heat sink material due to its good thermal conductivity and capacity for absorbing thermal energy. The calculations made in the exercise showcase the effectiveness of heat sinks in managing the thermal load of electronic devices. Heat sinks are critical components in preventing thermal overload and extending the life of electronic devices, from small computer processors to large industrial machines.