Question

Two parallel black disks are positioned coaxially with a distance of \(0.25 \mathrm{~m}\) apart in a surrounding with a constant temperature of \(300 \mathrm{~K}\). The lower disk is \(0.2 \mathrm{~m}\) in diameter and the upper disk is \(0.4 \mathrm{~m}\) in diameter. If the lower disk is heated electrically at \(100 \mathrm{~W}\) to maintain a uniform temperature of \(500 \mathrm{~K}\), determine the temperature of the upper disk.

Short answer

Answer: The temperature of the upper disk is approximately \(439.66\,\text{K}\).

Step-by-step solution

Identify the known variables

In this problem, we are given: - Distance between the disks: \(0.25\,\text{m}\) - Temperature of the surroundings: \(300\,\text{K}\) - Diameters of the disks: \(0.2\,\text{m}\) (lower) and \(0.4\,\text{m}\) (upper) - Temperature of the lower disk: \(500\,\text{K}\) - Power required to maintain the lower disk's temperature: \(100\,\text{W}\)

Write the equation for the heat transfer

Since the heat transfer is happening by radiation, we will use the Stefan-Boltzmann Law for the heat transfer equation: \(q = \epsilon\sigma A(T_1^4 - T_2^4)\) Here, \(q\) is the heat transfer rate, \(\epsilon\) the emissivity of the disk (assumed to be \(1\) for black disks), \(\sigma\) the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \,\text{W/m^2 K^4}\)), \(A\) the surface area of the disks, and \(T_1\) and \(T_2\) the temperatures of the lower and upper disks, respectively. Since the lower disk is being heated at a rate of \(100\,\text{W}\), this heat should be equal to the heat transfer between the two disks: \(q = 100\,\text{W}\)

Calculate the area of the disks

We have two disks with different diameters. We need to calculate the surface area of each disk. The area of a disk is given by: \(A = \pi r^2\) For the lower disk, the diameter is \(0.2\,\text{m}\), so the radius is \(0.1\,\text{m}\). For the upper disk, the diameter is \(0.4\,\text{m}\), and the radius is \(0.2\,\text{m}\). Thus, the surface areas of the lower and the upper disks are: \(A_1 = \pi (0.1)^2 = 0.01\pi\,\text{m^2}\) \(A_2 = \pi (0.2)^2 = 0.04\pi\,\text{m^2}\)

Write the equation for the heat transfer rate q

Substituting the given values and the calculated areas into the heat transfer equation, we get: \(100\,\text{W} = \epsilon\sigma (0.01\pi)(500^4 - T_2^4)\) Since the disks are black, \(\epsilon = 1\). Now we can solve for \(T_2\).

Solve for the temperature of the upper disk T_2

To find the temperature of the upper disk \(T_2\), first, plug in the values for \(\sigma\) and rearrange the equation: \(T_2^4 = 500^4 - \frac{100}{(5.67 \times 10^{-8})(0.01\pi)}\) Now, calculate the value inside the parenthesis and take the fourth root to find \(T_2\): \(T_2 = \sqrt[4]{500^4 - \frac{100}{(5.67 \times 10^{-8})(0.01\pi)}}\) Finally, compute the result numerically to obtain the temperature of the upper disk: \(T_2 \approx 439.66\,\text{K}\) The temperature of the upper disk is approximately \(439.66\,\text{K}\).

Key concepts

Stefan-Boltzmann Law

Understanding the Stefan-Boltzmann Law is critical when dealing with problems involving thermal radiation. This fundamental law reveals how an object's temperature is related to the amount of heat it radiates. It states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time (also known as the black body radiant exitance) is directly proportional to the fourth power of the black body's temperature. The mathematical expression of this law is represented as \( E = \sigma T^4 \) where \( E \) is the radiant exitance, \( T \) is the absolute temperature in kelvins, and \( \sigma \) is the Stefan-Boltzmann constant (\( 5.67 \times 10^{-8} \,\text{W/m^2 K^4} \) for a black body).

For practical applications where an object might not be a perfect black body, the equation incorporates emissivity \( \epsilon \) (with a value between 0 and 1) to account for its ability to emit thermal radiation compared to a black body. When we solve problems like the temperature between two discs, we use an adapted form of this law to identify the heat transfer rate. This adaptation is essential because, without it, estimations of temperature changes due to radiative heat transfer would be significantly off-mark, leading to incorrect solutions.

Heat Transfer Rate

The heat transfer rate, represented by \( q \), is a measure of the thermal energy flow per unit time. In the context of the Stefan-Boltzmann Law, it is used to calculate the net energy transfer by radiation between bodies or surfaces. It is important to note that the rate at which heat is transferred is influenced by the temperature difference between the bodies and their surface properties, such as their area and emissivity. The associated equation is written as \( q = \epsilon\sigma A(T_1^4 - T_2^4) \), where \( A \) is the area of the emitting surface, and \( T_1 \) and \( T_2 \) are the temperatures of the respective bodies involved in the heat exchange.

In problems involving radiative heat transfer between two objects, such as the textbook example with two disks, we have to consider the area of the smaller disk (since the heat must pass through it), and the power needed to maintain the temperature of one of the disks to find the temperature of the other one. This understanding allows one to quantify how changes in these variables can impact the equilibrium temperature, a critical skill for students studying thermodynamics.

Black Body Radiation

The concept of black body radiation addresses the idealistic radiative characteristics of a physical body that perfectly absorbs all incident electromagnetic radiation, irrespective of the frequency or angle of incidence. Such a body will also be a perfect emitter. In reality, no material perfectly fits this definition; however, black bodies are used as a reference model to understand the radiation of real objects.

When the problem mentions 'black disks', it's implying that these disks are ideal emitters and absorbers of radiation, with an emissivity (\( \epsilon \) value) of 1. This makes our calculations straightforward since we can use the Stefan-Boltzmann Law without having to adjust for a lower emissivity. This assumption is what allows us to use the law directly in our heat transfer rate calculations, as in the original exercise where we determine the temperature of an upper disk based on the known power and temperature of a lower disk.