Question

A small surface of area \(A=1 \mathrm{~cm}^{2}\) emits radiation as a blackbody at \(1800 \mathrm{~K}\). Determine the rate at which radiation energy is emitted through a band defined by \(0 \leq \phi \leq 2 \pi\) and \(45 \leq \theta \leq 60^{\circ}\), where \(\theta\) is the angle a radiation beam makes with the normal of the surface and \(\phi\) is the azimuth angle.

Short answer

Answer: The rate at which radiation energy is emitted through the given angular range is approximately 23,636.85 W.

Step-by-step solution

Determine the intensity of the blackbody radiation

To do this, we use the Stefan-Boltzmann law, which states that the intensity of blackbody radiation (power per unit area) is given by: \[I = \sigma T^{4}\] where \(I\) is the intensity, \(\sigma\) is the Stefan-Boltzmann constant (\(\sigma \approx 5.67 \times 10^{-8} \mathrm{W m^{-2} K^{-4}}\)) and \(T\) is the temperature. Given the temperature T = 1800 K, we can calculate the intensity of the blackbody radiation: \[I = \sigma (1800)^{4}\]

Calculate the intensity I

Plugging the values into the formula, we get: \[I = (5.67 \times 10^{-8})(1800)^{4} = 4.026 \times 10^{7} \mathrm{W m^{-2}}\] Now we need to convert this value to \(\mathrm{W cm^{-2}}\). Since \(1 \mathrm{m^{2}} = 10^{4} \mathrm{cm^{2}}\), we have: \[I = 4.026 \times 10^{7} \mathrm{W m^{-2}} \cdot \frac {1 \mathrm{m^{2}}} {10^{4} \mathrm{cm^{2}}} = 4.026 \times 10^{3} \mathrm{W cm^{-2}}\]

Calculate the solid angle

The solid angle \(\Omega\) is a measure of the angular area subtended by a region in space. The solid angle can be calculated using the formula: \[\Omega = \int_{\phi_1}^{\phi_2} \int_{\theta_1}^{\theta_2} \sin{\theta} \,d\theta \,d\phi\] where \(0 \leq \phi \leq 2\pi\), \(45 \leq \theta \leq 60^{\circ}\) in our case.

Integrate to find the solid angle

Now, we plug in the given values and perform the integration: \[\Omega = \int_{0}^{2\pi} \int_{45^{\circ}}^{60^{\circ}} \sin{\theta} \, d\theta \, d\phi\] \[= (2\pi) \int_{\pi/4}^{\pi/3} \sin{\theta} \, d\theta\] \[= 2\pi (-\cos{\theta}) \Big|_{\pi/4}^{\pi/3}\] \[= 2\pi (-\cos{\pi/3} + \cos{\pi/4})\] \[= 2\pi (\frac{\sqrt{2}}{2} + \frac{1}{2})\]

Calculate the total radiated power

Now, using the area of the surface A = 1 cm² we calculate the power per unit solid angle \(dP / d\Omega\): \[\frac{dP}{d\Omega} = IA = 4.026 \times 10^3 \mathrm{W cm^{-2}} \cdot 1 \mathrm{cm^{2}}\] Now we integrate the power over the solid angle to find the total radiated power \(P\): \[P = \int_{\Omega} \frac{dP}{d\Omega} \, d\Omega\] \[P = \frac{dP}{d\Omega} \cdot \Omega = (4.026 \times 10^{3} \mathrm{W}) \cdot (2\pi (\frac{\sqrt{2}}{2} + \frac{1}{2}))\]

Compute the total power

Finally, we calculate the total radiated power \(P\): \[P = 4.026 \times 10^{3} \times 2\pi (\frac{\sqrt{2}}{2} + \frac{1}{2}) \mathrm{W}\] \[P \approx 23,636.85 \mathrm{W}\] The rate at which radiation energy is emitted through the given angular range is approximately 23,636.85 W.

Key concepts

Understanding the Stefan-Boltzmann Law

When we talk about blackbody radiation, we're referring to the theoretical concept of an idealized object that perfectly absorbs and emits all radiation it comes into contact with. The Stefan-Boltzmann law is a cornerstone in the study of such radiation, defining the relationship between the radiated energy of a blackbody and its temperature.

The law is expressed with a simple yet powerful equation: \[I = \text{sigma} T^{4}\] Here, \(I\) represents the intensity of the emitted radiation in terms of power per unit area (Watts per square meter), \(\text{sigma}\) is the Stefan-Boltzmann constant (approximately \(5.67 \times 10^{-8} \mathrm{W m^{-2} K^{-4}}\)), and \(T\) is the absolute temperature in Kelvin. This equation highlights how the emitted energy increases rapidly with temperature - a critical concept for understanding phenomena such as stars' luminosity or Earth's heat balance.

To practically apply this law, one must take into account the area of the surface in question, as the total power radiated is the intensity times the surface area. This is a key point to reinforce when considering exercises like the one above, where knowing the surface area and temperature allows us to calculate the emitted power quite readily.

Solid Angle Integration in Radiation Analysis

The concept of a solid angle is analogous to the idea of an angle in two dimensions, but it is extended to three dimensions. A solid angle quantifies how 'large' an object appears to an observer located at the vertex of the angle. The solid angle is measured in steradians (sr), and a complete sphere encompasses a solid angle of \(4\pi\) steradians.

When analyzing radiative heat transfer, it's crucial to integrate over solid angles to find out how much radiation is going in a specific direction. This is described by the integral: \[\Omega = \int_{\phi_1}^{\phi_2} \int_{\theta_1}^{\theta_2} \sin{\theta} \,d\theta \,d\phi\] Here, \(\Omega\) is the solid angle, \(\theta\) is the polar angle, and \(\phi\) is the azimuthal angle in spherical coordinates.

In our textbook exercise, we are tasked to find the rate at which radiation is emitted within a specified angular band. By performing the integration with given limits, we determine the fraction of the total solid angle through which the radiation is being emitted. It's important to understand the integration process and the physical significance of its bounds, i.e., the range of angles \(\theta\) and \(\phi\), to comprehend the distribution of radiation over different directions.

Radiative Heat Transfer Calculations

Radiative heat transfer is a mode of heat exchange that does not require a medium; it can occur through a vacuum. This mode of heat transfer is the primary way energy moves through space, like from the Sun to the Earth. In our example, we're focusing on how much energy a surface emits as radiation into a specific range of angles.

To determine the total radiative power emitted through a certain solid angle, we multiply the power per unit solid angle (obtained from the Stefan-Boltzmann law) by the calculated solid angle from our integration. The formula looks as follows: \[P = \int_{\Omega} \frac{dP}{d\Omega} \, d\Omega\] In simple terms, we're summing up all the contributions of radiant energy within our angular constraints.

When faced with such exercises, visualizing the geometry of the problem and understanding the principles of solid angle integration are vital. These allow us to link the physical properties of blackbody radiation to the mathematical framework needed to solve real-world problems involving radiative heat transfer. Ensuring that students grasp the integration concept, especially interpreting the limits physically, is key for them to succeed in mastering these calculations.