Question

A 2-cm-diameter surface at \(1000 \mathrm{~K}\) emits thermal radiation at a rate of \(15 \mathrm{~W}\). What is the emissivity of the surface? Assuming constant emissivity, plot the rate of radiant emission, in \(\mathrm{W}\), for surface temperatures ranging from 0 to \(2000 \mathrm{~K}\). The Stefan- Boltzmann constant, \(\sigma\), is \(5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{4}\).

Short answer

The emissivity is approximately 0.827. Plot power versus temperature from the formula \(P = 0.827 \times 5.67 \times 10^{-8} \times 3.14 \times 10^{-4} \times T^4\) for temperatures 0 to 2000 K.

Step-by-step solution

Understand the formula for thermal radiation

The formula for thermal radiation emitted by a surface is given by \[ P = \text{Emissivity} \times \text{Stefan-Boltzmann constant} \times \text{Area} \times T^4 \] where \(P\) is the power emitted (in watts), the Stefan-Boltzmann constant \(σ\) is \(5.67 \times 10^{-8} \frac{W}{m^2 \times K^4}\), the area (\(A\)) of the surface is in square meters, and \(T\) is the absolute temperature in Kelvin.

Convert the diameter to area

The diameter of the surface is 2 cm. The area \(A\) of the surface is \[ A = \frac{\text{π} \times d^2}{4} \] Converting the diameter from cm to meters first, \[ d = 2 \text{ cm} = 0.02 \text{ meters} \] Then we calculate the area:\[ A = \frac{π \times (0.02)^2}{4} \] \[ A ≈ 3.14 \times 10^{-4} \text{ square meters} \]

Solve for Emissivity

The power emitted is given as 15 W, temperature \(T\) is 1000 K, the Stefan-Boltzmann constant \(σ\) is \(5.67 \times 10^{-8} \frac{W}{m^2 \times K^4}\), and the area \(A\) is already calculated. We need to solve for the emissivity (\(ε\)). Rearranging the formula: \[ ε = \frac{P}{σ \times A \times T^4} \]Substituting the values:\[ ε = \frac{15}{(5.67 \times 10^{-8}) \times (3.14 \times 10^{-4}) \times (1000)^4} \] \[ ε ≈ 0.827 \]

Plot the rate of radiant emission

Now, let's use the same formula to find the power emitted for temperatures ranging from 0 to 2000 K. The formula becomes: \[ P = ε \times σ \times A \times T^4 \] Already having ε ≈ 0.827, σ = 5.67 × 10^{-8}, and A = 3.14 × 10^{-4}, we can calculate P for a series of temperatures T. Create a table of T from 0 to 2000 K (e.g., in steps of 100 K) and calculate P for each T. Plot these points on a graph with Temperature (K) on the x-axis and Power (W) on the y-axis.

Key concepts

Thermal Radiation

Thermal radiation is the process by which a body emits energy in the form of electromagnetic waves due to its temperature. All objects with a temperature above absolute zero emit thermal radiation. The amount and type of radiation depend on the temperature and properties of the emitting surface. At higher temperatures, more energy is emitted, and the peak wavelength of the radiation shifts to shorter wavelengths.

This radiation is an important factor in areas such as heat transfer, astrophysics, and climate science. The body's temperature is crucial because it strongly influences the radiation's intensity and spectrum.

Stefan-Boltzmann Law

The Stefan-Boltzmann law is a fundamental principle that quantifies the power emitted by a perfect black body (an idealized object that absorbs all incident radiation). According to the law, the power emitted per unit area is proportional to the fourth power of the absolute temperature:\[ P = \text{Emissivity} \times \text{Stefan-Boltzmann constant} \times A \times T^4 \]where:

• \(P\) is the total power emitted (in watts, W),
• Emissivity (\(ε\)) characterizes how well a surface emits thermal radiation compared to a black body,
• The Stefan-Boltzmann constant (\(σ\)) is \(5.67 \times 10^{-8} \frac{\text{W}}{\text{m}^2 \times \text{K}^4}\),
• \(A\) is the surface area (in square meters,\(m^2\)),
• \(T\) is the absolute temperature (in Kelvin,\(K\)).

The law serves as a foundation for understanding various practical applications, including stellar physics and thermal management in engineering systems.

Emissivity Calculation

The emissivity (\(ε\)) of a surface is a measure of its efficiency in emitting thermal radiation compared to a perfect black body. It ranges from 0 to 1, with 1 being a perfect emitter. Calculating emissivity involves rearranging the Stefan-Boltzmann equation:\[ ε = \frac{P}{σ \times A \times T^4} \]Here's how the calculation proceeds step-by-step:

• First, convert measurements to consistent units. For example, if the diameter is given in centimeters, convert it to meters for area (\(A\)) calculation.
• Next, determine the surface area using the formula:\[ A = \frac{π \times d^2}{4} \]For a surface with a diameter of 2 cm:\( d = 0.02 \text{ meters}\)\( A ≈ 3.14 \times 10^{-4} \text{ square meters}\)
• Substitute all known values into the rearranged formula to solve for \(ε\).

Using this method, you can accurately determine the emissivity of any surface given proper measurements.

Power Emitted

Power emitted (\(P\)) refers to the amount of thermal radiation energy given off by a surface per unit time. For any temperature (\(T\)), the power emitted can be calculated by:\[ P = ε \times σ \times A \times T^4 \]In the given example, we calculated the emissivity \(ε ≈ 0.827\). With this value, let's find power emitted over a temperature range:

• Choose a set of temperatures, say from 0 to 2000 K in steps of 100 K.
• For each temperature, plug the values into the formula to calculate \(P\).
• Create a series of points and plot them on a graph (Temperature on the x-axis, Power on the y-axis).

This graphical representation helps visualize how dramatically power emission increases with temperature, following the \(T^4\) relationship.