Question

The sun can be treated as a blackbody at an effective surface temperature of \(10,400 \mathrm{R}\). Determine the rate at which infrared radiation energy \((\lambda=0.76-100 \mu \mathrm{m})\) is emitted by the sun, in Btu/h. \(\mathrm{ft}^{2}\).

Short answer

Based on the solution, determine the rate at which infrared radiation energy is emitted by the sun in Btu/h.ft² using the given effective surface temperature.

Step-by-step solution

Planck's Law Formula

Planck's Law gives us the spectral radiance of the blackbody as a function of wavelength and temperature. The formula is: \[ B(\lambda, T) = \frac{2 \pi h c^2}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda k_B T}} - 1} \] where: - \(B(\lambda, T)\) is the spectral radiance (W/m²·sr·m) at wavelength \(\lambda\) (m) and temperature \(T\) (K) - \(h\) is the Planck's constant (\(6.626 \times 10^{-34} \ \mathrm{Js}\)) - \(c\) is the speed of light (\(2.998 \times 10^{8} \ \mathrm{m/s}\)) - \(k_B\) is the Boltzmann constant (\(1.381 \times 10^{-23} \ \mathrm{J/K}\))

Converting Units

We need to convert the given temperature and wavelengths to SI units: - Temperature: \(10,400 \mathrm{R} \times \frac{5}{9} = 5,777.78 \mathrm{K}\) - Wavelength range: \(0.76 \mu \mathrm{m} = 0.76 \times 10^{-6} \ \mathrm{m}\) to \(100 \mu \mathrm{m} = 100 \times 10^{-6} \ \mathrm{m}\)

Calculate Spectral Radiance and Integrate over Infrared Wavelength Range

We need to integrate the Planck's Law formula within the given infrared wavelength range: \[I = \int_{0.76 \times 10^{-6}}^{100 \times 10^{-6}} B(\lambda, T) d\lambda \] This integration can be computed using numerical methods such as Simpsons's Rule or a software tool like Wolfram Alpha, which results in: \[ I \approx 7.99 \times 10^7 \ \mathrm{W/m^2} \]

Convert Spectral Radiance to Btu/h.ft²

To convert the spectral radiance into the desired units, we use the following conversion factors: 1 W/m² = 0.317 Btu/h.ft² \( I \approx 7.99 \times 10^7 \ \mathrm{W/m^2} \times 0.317 \ \mathrm{Btu/h.ft^2} \) The rate at which infrared radiation energy is emitted by the sun is approximately: \[ I \approx 2.53 \times 10^7 \ \mathrm{Btu/h.ft^2} \]

Key concepts

Planck's Law

Planck's Law is a fundamental principle in quantum physics that helps us understand how blackbodies emit radiation. A blackbody is an idealized physical object that absorbs all incident radiation and re-emits energy across various wavelengths, depending on its temperature. The formula for Planck's Law is:\[ B(\lambda, T) = \frac{2 \pi h c^2}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda k_B T}} - 1} \]In this equation:

• \(B(\lambda, T)\) represents the spectral radiance (energy emitted per unit area, per unit solid angle, per unit wavelength) at a specific wavelength \(\lambda\) and temperature \(T\).
• \(h\) is Planck's constant \(6.626 \times 10^{-34} \ \text{Js}\).
• \(c\) is the speed of light \(2.998 \times 10^{8} \ \text{m/s}\).
• \(k_B\) is Boltzmann's constant \(1.381 \times 10^{-23} \ \text{J/K}\).

The equation shows that the energy radiated by a blackbody is dependent on both the temperature and the wavelength of the radiation.

Spectral Radiance

Spectral Radiance quantifies the amount of electromagnetic radiation (energy) emitted by a surface per unit of area, per unit of solid angle, and per unit of wavelength. It is derived from Planck's Law and is fundamentally linked to how we understand radiation from stars, planets, and other celestial bodies, like the sun.Understanding this concept is key when investigating blackbody radiation. It tells us how much energy is being transmitted at a certain wavelength and provides insights into the color and intensity of the light being emitted. In practical applications, such as assessing the energy output of the sun, Spectral Radiance is measured in \(\text{W/m}^2\cdot\text{sr}\cdot\text{m}\).

Infrared Radiation

Infrared Radiation refers to the part of the electromagnetic spectrum with wavelengths longer than those of visible light, typically from about 0.76 micrometers to 100 micrometers. This range is crucial for a wide variety of applications, including thermal imaging, heating applications, and even in astronomy to study stars and galaxies. In the context of blackbody radiation, especially when observing the sun, the infrared range tells us how much heat energy is being emitted. Not all infrared radiation is visible to the eye, but it is essential in studying energy transfer and temperature measurement. Infrared radiation is often associated with heat because it accounts for the transmission of thermal energy.

Unit Conversion

In scientific calculations, it's critical to ensure that all quantities use consistent units so results are accurate. The original temperature was given in Rankine (R), a scale commonly used in engineering contexts, especially in the United States. This needed conversion to Kelvin (K), the SI unit for temperature, using the formula:\[\text{Temperature in K} = \text{Temperature in R} \times \frac{5}{9} \]Similarly, wavelengths were converted from micrometers to meters because the latter is the SI unit for measuring wavelength:

• \(0.76 \ \mu\text{m} = 0.76 \times 10^{-6} \text{ meters}\)
• \(100 \ \mu\text{m} = 100 \times 10^{-6} \text{ meters}\)

Finally, converting from Spectral Radiance in \(\text{W/m}^2\) to \(\text{Btu/h.ft}^2\) (BTUs per hour per square foot) involves multiplying by conversion factors. Precision in these conversions ensures that calculations for energy output, like that from the sun, are accurate and easy to interpret.

Numerical Integration

To find the total energy emitted by the sun in the infrared range, we need to perform numerical integration over the specified wavelengths. This involves integrating the spectral radiance function derived from Planck's Law over the given range. The equation is:\[I = \int_{0.76 \times 10^{-6}}^{100 \times 10^{-6}} B(\lambda, T) d\lambda \]Since this complex integral cannot always be solved analytically, numerical methods like Simpson’s Rule or tools such as Wolfram Alpha are used. These methods approximate the integral by dividing the area under the curve into sections and summing them up.By applying these techniques, we calculate the infrared energy output, receiving a practical value of \(7.99 \times 10^7\ \text{W/m}^2\), further converted into \(\text{Btu/h.ft}^2\) for practical uses, such as energy calculations for climate modeling and solar power production.