Question

Daylight and candlelight may be approximated as a blackbody at the effective surface temperatures of \(5800 \mathrm{~K}\) and \(1800 \mathrm{~K}\), respectively. Determine the radiation energy (in \(\mathrm{W} / \mathrm{m}^{2}\) ) emitted by both lighting sources (daylight and candlelight) within the visible light region \((0.40\) to \(0.76 \mu \mathrm{m})\).

Short answer

Question: Determine the radiation energy emitted by daylight and candlelight within the visible light region (0.40 to 0.76 micrometers). Answer: To find the radiation energy emitted by daylight and candlelight within the visible light region, we need to use Planck's Law and Stefan-Boltzmann Law. First, calculate the total energy emitted by daylight and candlelight using the Stefan-Boltzmann Law. Next, calculate the ratio of energy emitted in the visible light region compared to the total energy by finding the integral of Planck's Law over the visible light range divided by total energy from Stefan-Boltzmann Law using numerical methods. Finally, multiply the calculated ratio with the total energy emitted by daylight and candlelight to find the radiation energy within the visible light region.

Step-by-step solution

Identify the Constants

First, we identify the constants provided in the exercise: - Effective surface temperature of daylight (T1): \(5800 \mathrm{~K}\) - Effective surface temperature of candlelight (T2): \(1800 \mathrm{~K}\) - Range of visible light: 0.40 to 0.76 \(\mu m\) (0.40 × 10⁻⁶ to 0.76 × 10⁻⁶ m)

Introduce Planck's Law

Planck's Law describes the intensity of radiation emitted by a blackbody at a particular temperature and wavelength. The equation is as follows: $$ I(\lambda, T) = \frac{2\pi h c^2}{\lambda^5} \cdot \frac{1}{e^{\frac{h c}{\lambda k_\mathrm{B} T}} - 1}, $$ where - \(I(\lambda, T)\): Radiation intensity at wavelength \(\lambda\) and temperature \(T\) - \(h\): Planck's constant (\(6.626 \times 10^{-34} \mathrm{J} \cdot \mathrm{s}\)) - \(c\): Speed of light (\(2.998 \times 10^{8} \mathrm{m/s}\)) - \(k_\mathrm{B}\): Boltzmann constant (\(1.381 \times 10^{-23} \mathrm{J/K}\)) - \(\lambda\): Wavelength - \(T\): Temperature

Introduce Stefan-Boltzmann Law

The Stefan-Boltzmann Law helps us calculate the total energy emitted by a blackbody across all wavelengths. The equation is as follows: $$ E(T) = \sigma \cdot T^4, $$ where - \(E(T)\): Total energy emitted by a blackbody at temperature \(T\) - \(\sigma\): Stefan-Boltzmann constant (\(5.670 \times 10^{-8} \mathrm{Wm^{-2}K^{-4}}\)) - \(T\): Temperature

Calculate Total Energy Emitted by Daylight and Candlelight

Using the Stefan-Boltzmann Law, we calculate the total energy emitted by daylight and candlelight: $$ E_{\text{daylight}} = \sigma \cdot T_1^4 = (5.670 \times 10^{-8}) \cdot (5800)^4 = 6.422 \times 10^{7} \mathrm{W/m^2} $$ $$ E_{\text{candlelight}} = \sigma \cdot T_2^4 = (5.670 \times 10^{-8}) \cdot (1800)^4 = 1.310 \times 10^{6} \mathrm{W/m^2} $$

Calculate the Ratio of Energy Emitted in Visible Light Region

To find the radiation energy emitted by daylight and candlelight within the visible light region, we need to know the ratio of energy emitted in the visible light region compared to the total energy. To find this ratio, we need the integral of Planck's Law over the visible light range divided by total energy from Stefan-Boltzmann Law: $$ \text{Ratio} = \frac{\int_{\lambda_1}^{\lambda_2} I(\lambda, T) d\lambda}{E(T)} $$ This integral does not have an analytical solution, so we need to use numerical methods like Simpson's rule, trapezoidal rule, or a relevant numerical library to calculate the integral.

Calculate the Radiation Energy in Visible Light Region

Multiply the calculated ratio with the total energy emitted by daylight and candlelight to find the radiation energy within the visible light region: $$ E_{\text{daylight-visible}} = \text{Ratio}_1 \times E_{\text{daylight}} $$ $$ E_{\text{candlelight-visible}} = \text{Ratio}_2 \times E_{\text{candlelight}} $$ Now we have the radiation energy emitted by daylight and candlelight within the visible light region, which are the answers to this exercise.

Key concepts

Planck's Law

Planck's Law is vital for understanding blackbody radiation, which describes how perfect emitters, or blackbodies, radiate energy depending on their temperature and wavelength. The law provides a mathematical formula that expresses the radiation intensity \( I(\lambda, T) \) of a blackbody at a specific wavelength \( \lambda \) and temperature \( T \):
\[I(\lambda, T) = \frac{2\pi h c^2}{\lambda^5} \cdot \frac{1}{e^{\frac{h c}{\lambda k_\mathrm{B} T}} - 1}\]
The constants involved in the equation are:

• \( h \): Planck's constant, \( 6.626 \times 10^{-34} \mathrm{J} \cdot \mathrm{s} \)
• \( c \): Speed of light, \( 2.998 \times 10^{8} \mathrm{m/s} \)
• \( k_\mathrm{B} \): Boltzmann constant, \( 1.381 \times 10^{-23} \mathrm{J/K} \)

This equation helps to determine how different wavelengths emit different radiation intensities at different temperatures. Thus, it predicts the spectral distribution of the emitted radiation. It's particularly useful when assessing the energy output within specific wavelength ranges, such as the visible light emitted by daylight and candlelight.

Stefan-Boltzmann Law

The Stefan-Boltzmann Law plays a crucial role in quantifying the total energy radiated by a blackbody. Unlike Planck’s Law, which focuses on specific wavelengths, this law gives us the summed energy emission over all wavelengths. The formula is:
\[ E(T) = \sigma \cdot T^4 \]
Where:

• \( E(T) \): Total energy emitted by a blackbody at temperature \( T \)
• \( \sigma \): Stefan-Boltzmann constant, \( 5.670 \times 10^{-8} \mathrm{Wm^{-2}K^{-4}} \)

This equation illustrates that the total energy emitted is directly proportional to the fourth power of the blackbody's temperature, \( T \).The implication of the Stefan-Boltzmann Law is significant in understanding why hotter objects like the sun (daylight) radiate more energy than cooler objects like a candle (candlelight). For instance, calculating the emitted energy for daylight compared to candlelight at respective effective temperatures of 5800 K and 1800 K highlights the vast difference in energy output.

Radiation Intensity

Radiation intensity refers to the energy emitted by a body per unit surface area per unit wavelength interval, often expressed in terms of watts per square meter per micrometer (W/m²/µm). It's a key concept in understanding how radiation is distributed across different wavelengths.
In the context of the original exercise, radiation intensity assists in calculating the energy emitted by daylight and candlelight within the visible light spectrum (0.40 to 0.76 µm). Before evaluating the radiation energy in visible regions, one combines both Planck's and Stefan-Boltzmann laws to understand the full range of emission:

• Planck's law helps to calculate intensity at specific wavelengths.
• Stefan-Boltzmann law quantifies total emission.

By comparing the intensity distribution and the total emission, you can find the energy within the visible range by integrating the intensity over the visible wavelengths. This mathematical process highlights why approximations or numerical methods are necessary, as exact solutions often don’t exist analytically. Moreover, understanding radiation intensity enhances the ability to reply on numerical tools to explore real-world applications such as designing light sources and enhancing energy efficiency.