## Radiant Energy Formula

## What is Radiant Energy Formula?

**Radiant energy** is the energy carried by electromagnetic waves, such as light and heat. The fundamental formula for radiant energy emitted by a black body in thermal equilibrium is given by

**πΈβπβ΄Β**

or

**E=π πβ΄Β **

- πΈ represents the radiant energy emitted per unit area
- π is the absolute temperature of the black body.
*Ο*is the Stefan-Boltzmann constant.

This formula signifies that the radiant energy emitted by a black body per unit area is directly proportional to the fourth power of its absolute temperature.

The radiant energy formula was derived by Josef Stefan in 1879 based on experimental data, and later theoretically derived from thermodynamic principles by Ludwig Boltzmann in 1884. Hence, the law is known as the Stefan-Boltzmann Law. To derive this formula, consider a black body, an idealized physical body that absorbs all incident electromagnetic radiation. The total power per unit area (P) emitted by a black body is directly proportional to the fourth power of the absolute temperature, which is mathematically expressed as

**π=ππβ΄**

*Ο*is the Stefan-Boltzmann constant, approximately 5.67Γ10β»βΈβW mβ»Β²Kβ»β΄.

This derivation involves assumptions of the black body as an ideal emitter and utilizes Planck’s law of black body radiation, integrating over all wavelengths of the emitted radiation.

## Applications of Radiant Energy Formula

**Climate Science**: Scientists use this law to understand Earth’s climate balance. They calculate how much energy the Earth receives from the sun and how much it radiates back into space.**Astronomy**: Astronomers estimate the temperature of stars by measuring the radiant energy they emit. This helps in understanding the life cycle of stars and the dynamics of galaxies.**Engineering**: In thermal engineering, this law assists in designing systems like solar panels and furnaces, ensuring they operate efficiently at desired temperatures.**Building Design**: Architects apply this formula to enhance building insulation. By understanding how materials emit radiant energy, they design buildings that maintain more stable internal temperatures.**Medical Technology**: It is utilized in medical devices such as infrared cameras and therapeutic heat lamps, which rely on the emission of radiant energy.

## Example Problems on Radiant Energy Formula

### Example 1: Calculating Radiant Energy from the Sun

**Problem:** Estimate the radiant energy emitted per square meter from the surface of the Sun, given its surface temperature is approximately 5,778 K.

**Solution:** Using the formula **πΈ = ππβ΄**

πΈ = (5.67Γ10β»βΈβW mβ»Β²Kβ»β΄) x (5778βK)β΄

πΈ= 5.67Γ10β»βΈΓ1.115Γ1014

πΈ β 6.33Γ10βΆβW mβ»Β²

**Explanation:** This calculation shows that the Sun emits about 6.33 million watts per square meter, demonstrating the immense power output of our star.

### Example 2: Comparing Earth and Moon Radiation

**Problem:** If the Earth and the Moon have average surface temperatures of 288 K and 220 K respectively, compare their radiant energies per unit area.

**Solution:** First, calculate for Earth:

*E*ββα΅£βββ= *Ο* (288K)β΄ β= 5.67 Γ 10β»βΈ Γ6.894Γ10βΉ

*E*ββα΅£ββ ββ 390 W mβ»Β²

Then, calculate for Moon:

*E*βββββ = *Ο* (220K)β΄

πΈββββ = 5.67Γ10β»βΈΓ2.336Γ109

πΈββββ β 132 βW mβ»Β²

**Explanation:** Earth emits about 390 watts per square meter, almost three times as much as the Moon, which emits about 132 watts per square meter. This difference is primarily due to Earth’s higher surface temperature.

### Example 3: Energy Loss from a Hot Object

**Problem:** A metal ball at 500 K loses energy by radiation. Calculate the radiant energy loss per square meter.

**Solution:** πΈ = π (500βK)β΄

πΈ = 5.67 Γ 10β»βΈ Γ 6.25 Γ 10ΒΉβ°

πΈ β 3545βW mβ»Β²

**Explanation:** The metal ball emits approximately 3545 watts per square meter due to its high temperature, showcasing how objects lose heat by emitting radiant energy.

### Example 4: Impact of Temperature Increase

**Problem:** If the temperature of an object increases from 300 K to 600 K, how does its radiant energy emission change?

**Solution:** Calculate the initial and final emissions:

πΈα΅’ββα΅’ββ = π (300βK)β΄

πΈΥ’α΅’βββ = π (600βK)β΄

πΈα΅’ββα΅’ββ β 459βW mβ»Β²

πΈΥ’α΅’βββ β 7344βW mβ»Β²

**Explanation:** By increasing the temperature, the radiant energy output increases by a factor of (600 / 300)β΄=16. This example illustrates the sensitivity of radiant energy emission to temperature changes.

## FAQs

## How Do You Calculate Radiant Energy?

Calculate radiant energy using the Stefan-Boltzmann law: πΈ=ππβ΄, where *Ο* is the constant and *T* is temperature.

## What Is the Formula for Energy?

The general energy formula depends on context, but for radiant energy, it’s πΈ=ππβ΄, based on the Stefan-Boltzmann law.

## What Is the Equation for Radiant Power?

The equation for radiant power is π=ππβ΄, highlighting how power emitted relates to the fourth power of temperature.