Energy carried from one system to another by light and other electomagnetic waves is called radiation. Unlike conduction or convection, energy exchange between two bodies by radiation can occur even across a vacuum between the two bodies. A spectacular evidence is the radiation from far away stars reaching the Earth.

The radiation mode of heat transfer is based on the fact that all materials above absolute zero temperature ($-273.15^{\circ}$C or zero degree kelvin) emit electromagnetic waves. We use a perfect emitter, called black body, to model radiation emitted by a body.

The total radiation emitted by an ideal black body at a given temperature $T$ is given by the Stefan-Boltzmann law in terms of the power radiated per unit surface area, the power flux.

$$\left(\dfrac{\text{Power}}{\text{Area}}\right) = \sigma\, T^4,\label{eq-stefan-boltzmann-law-blackbody}\tag{23.3.1}$$

where $T$ is in kelvin scale and $\sigma$ is a universal constant called the Stefan-Boltzmann constant with magnitude

$$\sigma = 5.7\times 10^{-8} \text{W/m}^2\text{K}^4.\tag{23.3.2}$$

The power flux radiated by a real body is usullay smaller than given by Eq. (23.3.1). We take care of this by multiplying the result of Eq. (23.3.1) by a factor, called the emissivity factor, denoted by $\epsilon \text{.}$

$$\left(\dfrac{\text{Power}}{\text{Area}}\right) = \epsilon\,\sigma\, T^4,\label{eq-stefan-boltzmann-law-all}\tag{23.3.3}$$

with emissivity in the range, $0 \le \epsilon \le 1\text{.}$ Similar to emissivity of a body is absorptivity, $\alpha\text{.}$ The ideal blackbody absorbs all radiation incident upon it, but real objects absorb only a part of it. The absorptivity is defined by

$$\text{Absorptivity, }\ \alpha = \dfrac{\text{Radiation absorbed}}{\text{Radiation incident}}.\tag{23.3.4}$$

To simplify calculations we often assume that the absorptivity of a material is equal to the emissivity.

$$\text{Simplifying assumption: }\ \ \alpha = \epsilon.\tag{23.3.5}$$

This relation is strictly true only for an ideal black body, where both are equal to 1, but for most materials, this relation can be taken as a starting point of analysis. The absorptivity and emissivity are experimental quantities which we normally find by empirical means for each material.

The Sun radiates $4\times10^{26}\text{ J}$ of energy per second. Assuming that the radiation from the Sun is very close to that expected from a blackbody, and the radius of Sun to be $7.0 \times 10^8\text{ m}\text{,}$ find the surface temperature of Sun.

Hint

Use Stefan-Boltzmann law.

$5,800\text{K}\text{.}$

Solution

The amount of energy radiated per unit time per unit area, i.e., Power/Area, is given by the Stefan-Boltzmann law.

\begin{equation*} \left(\dfrac{P}{A}\right)_{\text{emitted}} = \sigma T^4. \end{equation*}

Here, the area is equal to the surface area of a sphere, which is $4\pi R^2\text{.}$ Solving for $T\text{,}$ we find

\begin{equation*} T = \left( \dfrac{P}{4\pi R^2 \sigma}\right)^{1/4}. \end{equation*}

Now, we put values of $p\text{,}$ $R$ and $\sigma$ to obtain the following temperature at the surface of Sun.

\begin{equation*} T = \left( \dfrac{4.0\times 10^{26}\text{W}}{4\pi (7.0\times 10^8 \text{m})^2 (5.7\times 10^{-8} \text{W/m}^2\text{K}^4)}\right)^{1/4} = 5,800\text{K}. \end{equation*}