This article aims to find the surface temperature. According to Stefan Boltzmann’s law, the amount of radiation emitted per unit time from region $A$ of a black body at absolute temperature represented by $T$ is directly proportional to the fourth power of temperature.
\[\dfrac{u}{A}=\sigma T^{4}\]
where $\sigma$ is the Stefan constant $\sigma=5.67 \times 10^{-8} \dfrac{W}{m^{2}. {K}^{4}}$ is derived from other known constants. A non-blackbody absorbs and therefore emits less radiation, given by the equation.
For such a body,
\[u=e\sigma A T^{4}\]
where $\varepsilon$ is the emissivity (equal to absorptivity) that lies between $0$ and $1$.For a real surface, the emissivity is a function of temperature, radiation wavelength, and direction, but a useful approximation is a diffuse gray surface where $\varepsilon$ is considered constant. With ambient temperature $T_{0}$, the net energy radiated by area $A$ per unit time.
\[\Delta u = u – u_{o} = e\sigma A (T^{4} – T_{0}^{4})\]
Stefan Boltzmann’s law relates the temperature of a blackbody to the amount of energy it emits per unit area. The law states that;
Total energy emitted or radiated per unit surface area of a blackbody at all wavelengths per unit time is directly proportional to $4$ power of the thermodynamic temperature of the blackbody.
Law of Conservation of Energy
Law of conservation of energy says that energy cannot be created or destroyed — only converted from one form of energy to another. This means the system always has the same energy unless it is added from outside. This is particularly confusing in case of non-conservative forces, where energy is converted from mechanical to thermal energy, but the total energy remains the same. The only way to use power is to convert energy from one form to another.
Thus, the amount of energy in any system is given by the following equation:
\[U_{T}=U_{i}+W+Q\]
- $U_{T}$ is the total internal energy of the system.
- $U_{i}$ is the initial internal energy of system.
- $W$ is the work done by or on system.
- $Q$ is the heat added to or removed from the system.
Although these equations are extremely powerful, they can make it difficult to understand the power of statement. The takeaway message is that it is not possible to create energy out of anything.
Expert Answer
Given data
- Probe diameter: $D=0.5\:m$
- Electronics heat rate: $q=E_{g}=150W$
- Probe surface emissivity: $\varepsilon=0.8$
Use conservation of energy law and Stefan-Boltzmann’s
\[-E_{o}+E_{g}=0\]
\[E_{g}=\varepsilon\pi D^{2}\sigma T_{s}^{4}\]
\[T_{s}=(\dfrac{E_{g}}{\varepsilon \pi D^{2} \sigma})^{\dfrac{1}{4}}\]
\[T_{s}=(\dfrac{150W}{0.8\pi (0.5)^{2}\times 5.67\times 10^{-8}})^{\dfrac{1}{4}}\]
\[T_{s}=254.7K\]
The surface temperature is $254.7K$.
Numerical Result
The surface temperature is $254.7K$.
Example
A spherical probe with a diameter of $0.6\: m$ contains electronics that dissipate $170\: W$. If the surface of the probe has an emissivity of $0.8$ and the probe does not receive radiation from other surfaces, e.g., from the Sun, what is its surface temperature?
Solution
Given data in the example
Probe diameter: $D=0.7\:m$
Electronics heat rate: $q=E_{g}=170W$
Probe surface emissivity: $\varepsilon=0.8$
Use conservation of energy law and Stefan-Boltzmann’s
\[E_{g}=\varepsilon\pi D^{2}\sigma T_{s}^{4}\]
\[T_{s}=(\dfrac{E_{g}}{\varepsilon \pi D^{2} \sigma})^{\dfrac{1}{4}}\]
\[T_{s}=(\dfrac{170W}{0.8\pi (0.7)^{2}\times 5.67\times 10^{-8}})^{\dfrac{1}{4}}\]
\[T_{s}=222K\]
The surface temperature is $222K$.