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 $222K$.