7.1 - Emissivity
- 1The concept of grey bodies as approximations to black bodies
- 2Defining emissivity and its relationship to the power emitted by a grey body
- 3The modified Stefan-Boltzmann law for grey bodies
- 4Typical emissivity values for common substances
- 5The wavelength dependence of emissivity
Grey bodies as approximations to black bodies
Black bodies, have a characteristic emission spectrum with a maximum intensity at a specific wavelength ($\lambda_{max}$) for a given temperature (T). This relationship is described by Wien's displacement law:
$\lambda_{max}T = 2.9 \times 10^{-3} \text{ m K}$
The total power output of a black body is related to its surface area (A) and absolute temperature (T) through the Stefan-Boltzmann law:
$P = \sigma AT^4$
In reality, objects can behave similarly to black bodies without being perfect. These approximations are called grey bodies. At a given temperature, a grey body emits less energy per second than a perfect black body with the same dimensions.
Emissivity - Quantifying the difference between grey and black bodies
The emissivity (e) is a measure of how much less power a grey body emits compared to a perfect black body.
$e = \frac{\text{power emitted by a radiating object}}{\text{power emitted by a black body with the same dimensions and at the same temperature}}$
Emissivity does not have a unit, because it is a ratio.
For a grey body with emissivity (e), the modified Stefan-Boltzmann law becomes:
$P_e = e\sigma AT^4$
Where:
- P_e_ = Power emitted by radiating object (W)
- e = emissivity
- A = Area (m^2^)
- T = Temperature (K)
Key points about emissivity:
- A perfect black body has an emissivity of 1.
- An object that completely reflects radiation without any absorption has an emissivity of 0.
- All real objects have an emissivity between 0 and 1.
Worked example - calculating the power emitted by a grey body
Calculate the power emitted by a brick wall with a surface area of 10 m² at a temperature of 300 K. The emissivity of brick is 0.90.
Step 1: Formula
$\text{P}_\text{e}\text{ = e }\sigma \text{A T}^4$
Step 2: Identify known values
- e = 0.90
- $\sigma$ = 5.67 $\times 10^{-8} \text{ W m}^{-2} \text{ K}^{-4}$
- A = 10 $\text{ m}^2$
- T = 300 K
Step 3: Substitution and correct evaluation
$P_e = 0.90 \times 5.67 \times 10^{-8} \times 10 \times 300^4$
$P_e = 4133.43 \text{ W}$
Typical emissivity values
The table below shows typical emissivity values at visible wavelengths for some common substances:
| Substance | Emissivity |
|---|---|
| Brick | 0.90 |
| Glass | 0.95 |
| Ice | 0.97 |
| Polished Silver | 0.02 |
| Snow | 0.8 - 0.9 |
It's important to note that emissivity values are wavelength-dependent. Surprisingly, snow, which appears white and reflective, is an effective emitter (and absorber) at infrared wavelengths.