Emissivity Measurement Apparatus Experiment: Stefan-Boltzmann Law, Procedure and Calculations

Aim of the Experiment

• To determine the emissivity of a given surface by comparing its radiation with that of a blackbody at the same temperature
• To verify the Stefan-Boltzmann law of thermal radiation

Theory

Thermal radiation is electromagnetic energy emitted by a surface due to its temperature. A blackbody is an ideal surface that absorbs all incident radiation and emits the maximum possible radiation at any temperature. The Stefan-Boltzmann law gives the total emissive power of a blackbody:

E_b = σ T⁴

Where:
E_b = emissive power of blackbody (W/m²)
σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m²K⁴
T = absolute surface temperature (K)

For a real surface (grey body), the actual emissive power is:

E = ε σ T⁴

Where ε (0 < ε ≤ 1) is the emissivity of the surface, the ratio of radiation emitted by the real surface to that of a blackbody at the same temperature.

Emissivity depends on the material, surface finish, and temperature. Polished metals typically have ε = 0.05–0.15; matte black surfaces have ε ≈ 0.95–0.98.

Apparatus

• Emissivity measurement apparatus with two identical copper discs, one coated matte black (reference blackbody, ε ≈ 1), one with the test surface finish
• Heaters embedded in each disc with individual power controllers
• K-type thermocouples at disc surfaces and ambient
• Digital temperature indicator with multi-channel selector
• Enclosure to minimise convection effects and maintain radiation-only heat transfer
• Dimmer stats / power controllers for each heater

Procedure

1. Switch on both heaters and set them to the same power input.
2. Allow the system to reach steady state, typically 45–60 minutes. Steady state is achieved when temperature readings remain constant (change < 0.5°C over 5 minutes).
3. At steady state, record:
   • T₁ = temperature of blackbody (black disc) surface (K)
   • T₂ = temperature of test surface disc (K)
   • T_a = ambient temperature (K)
4. Both discs receive the same power input. At steady state, the blackbody disc reaches a lower temperature than the test disc, because the blackbody emits more radiation to the surroundings.
5. Calculate emissivity using the formula derived below.
6. Repeat for different power settings to verify consistency.

Derivation of Emissivity Formula

At steady state, heat input (Q) = heat radiated to surroundings for each disc:

For blackbody disc: Q = σ A (T₁⁴ − T_a⁴) × 1 (since ε_b = 1)

For test disc: Q = σ A (T₂⁴ − T_a⁴) × ε

Since Q is the same for both (same power input, same disc geometry):

σ A (T₁⁴ − T_a⁴) = ε σ A (T₂⁴ − T_a⁴)

ε = (T₁⁴ − T_a⁴) / (T₂⁴ − T_a⁴)

Observation Table

Expected Results

• For a polished aluminium test surface: ε ≈ 0.05–0.10
• For a painted or anodised surface: ε ≈ 0.80–0.95
• The blackbody disc will always be at a lower temperature than the test disc when both receive the same power, confirming higher emissivity = more radiation loss at the same temperature.

Precautions

• Ensure the apparatus is shielded from drafts, convection effects can corrupt the radiation-only measurement.
• Allow full thermal steady state before recording temperatures.
• Both discs must receive identical power, verify with a wattmeter.
• Temperatures must be recorded in Kelvin for the T⁴ formula.

Viva Questions

1. What is a blackbody? Does a blackbody exist in nature?
2. Distinguish between emissivity, absorptivity, and reflectivity.
3. State Kirchhoff’s law of radiation and its significance.
4. Why does a polished surface have lower emissivity than a matte surface?
5. What is the Stefan-Boltzmann constant and its units?