Thermal conductivity is a measure of a material's ability to conduct heat. It indicates how efficiently heat is transferred through a material. This property is crucial in many applications, including building insulation, electronic device cooling, and material science.

Symbol and Units

• Symbol: \( k \) or \( \lambda \)
• SI Unit: Watts per meter-kelvin (W/(m·K))
• Other Units: British thermal units per hour per foot per degree Fahrenheit (BTU/(h·ft·°F))

Formula

The thermal conductivity \( k \) is calculated using the formula:

\( Q = \frac{k \cdot A \cdot \Delta T}{d} \)

where:

• Q is the rate of heat transfer (W),
• k is the thermal conductivity (W/(m·K)),
• A is the cross-sectional area through which heat is conducted (m²),
• ΔT is the temperature difference across the material (K),
• d is the thickness of the material (m).

Key Points

• Thermal conductivity measures how well heat is conducted through a material.
• Materials with high thermal conductivity transfer heat more efficiently than those with low thermal conductivity.
• Thermal conductivity can be affected by the material's temperature, structure, and phase.

Applications

• Construction: Selecting materials for insulation to improve energy efficiency in buildings.
• Electronics: Managing heat dissipation in electronic components to prevent overheating.
• Manufacturing: Designing products that require precise thermal management, such as heat exchangers and thermal barriers.

Material Variance

• High Thermal Conductivity: Metals like copper and aluminum, which are used in heat sinks and conductors.
• Low Thermal Conductivity: Insulating materials like fiberglass and polystyrene, used to reduce heat transfer in walls and roofs.

Examples

Example 1: Calculate the rate of heat transfer through a material with thermal conductivity \( k = 0.5 \text{ W/(m·K)} \), thickness \( d = 0.1 \text{ m} \), area \( A = 2 \text{ m}^2 \), and temperature difference \( \Delta T = 10 \text{ K} \):

Q = \(\frac{0.5 \cdot 2 \cdot 10}{0.1} = 100 \text{ W}\)

Example 2: Convert thermal conductivity from W/(m·K) to BTU/(h·ft·°F):

1 W/(m·K) ≈ 0.5779 BTU/(h·ft·°F).

If \( k = 2 \text{ W/(m·K)} \), then \( k \approx 2 \times 0.5779 = 1.156 \text{ BTU/(h·ft·°F)} \).