Thermal Properties of Materials

Thermal properties govern heat transfer through building envelopes, HVAC equipment, and distribution systems. Understanding these fundamental material characteristics enables accurate load calculations, proper insulation specification, and effective thermal storage design.

Thermal Conductivity (k)

Thermal conductivity quantifies a material’s ability to conduct heat. Materials with high conductivity rapidly transfer thermal energy, while low-conductivity materials resist heat flow and function as insulators.

Definition and Units

Thermal Conductivity (k): Heat transfer rate through unit thickness per unit area per unit temperature difference.

Conversion: 1 Btu·in/(hr·ft²·°F) = 0.1442 W/(m·K)

Fourier’s Law

Heat conduction through a homogeneous material follows Fourier’s law:

q = -k·A·(dT/dx)

Where:

• q = heat transfer rate (W or Btu/hr)
• k = thermal conductivity (W/(m·K) or Btu·in/(hr·ft²·°F))
• A = cross-sectional area (m² or ft²)
• dT/dx = temperature gradient (K/m or °F/ft)

For steady-state conduction through a plane wall:

q = k·A·(T₁ - T₂)/L

Where:

• T₁, T₂ = surface temperatures (K or °F)
• L = material thickness (m or ft)

Thermal Conductivity of Building Materials

Structural Materials:

Insulation Materials:

Roofing and Finish Materials:

Temperature Dependence

Thermal conductivity varies with temperature. For most building materials over typical HVAC operating ranges (-40°F to 150°F or -40°C to 65°C), linear approximation suffices:

k(T) = k₀[1 + β(T - T₀)]

Where:

• k₀ = conductivity at reference temperature T₀
• β = temperature coefficient (typically 0.001-0.003 K⁻¹)
• T = material temperature

Moisture Effects

Moisture dramatically increases thermal conductivity. Water has k = 0.6 W/(m·K), roughly 20 times higher than dry insulation. Moisture content by volume increases effective conductivity:

k_effective = k_dry + k_water × moisture_fraction

For fibrous insulation at 5% moisture by volume, conductivity can increase 50-100%. Vapor retarders and proper drainage prevent moisture accumulation.

Thermal Resistance (R-value)

Thermal resistance quantifies a material’s ability to resist heat flow. R-value is the inverse of thermal conductance.

Definition and Calculation

R-value for Homogeneous Material:

R = L/k

Where:

• R = thermal resistance (m²·K/W or hr·ft²·°F/Btu)
• L = thickness (m or in)
• k = thermal conductivity

Heat Flow Through Resistance:

q = A·(T₁ - T₂)/R

Series Resistances

For layered assemblies, resistances add in series:

R_total = R₁ + R₂ + R₃ + ... + R_n

Complete Wall Assembly:

R_total = R_interior_film + R_wall_layers + R_exterior_film

Typical surface film resistances:

• Interior wall (still air): 0.68 hr·ft²·°F/Btu (0.12 m²·K/W)
• Exterior wall (15 mph wind): 0.17 hr·ft²·°F/Btu (0.03 m²·K/W)
• Interior ceiling (heat flow up): 0.61 hr·ft²·°F/Btu (0.11 m²·K/W)
• Interior floor (heat flow down): 0.92 hr·ft²·°F/Btu (0.16 m²·K/W)

Overall Heat Transfer Coefficient (U-factor)

U-factor represents the overall heat transmission:

U = 1/R_total

Units: W/(m²·K) or Btu/(hr·ft²·°F)

Lower U-factors indicate better insulation performance. Energy codes specify maximum U-factors for envelope components.

Specific Heat Capacity (c_p)

Specific heat quantifies thermal energy storage per unit mass per degree temperature change.

Definition and Units

Conversion: 1 Btu/(lb·°F) = 4186.8 J/(kg·K)

Sensible Heat Equation

Energy required to change material temperature without phase change:

Q = m·c_p·ΔT

Where:

• Q = sensible heat (J or Btu)
• m = mass (kg or lb)
• c_p = specific heat (J/(kg·K) or Btu/(lb·°F))
• ΔT = temperature change (K or °F)

Specific Heat of Common Materials

Building Materials:

HVAC Fluids:

Volumetric Heat Capacity

For thermal mass calculations, volumetric heat capacity proves more useful:

C_v = ρ·c_p

Where:

• C_v = volumetric heat capacity (Btu/(ft³·°F) or kJ/(m³·K))
• ρ = density (lb/ft³ or kg/m³)
• c_p = specific heat

Thermal Diffusivity (α)

Thermal diffusivity characterizes transient heat conduction, indicating how rapidly temperature changes propagate through a material.

Definition and Calculation

α = k/(ρ·c_p)

Where:

• α = thermal diffusivity (m²/s or ft²/hr)
• k = thermal conductivity
• ρ = density
• c_p = specific heat

Units:

• SI: m²/s (typical range: 10⁻⁷ to 10⁻⁴ m²/s)
• I-P: ft²/hr (typical range: 0.01 to 50 ft²/hr)

Physical Interpretation

High thermal diffusivity means:

• Rapid temperature response to boundary condition changes
• Low thermal mass effect
• Temperature quickly equalizes throughout material

Low thermal diffusivity means:

• Slow temperature response
• High thermal mass effect
• Temperature gradients persist longer

Thermal Diffusivity Values

Application in Transient Analysis

Penetration Depth for cyclic temperature variations:

δ = √(α·t)

Where:

• δ = penetration depth (m or ft)
• t = time period (s or hr)

For 24-hour cycle (diurnal temperature swing):

Thermal Time Constant for slab or wall:

τ = L²/(π²·α)

Where:

• τ = time constant (s or hr)
• L = thickness (m or ft)

This determines lag time for temperature changes to propagate through the material.

Thermal Mass

Thermal mass stores sensible heat, moderating indoor temperature swings and reducing peak cooling/heating loads.

Effective Thermal Mass

Materials exposed to conditioned space provide thermal storage. The effective thermal mass depends on:

• Volumetric heat capacity (ρ·c_p)
• Surface area exposed to indoor air
• Thermal coupling (convective heat transfer coefficient)
• Diurnal temperature swing penetration depth

Effective Heat Capacity:

C_eff = ρ·c_p·V_eff

Where V_eff = A·δ, with A = surface area and δ = penetration depth.

Thermal Mass Benefits

Load Shifting: Absorbs heat during day, releases at night, reducing peak cooling demand.

Temperature Stabilization: Dampens indoor temperature swings from solar gains and outdoor temperature variations.

Quantification: Thermal decrement factor and time lag from ASHRAE Handbook of Fundamentals, Chapter 18 (Heat, Air, and Moisture Control in Building Assemblies).

Thermal Effusivity (e)

Thermal effusivity characterizes transient surface temperature response to heat flux, critical for contact thermal comfort and short-term heat storage.

Definition

e = √(k·ρ·c_p)

Units: J/(m²·K·s^0.5) or Btu/(ft²·°F·hr^0.5)

Physical Significance

High effusivity materials:

• Feel cool to touch (extract heat rapidly from skin)
• Rapidly absorb/release heat at surface
• Metal, concrete, stone

Low effusivity materials:

• Feel warm to touch (extract heat slowly)
• Slowly absorb/release surface heat
• Wood, carpet, insulation

Contact Temperature

When two semi-infinite bodies at different temperatures contact, the interface temperature is:

T_interface = (e₁·T₁ + e₂·T₂)/(e₁ + e₂)

This explains why metal feels colder than wood at the same temperature: metal’s high effusivity extracts heat rapidly from skin.

Emissivity (ε)

Emissivity quantifies thermal radiation emission relative to an ideal blackbody.

Definition and Range

Emissivity (ε): Ratio of actual emitted radiation to blackbody radiation at same temperature.

Range: 0 ≤ ε ≤ 1

• ε = 1: Perfect blackbody (theoretical maximum)
• ε = 0: Perfect reflector (no emission)

Stefan-Boltzmann Law

Radiation heat transfer from a surface:

q = ε·σ·A·(T_s⁴ - T_surr⁴)

Where:

• σ = Stefan-Boltzmann constant = 5.67×10⁻⁸ W/(m²·K⁴) = 0.1714×10⁻⁸ Btu/(hr·ft²·R⁴)
• T_s = surface absolute temperature (K or R)
• T_surr = surrounding surface absolute temperature (K or R)

Emissivity Values

Building Surfaces:

Radiant Barriers

Low-emissivity surfaces (ε < 0.1) function as radiant barriers, reducing radiative heat transfer across air spaces. Applications include:

Attic Radiant Barriers: Aluminum foil facing reduces radiant gain to ceiling insulation, lowering cooling loads in hot climates. Per ASHRAE, effective R-value increase: R-3 to R-6 depending on configuration.

Reflective Insulation Systems: Multiple low-emissivity surfaces separated by air spaces achieve high thermal resistance despite minimal material thickness.

Effective R-value of Air Space with radiant barrier (3.5 in horizontal, heat flow up, winter):

• Without radiant barrier (ε = 0.90 both surfaces): R-0.84 hr·ft²·°F/Btu
• With one radiant barrier (ε = 0.05): R-2.86 hr·ft²·°F/Btu
• With two radiant barriers (ε = 0.05 both): R-4.55 hr·ft²·°F/Btu

Design Considerations

Load Calculation Implications

Transmission Loads: U-factor (inverse of R-value) directly determines conduction heat gain/loss:

Q_transmission = U·A·(T_outdoor - T_indoor)

Material selection affects:

• Peak heating/heating loads
• Annual energy consumption
• Equipment sizing

Thermal Bridging

Discontinuities in insulation create thermal bridges, locally increasing heat transfer. Common bridges:

• Steel studs in walls (reduce effective R-value by 20-50%)
• Concrete balconies penetrating envelope
• Window frames and mullions
• Parapets and ledgers

Effective U-factor accounting for thermal bridging:

U_eff = (U_clear·A_clear + U_bridge·A_bridge)/(A_clear + A_bridge)

Moisture Control

Materials must remain dry to maintain thermal performance. Water condensation occurs when material temperature drops below dewpoint. Vapor retarders, air barriers, and continuous insulation prevent moisture accumulation.

Critical Condensation Plane: Location within assembly where temperature equals dewpoint. Proper vapor retarder placement keeps this plane outside moisture-sensitive materials.

Thermal Mass Strategy

Effective thermal mass requires:

• Exposure: Material surface exposed to indoor air, not covered by carpet or insulation
• Thickness: Sufficient to engage penetration depth (4-6 in concrete optimal)
• Location: Internal mass most effective (exterior insulation with interior mass)
• Climate: Most beneficial in climates with large diurnal temperature swings

Material Selection Criteria

Envelope Insulation:

• Target R-value per code requirements (IECC, ASHRAE 90.1)
• Compatibility with assembly (compression, moisture resistance)
• Cost-effectiveness ($/R-value)

Thermal Storage:

• High volumetric heat capacity (concrete, masonry, water)
• Appropriate thickness for diurnal cycle
• Low resistance between mass and conditioned space

Equipment and Piping:

• Minimize conduction losses (insulate hot/cold surfaces)
• Consider thermal expansion (α affects dimensional change)
• Efficiency implications (insulation reduces standby losses)

Code and Standards References

ASHRAE Standards:

• ASHRAE Handbook—Fundamentals, Chapter 26: Heat, Air, and Moisture Control in Building Assemblies—Climatic Data
• ASHRAE Handbook—Fundamentals, Chapter 33: Properties of Materials
• ASHRAE Standard 90.1: Energy Standard for Buildings (prescriptive envelope requirements)
• ASHRAE Standard 90.2: Energy-Efficient Design of Low-Rise Residential Buildings

Test Methods:

• ASTM C177: Steady-State Heat Flux Measurements (guarded hot plate)
• ASTM C518: Steady-State Thermal Transmission Properties (heat flow meter)
• ASTM C1549: Solar Reflectance (for cool roof materials)
• ASTM E1980: Calculation of Solar Reflectance Index

Building Codes:

• International Energy Conservation Code (IECC): Prescriptive envelope insulation requirements by climate zone
• International Building Code (IBC): Fire resistance ratings affected by material thermal properties
• Local Amendments: Jurisdiction-specific envelope performance requirements

Practical Application Examples

Example 1: Wall Assembly R-Value

8-inch concrete block wall with exterior insulation:

Layers (exterior to interior):

1. Exterior film resistance: R-0.17
2. 2-inch XPS insulation (k = 0.22): R = 2.0/0.22 = 9.09
3. 8-inch concrete block (k = 4.0): R = 8.0/4.0 = 2.00
4. Interior film resistance: R-0.68

R_total = 0.17 + 9.09 + 2.00 + 0.68 = 11.94 hr·ft²·°F/Btu
U = 1/11.94 = 0.084 Btu/(hr·ft²·°F)

Example 2: Thermal Mass Storage Capacity

1000 ft² concrete slab, 4 inches thick:

Volume = 1000 ft² × (4/12) ft = 333.3 ft³
Mass = 333.3 ft³ × 145 lb/ft³ = 48,333 lb
Temperature swing = 75°F - 70°F = 5°F

Q = m·c_p·ΔT = 48,333 lb × 0.22 Btu/(lb·°F) × 5°F
Q = 53,166 Btu = 4.43 ton-hours

This thermal storage can shift peak cooling loads by several hours.

Example 3: Heat Loss Through Pipe Insulation

4-inch steel pipe carrying 180°F water, ambient 70°F:

Pipe parameters:

• Outside diameter: 4.5 in
• Insulation: 1.5-inch fiberglass (k = 0.27)
• Pipe length: 100 ft

R_insulation = (d_o/2)·ln(r_outer/r_inner)/k
R_insulation = (4.5/2)·ln(3.75/2.25)/0.27 = 3.63 hr·ft²·°F/Btu (per ft²)

Surface area = π·d_o·L = π·(7.5/12 ft)·100 ft = 196 ft²

q = A·ΔT/R = 196 ft²·(180-70)°F / 3.63 = 5,934 Btu/hr

Without insulation, heat loss would be approximately 50,000 Btu/hr, demonstrating 88% reduction.

Last Updated: 2026-01-12 Cross-References:

• Material Properties Overview
• Heat Transfer Fundamentals
• Insulation Systems
• Load Calculation Methods
• Envelope Design