Skip to the content
Find a formula

Formula of Thermal Conductivity

Heat and thermodynamics formulasScalarClass 11Updated 1 Nov 2026

k=QLA ΔT tk = \frac{QL}{A\,\Delta T\,t}
SI unit
W/(m K)
Dimensions
[M¹ L¹ T⁻³ K⁻¹]
where
  • k thermal conductivity W/(m K)
  • Q heat conducted J
  • L thickness of the slab m
  • A area of the face m²
  • ΔT temperature difference across the slab K
  • t time s

The formula of thermal conductivity is k=QLAΔT tk = \frac{QL}{A\Delta T\,t}, rearranged from the law of heat conduction Qt=kAΔTL\frac{Q}{t} = \frac{kA\Delta T}{L}. It measures how much heat a material lets through each second, per square metre of face, per metre of thickness and per kelvin of temperature difference.

Metals have large k and feel cold to touch because they draw heat from your hand quickly. Wood, wool, glass and still air have small k and are used as insulators. The quantity is part of the Class 11 chapter on thermal properties of matter and also turns up in AP Physics 2.

Quick facts

  • Formula

    k=QLAΔT tk = \frac{QL}{A\Delta T\,t}

  • Rate of flow

    H=Qt=kAΔTLH = \frac{Q}{t} = \frac{kA\Delta T}{L}

  • SI unit

    W/(m K), also J/(s m K)

  • Dimensions

    [M¹ L¹ T⁻³ K⁻¹]

  • Thermal resistance

    Rth=LkAR_{th} = \frac{L}{kA}, unit K/W

  • Type

    Scalar property of a material

Each quantity in the formula of thermal conductivity

Symbols in k = QL / (AΔT t)
SymbolMeaningSI unit
kkThermal conductivity of the materialW/(m K)
QQHeat conducted through the slabJ
LLThickness, the length heat travelsm
AAArea of the face, at right angles to the flowm²
ΔT\Delta THot face temperature minus cold face temperatureK
ttTime of flows

The ratio ΔTL\frac{\Delta T}{L} is called the temperature gradient. Heat flows faster through a thin slab with a large gradient, a wide face and a material with high k.

Problems with full working

Problem 1: finding k for a slab

In 60 s, 3000 J of heat passes through a slab 2.0 cm thick with a face of 0.50 m². The faces differ by 10 K. Find k.

L = 0.020 m.

k=(3000)(0.020)(0.50)(10)(60)=60300=0.20 W/(m K)k = \frac{(3000)(0.020)}{(0.50)(10)(60)} = \frac{60}{300} = 0.20\ \text{W/(m K)}

Problem 2: heat loss through a window

A glass pane of area 1.5 m² and thickness 4.0 mm has k = 0.80 W/(m K). The inside face is 15 K warmer than the outside. Find the rate of heat loss.

H=(0.80)(1.5)(15)4.0×10−3=4.5×103 WH = \frac{(0.80)(1.5)(15)}{4.0\times10^{-3}} = 4.5\times10^{3}\ \text{W}

Real windows lose much less, because thin layers of still air on each face add extra resistance.

Problem 3: a copper rod with area in cm²

A copper rod 0.50 m long, whose cross section measures 1.0 cm², has its ends at 100 °C and 20 °C. Take k = 400 W/(m K). Find the heat conducted in 5 minutes.

A = 1.0 × 10⁻⁴ m², ΔT = 80 K, t = 300 s.

H=(400)(1.0×10−4)(80)0.50=6.4 WH = \frac{(400)(1.0\times10^{-4})(80)}{0.50} = 6.4\ \text{W}

Q = 6.4 × 300 ≈ 1.9 × 10³ J.

Typical values

Approximate thermal conductivities at room temperature
Materialk in W/(m K)
Silverabout 420
Copperabout 400
Aluminiumabout 200 to 240
Steelabout 15 to 50
Glassabout 0.8 to 1
Waterabout 0.6
Woodabout 0.1 to 0.2
Air (still)about 0.026

The spread is huge: copper conducts roughly fifteen thousand times better than still air. Insulating materials such as foam and wool work mainly by trapping air.

Units and dimensions

Rearranging the law gives k=HLAΔTk = \frac{H L}{A\Delta T}, so its unit is W m/(m² K) = W/(m K). Writing the watt as kg m² s⁻³ leads to

[k]=[ML2T−3][L][L2][K]=[MLT−3K−1][k] = \frac{[M L^2 T^{-3}][L]}{[L^2][K]} = [M L T^{-3} K^{-1}]

The dimensional formula of thermal conductivity page goes through this one step at a time. In older problems you may meet k in cal/(s cm °C); multiply by about 418.6 to convert to W/(m K).

Slabs in series and parallel

Treat Rth=LkAR_{th} = \frac{L}{kA} like an electrical resistance, with heat flow H in place of current and ΔT in place of voltage: H=ΔTRthH = \frac{\Delta T}{R_{th}}. Slabs stacked face to face simply add their thermal resistances, Rth=R1+R2R_{th} = R_1 + R_2. Slabs side by side combine like parallel resistors. This is the same pattern as the resistance formula.

For two layers of equal thickness in series, the effective conductivity is 2k₁k₂ / (k₁ + k₂), not their simple average.

Mistakes to avoid

  • Using thickness in cm or mm without converting to m.
  • Forgetting the time when Q is given in joules rather than H in watts.
  • Using the area of the edge instead of the face the heat crosses.
  • Treating k as a property of the object; it belongs to the material.

The thermodynamics formula sheet places conduction alongside the gas laws, and Newton's law of cooling compares it with cooling to the surrounding air.

Frequently asked questions

What is the formula of thermal conductivity?

k = QL / (AΔT t). In terms of heat flow rate, H = kAΔT / L.

Which unit is thermal conductivity measured in?

Watt per metre per kelvin, W/(m K). Its dimensions are [M¹ L¹ T⁻³ K⁻¹].

Which material has the highest thermal conductivity?

Among common metals, silver is highest, closely followed by copper. Some special materials such as diamond conduct heat even better.

Does thermal conductivity depend on the size of the object?

No. It is a property of the material. The heat flow depends on size, but k does not.

What is thermal resistance?

It is R = L / (kA), measured in K/W. A larger thermal resistance means less heat flows for the same temperature difference.