Given:

Thickness of air, Δx=1mm=0.001m

Area of the layer of air, A=1m2

Temperature difference, ΔT=20°C

Thermal conductivity of air, kt=0.026 J.s-1m-1.K-1

Formula Used:

Rate of heat conduction:

  QΔt=−ktAΔTΔx

Where Q is the amount of heat, and Δt is the time interval

Calculation:

The rate of heat conduction through a layer of still air is

  QΔt=−ktAΔTΔx=−0.026×1×200.001=−520watts

Conclusion:

  QΔt=−520watts

The negative sign indicates that heat flows down the temperature gradient.