0:00 / 0:00

Heat Transfer Through Conduction

  • Heat transfer through a material is called conduction.
  • Heat current in conduction has the following relation:
H=dQdt=kAΔTLH=\dfrac{dQ}{dt}=kA\dfrac{\Delta T}{L}

  • kk is the thermal conductivity of the material
  • The larger kk, the better conductor it is
  • LL is the thickness of the material
  • The thicker the material is, the less heat is transferred
  • The more area of material that can conduct heat, the more heat that is transferred
  • ΔT\Delta T is the temperature difference on both sides of the conductor
  • ΔTL\frac{\Delta T}{L} is the temperature difference per unit length and is called the temperature gradient
0:00 / 0:00

Example: Heat Transfer

The glass of a car window is rectangular in shape and has dimensions of 40 cm by 60 cm and is 3cm thick. How much heat will be transferred by conduction in 10.0 s through the glass if there is a temperature difference of 3.5º between the inner and outer surfaces? (kglass=0.84 J/m s ºC)

Solution:

A=40×60cm2A=40\times60cm^{2}

=2400cm2=0.24m2=2400cm^{2}=0.24m^{2}

L=3cmL=3cm

0.03m0.03m



ΔQ=kAΔTΔtLthickness\Delta Q=\dfrac{kA\Delta T\Delta t}{L}_{{\large \rightsquigarrow}\hspace{0.2cm}\text{thickness}}

=(0.84J/ms°C)(0.24m2)(3.5°C)(10s)0.03m=\dfrac{(0.84J/ms\degree C)(0.24m^2)(3.5\degree C)(10s)}{0.03m}

=235.2J=235.2 J
Extra Practice