Collection of Solved Problems in Physics

Heat Loss Through a Glass Pane

Task number: 1277

• Hint

  What variables does the amount of heat that passes through a pane of glass depend on? Does it depend directly or inversely? The constant of proportionality is called the thermal conductivity coefficient.

• Thermal conduction equation

  The thermal conduction equation is:

  \[Q=\lambda\frac{S\mathrm{\Delta}t}{d}\tau\]

  where Q is the lost heat, λ is the thermal conductivity coefficient (characterizing the material), whose value for glass is listed in The Handbook of Chemistry and Physic, S is the surface area, d is the thickness of material, through which the heat passes, Δt is the temperature difference between both sides of the material and τ is time.

  The equation assumes that the temperature difference between both sides remains constant.

• Analysis

  The heat that passes through the pane of glass per a unit of time is directly proportional to the temperature differences and the surface area of the glass pane, and inversely proportional to the thickness of the glass. The constant of proportionality is the thermal conductivity coefficient.

• Numerical values

  d = 4 mm = 4·10−3m	the thickness of the glass pane
  S = 2 m2	the surface area of the glass pane
  t1 = −10 °C	the outdoor temperature
  t2 = 20 °C	the room temperature
  τ = 3600 s	the time, for which we measure how much heat passes through the glass pane
  Q = ?	the heat lost through the glass pane within one hour

  From The Handbook of Chemistry and Physics:

  λ = 1,0 W m−1K−1

  the thermal conductivity coefficient of glass

• Solution

  The heat that passes through the glass pane is directly proportional to the surface area S of the pane, the temperature difference Δt and time τ, and inversely proportional to the pane thickness d. The constant of proportionality is the thermal conductivity coefficient λ of the material. Therefore, we can write:

  \[Q=\frac{\lambda S\tau\mathrm{\Delta}t}{d}=\frac{\lambda S\tau\left(t_2-t_1\right)}{d}\,\mathrm{.}\]

  Numerical insertion:

  \[Q=\frac{\lambda S\tau\left(t_2-t_1\right)}{d}=\frac{1.0\cdot{2}\cdot{3600}\left(20+10\right)}{4\cdot{10^{-3}}}\,\mathrm{J}\dot=\] \[\dot=54\cdot{10^6}\,\mathrm{J}=54\,\mathrm{MJ}\]

• Answer

  The amount of heat that passes through a glass pane within one hour is approximately 54 MJ.

• For information

  If we had windows made of glass with properties stated in this task, the heat losses would be enormous. The heat losses are nowadays reduced by using various methods, for example doubling the glass. The layer of air between the two panes has much lower value of the thermal conductivity coefficient than the glass itself, hence loosing less heat. The space between two glass panes can also be filled with other gases, such as argon, xenon or krypton; all of them have lower values of thermal conductivity coefficient.

  Another method is using a thin layer of metal coating or bonding a foil to reflect the heat back into the room.

  The manufacturers characterize the properties of window panes with the coefficient U₀, which indicates, how much heat passes through a window with surface area of 1 m² at a temperature difference 1 °C per a unit of time. Its unit is therefore Wm⁻²K⁻¹. We obtain the coefficient U₀ by dividing the thermal conductivity coefficient λ by the thickness d of the pane.

  For glass from this task we get the value of the coefficient U₀:

  U₀ = 250 W m⁻²K⁻¹.

  The manufacturers of window panes indicate that the value of U₀ of certain types of windows is less than 1.15  W m⁻²K⁻¹, which differs significantly from the value for the glass in our task.