I'm trying to understand how modern window films or coating (so-called "Low-E" coatings) can improve the "R" value of the window just by reflecting internal room radiant energy (presumably long-wave IR or black-body radiation) back into the room.

I presume that a room containing air and various objects at "room temperature" will emit a certain amount of IR and some percentage of that will pass through the window aperture, but glass with a low-e coating will reflect some of that IR back into the room.

My "gut" (or rather, my hand - on a cold window surface in the winter) tells me that the window does an excellent job of presenting a cold surface to the air circulating in the room, and a lot of heat is carried out of the room through thermal transfer, which does not involve the low-e coating.

Since I don't have a good "feel" for how energetic or intense the long-wave ambient room IR is, I really can't appreciate what the low-e coating is doing in terms of improving the window's total insulative property. Nor can I get some sort of idea as to what my primary question is here - which is to get a rough idea if the ambient room IR amounts to 5% or 50% of the total heat energy passing through a window in the winter.

Bonus question: Does the air in the room emit IR (enough to be of any significance) or would it be just the more dense materials / objects / surfaces?

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    $\begingroup$ Re, "my hand - on a cold window..." Was that a single pane window? double pane? triple pane? Most "Low-E" windows will also be double or triple pane assemblies. $\endgroup$ Aug 23, 2019 at 1:18
  • $\begingroup$ I am referring to the now typical dual-pane IGU (Insulated glass unit) that has a steel or composite spacer and is sealed (air tight?) and usually filled with an inert, dry gas (argon, etc). Although my question about just what a low-e coating is doing in terms of reflecting interior IR back into a room could apply to a single-pane or an IGU (the IGU would just have lower thermal heat losses but it's not the thermal heat-transfer phenomena that I'm asking about). $\endgroup$ Aug 23, 2019 at 1:32

2 Answers 2


As a first rough estimate we can assume the indoor and outdoor matter behave as perfect black bodies and are each at thermodynamic equilibrium with constant temperature. Then it follows from the Stefan–Boltzmann law that the heat loss is equal to

$P = A \epsilon \sigma\left( (T+\delta T)^4 - T^4\right) \approx A\epsilon\times 103 [W/m^2] $,


  • $A$: surface area of the window in $m^2$
  • $\epsilon$: emissivity of the black bodies times the proportion of the black-body radiation which passes through the window
  • $\sigma\approx 5.67037\times 10^{-8} [W/(m^2*K^4)]$: Stefan–Boltzmann constant
  • $T = 273.15[K]$: outdoor temperature assumed to be 0°C
  • $T+\delta T=293.15[K]$: indoor temperature assumed to be 20°C

Under these assumptions, and 100% transmission of the EV spectrum of the glass, the power loss is equal to 103W per $m^2$ window area for the 20K difference in temperatures, or about $5W/(m^2 K)$.

This is higher than the thermal transmittance (or U-values which count the losses due all of conduction, convection and radiation) of most current windows, e.g.:

  • Single glazing: $U_g = 5.7\, W/(m^2 K)$
  • Triple glazing: $U_g = 0.4 - 0.8\, W/(m^2 K)$

Note, direct sunlight will also contribute to the energy transfer and can be very substantial depending on the angle and absorption/reflection of the glass and indoor material and can reach $800\,W/m^2$ (assuming the glass transmits 80% of the solar spectrum, 90 degree angle, and black indoor materials).


I'm trying to understand how modern window films or coating (so-called "Low-E" coatings) can improve the "R" value of the window

Do you have a reference for that statement?

In winter these will let in sunlight (visible range) while keeping in thermal radiation. They might say this is equivalent to increasing the "R value", but in summer it also keeps the interior warmer, which is like a decrease in "R value".

You can calculate heat conduction using the equations here (assuming good air circulation at the window) and radiation using the Stefan–Boltzmann law (assuming a small window compared to room dimension) (Note that you have to subtract the radiation coming in from the lower temperature exterior).

Bonus question: Assuming the air and objects are all at the same temperature, it doesn't matter.

  • $\begingroup$ All I know is that somehow they've created an R rating that takes into account whether or not the window has low-e coated glass. This R rating takes into account the basic thermal resistance of the IGU (ie glass thickness, pane-spacing, type of gas used between the panes). So you take that basic thermal R value and then increase it when a low-e coating is applied (based on what math or physics I don't know) I think this new R value is more for marketing purposes. -> My primary question is still - what portion of lost room heat is radiant vs thermal. $\endgroup$ Aug 24, 2019 at 13:17
  • $\begingroup$ Without a link to to the "R value" claims I don't think anyone can answer. $\endgroup$ Aug 24, 2019 at 14:55
  • $\begingroup$ My question is not about how the window industry comes up with an R value for windows. My question is: Given a room containing objects and air at "room temperature" - say 300 kelvin, then how much of the room's heat energy will be radiated (through long-wave IR) out a window (size is 1 m x 1.5 m) vs how much heat energy will be conducted through the window (given an exterior temperature of, say, 275 K and a window with a thermal heat loss of 480 watts x 25 (degrees) = 12,000 watts (single pane, 3 mm thick). $\endgroup$ Aug 24, 2019 at 20:03

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