I am trying to understand how heat is lost from a glass aquarium tank maintained at some temperature above ambient. If we imagine a direct interface between the body of water and the surrounding air, I think heat transfer would be entirely due to radiation and convection. When a glass pane is placed in-between the two, heat transfer is due (entirely?) to conduction. What I am having trouble understanding, is how the heat lost by conduction through the glass can be so significant but as soon as that glass is removed the heat loss by radiation and convection is so trivial. It seems that the presence of the glass pane encourages heat-loss? Here is what I am considering: A body of water with surface area 1m x 0.25m , or 0.25 m^2 at temperature 29C. The surrounding environment temperature of 25C. A glass pane of thickness 0.01 m. Heat loss by conduction: Watts = (k * A * (Th - Tl)) / d , where k for glass is 1.1 Watts/m*K = (1.1 * 0.25 * (29 - 25)) / 0.01 = 110 Watts of heat flow. When the pane is removed: Heat loss by radiation: Watts = (em * sbc * A * (Th^4 - Tl^4)), where em for water is 0.95 and sbc = 5.6703x10^(-8) W/m^2*K^4 = 0.95 * 0.000000056703 * 0.25 * ( (29^4) - (25^4) ) = 0.0043 Watts Heat loss by convection: Watts = (h * A * (Th - Tl)), where h for water is taken to be 20 W/m^2*K = 100 * 0.25 * (29 - 25) = 20 Watts Here it appears that when a glass pane is present 110 watts of heat are lost, but when it is removed only 20.0043 watts are lost. Can anyone explain this to me? Thank you for your time.