Glass thickness?

Consider a double-paned window consisting of two panes of glass, each with a thickness of 0.500 cm.?

• Consider a double-paned window consisting of two panes of glass, each with a thickness of 0.500 cm and an area of 0.720 m2, separated by a layer of air with a thickness of 1.75 cm. The temperature on one side of the window is 0.00° C; the temperature on the other side is 20.0° C. In addition, note that the thermal conductivity of glass is roughly 36 times greater than that of air. (a) Approximate the heat transfer through this window by ignoring the glass. That is, calculate the heat flow per second through the 1.75 cm of air with a temperature difference of 20.0° C. (The exact result for the complete window is 19.0 J/s.) Express your answer with three significant digits. (b) Use the approximate heat flow found in part (a) to find an approximate temperature difference across each pane of glass. (The exact result is 0.157° C.) Express your answer with three significant digits.

• Answer:

  Qdot = kappa * A *delta(T)/D For air, kappa = 0.025 W / m C, so Qdot = 0.025 J/s / (m C) * 0.720m^2 * 20.0C/(0.0175m) = 20.6 J/s (b) That same heat flow is through a pane, so 20.6 J/s = 36*0.025W/(m C) * 0.720m^2 * Delta(T) / (0.005 m) => Delta(T) = 0.157 C