Question

The windows on the Apollo Command Module were double-paned tempered silica glass with an additional amorphous-fused silicon layer outside. Let's consider the heat flow through one of these windows as though it was a regular double-paned window that we use down here on Earth. A side window on the Apollo command module was a 13-in x 13-in square, consisting of two layers of glass separated by 0.10-in of air. Neglect the heat flow through the glass, and only consider the heat flow through the thickness of the air. Calculate the heat flow through this window over a span of 4.00 hours, given that the temperature inside the command module is kept at a constant 22.0 °C and the temperature of outer space is -454 °F.

Answer 1

y

Area (A) = 13 inch times 13 inch = 169 times (0.0254 m)^2 = 0.109 m^2 Thickness of air = 0.1 inch = 0.1 times 0.0254 m = 0.00254 m time taken (t) = 4.00 hr = 14,400 second temp of room T_1 = 22.0 degree c = 295.15 k temp of T_2 = -454 degree F = 3.15 k Thermal conductivity air = 26.02 times 10^-3 w/m - k. therefore Heat flow delta Q = KA/d times delta T (t) = 26.02 times 10 times 0.109/0.00254 times 292 times 14,400 = 4,695,106.16 J delta Q = 4.695 times 10^6 J