Question

A picture window has dimensions of 1.40 m

Answer 1

Rate of heat loss = dQ/dt = 0.80 x (1.40 x 2.50)m² x (21 + 18) / 0.0055 = 19854.54 W

At steady state, the the temperature gradient (and hence the flux) between two "reservoirs" held at constant temperatures is a constant.

Here, the thing to recognize is that once one adds the layer of paper (and lets the system come to a thermal steady state) the heat flux through the glass must be equal to the heat flux through the paper at steady state, otherwise, there would be a buildup (or loss) of thermal energy at the interface between the glass and paper, and the temperature would be changing, which is contrary to the assumption of steady state.

Let T_interface be the temperature at the glass-paper interface. Then:

q = -k_glass*(T_interior - T_interface)/(5.50mm) = -k_paper*(T_interface - T_outside)/(0.750mm)

Plugging in the values for the thermal conductivities for the glass and paper, as well as the inside and outside temperatures, we get:

(0.8W/(m*C))*(21 - T_interface)/(5.50mm) = (0.050W/(m*C))*(T_interface +18)/(0.750mm)

(Note that in this case the difference between C and kelvins doesn't matter, because we are dealing with changes in temperature, not absolute temperatures)

Solving for T_interface, we get:

T_interface = 8.74C

Now use this to calculate the heat flux, and multiply that by the area of the window to get the total heat loss. You can either use the flux through the glass or the flux through the paper to calculate the heat flux because these quantities are equal.

q = (0.8W/(m*