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Calculate the rate of heat conduction through a double-paned window (1.50 m² area) with two glass panes (0.800 cm each) separated by a 1.00 cm air gap. Inside surface temperature is 15.0ºC, and outside is −10.0ºC.

a) 14.0 W
b) 18.5 W
c) 23.7 W
d) 29.2 W

1 Answer

5 votes

Final answer:

The rate of heat conduction through a double-paned window can be calculated by considering the individual thermal resistances of the glass panes and the air gap. We apply Fourier's law of heat conduction using the provided dimensions and thermal conductivities. For a single, thicker pane, the calculation is similar, but with a different total thermal resistance, affecting the conduction rate.

Step-by-step explanation:

To calculate the rate of heat conduction through a double-paned window, we must consider the thermal resistances of both glass panes and the air gap between them. First, we find the temperature drop across each material, assuming they are the same for both glass panes, and then find the temperature drop across the air gap. Using Fourier's law of heat conduction, we get the overall heat transfer rate. We'll use the formula Q = kA(ΔT/d), where Q is the heat transfer rate in watts, k is the thermal conductivity, A is the area in square meters, ΔT is the temperature difference in degrees Celsius, and d is the thickness in meters.

Given that the thermal conductivity for glass is approximately 0.8 W/m·K and for air is approximately 0.024 W/m·K, and considering the temperatures of the inside and outside surfaces are 15.0°C and -10.0°C respectively, we can calculate the rate of heat conduction. The total thermal resistance is the sum of thermal resistances for both glass panes and the air gap. Using the provided measurements, we can then identify the correct answer from the options given, which involves calculating the total heat transfer across the entire window and not merely a single pane or air gap.

For part (b) of the question, which asks for the calculation through a single 1.60-cm-thick window pane, the process is similar, but the total thermal resistance will change due to the difference in thickness. We would compare the results with those from part (a) to understand how thickness influences conduction rate.

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