(a) Calculate the rate of heat conduction through house walls that are 13.0cm thick and have an average thermal conductivity twice that of glass wool (fiberglass insulation). Assume there are no windows or doors (for simplicity). The surface area of the walls is 120m2, and their inside surface is at 18.0°C, while their outside surface is at 5.00°C (b) How many 1-kW room heaters would be needed to balance the heat transfer due to this conduction?
Added by James V.
Step 1
042 = 0.084 \, \text{W/m} \cdot \text{K} \)), \( A \) = surface area of the walls (120 m\(^2\)), \( \Delta T \) = temperature difference (18°C - 5°C = 13°C), \( d \) = thickness of the walls (13.0 cm = 0.13 m). Show more…
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(a) Calculate the rate of heat conduction (in W) through house walls that are 12.5 cm thick and that have an average thermal conductivity twice that of glass wool. Assume there are no windows or doors. The surface area of the walls is 150 m^2 and their inside surface is at 19.5°C, while their outside surface is at 5.00°C. (b) How many 1 kW room heaters would be needed to balance the heat transfer due to conduction?
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Compare the rate of heat conduction through a 13.0 -cm- thick wall that has an area of 10.0 ${m}^{2}$ and a thermal conductivity twice that of glass wool with the rate of heat conduction through a window that is 0.750 ${cm}$ thick and that has an area of 2.00 ${m}^{2}$ , assuming the same temperature difference across each.
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