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The thermal conductivity of a sheet of rigid, extruded insulation is reported to be k 0.029 W/m measured temperature difference across a 20-mm-thick sheet of the material is Ti-T2 = 10°C. K. The

User Fiveagle
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Final answer:

The question involves calculating the rate of heat conduction through materials with different thicknesses and thermal conductivities, assuming the same temperature difference, a topic covered in high school Physics.

Step-by-step explanation:

The question pertains to the concept of thermal conductivity and the rate of heat conduction through different materials with varying thermal conductivities and thicknesses. This topic relates to the Physics domain, particularly within the area of thermodynamics and heat transfer in materials. The given values allow for the comparison of heat transfer rates across different materials assuming a constant temperature difference.

To compare rates of heat conduction, we use the equation Q/t = κ * A * (T₂ - T₁) / d, where Q/t is the rate of heat transfer, κ is the thermal conductivity, A is the area, d is the thickness of the material, and (T₂ - T₁) represents the temperature difference across the material.

Using this formula, we can determine the rate of heat conduction for the wall and the window as described, taking into account their respective thicknesses, areas, and the thermal conductivities. By doing so, one can gauge the effectiveness of insulation provided by different materials, exemplified by the reference to R factors used for common household insulators.

User SQVe
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