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Fill in the Blank: If the slab is 2 cm thick and has a density of 2.7 g/cm³, the average value of (k⋅c) ∗throughout the medium is .

a) 1.35 g/cm²
b) 5.4 g/cm²
c) 2.7 g/cm²
d) 10.8 g/cm²

1 Answer

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Final answer:

The average value of (k⋅c) throughout the medium for a 2 cm thick slab with a density of 2.7 g/cm³ is 5.4 g/cm² (option b). For the 100-gram hollow cube made of foil with a surface density of 15 g/cm², the approximate side length is 1.05 cm (option a).

Step-by-step explanation:

The student's question pertains to calculating the surface density of a slab with a known density and thickness. In the first scenario given, we need to determine the average value of (k⋅c) throughout the medium for a slab that is 2 cm thick and has a density of 2.7 g/cm³. This calculation does not directly apply to thermal conductivity or specific heat capacity calculations, but seems to be a simple product of density and thickness. The correct answer would be obtained by multiplying the density by the thickness:

Density = 2.7 g/cm³
Thickness = 2 cm

Surface density (k⋅c) = Density × Thickness = 2.7 g/cm³ × 2 cm = 5.4 g/cm²

Hence, the surface density of the slab is 5.4 g/cm², which corresponds to option b).

To address the second given problem, where a piece of aluminum foil with known surface density is used to construct a hollow cube, the volume of the cube can be determined from its mass and surface density:

Surface Density = 15 g/cm²
Mass = 100 g

Volume = Mass / (Surface Density × 6)

(Considering all six sides of the cube)

Volume = 100 g / (15 g/cm² × 6) = 1.11 cm³

The side length s of the cube can be obtained by taking the cube root of the volume:

s = √₃(Volume) = √₃(1.11 cm³) ≈ 1.04 cm

Therefore, the approximate side length of the hollow cube is closest to 1.05 cm, which is option a).

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