Question

A \(2.5\)-m-high, 4-m-wide, and 20 -cm-thick wall of a house has a thermal resistance of \(0.025^{\circ} \mathrm{C} / \mathrm{W}\). The thermal conductivity of the wall is (a) \(0.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (b) \(1.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (c) \(3.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (d) \(5.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (e) \(8.0 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\)

Short answer

a) 0.8 W/m·K b) 1.0 W/m·K c) 1.2 W/m·K d) 1.5 W/m·K Answer: a) 0.8 W/m·K

Step-by-step solution

Identify the formula for thermal resistance

Recall that the formula for thermal resistance (\(R\)) of a flat object is given by: $$ R = \dfrac{d}{kA} $$ where \(d\) is the thickness of the material, \(k\) is the thermal conductivity, and \(A\) is the surface area of the material.

Gather the given information

The problem provides us with the dimensions and thermal resistance of the wall. We have: - Thermal resistance (\(R\)) = \(0.025^{\circ} \mathrm{C} / \mathrm{W}\) - Height of the wall = \(2.5~\mathrm{m}\) - Width of the wall = \(4~\mathrm{m}\) - Thickness of the wall (\(d\)) = \(20~\mathrm{cm}=0.2~\mathrm{m}\)

Calculate the surface area (A)

The surface area (A) can be calculated as follows: $$ A = Height \cdot Width = 2.5~\mathrm{m} \times 4~\mathrm{m} = 10~\mathrm{m}^2 $$

Solve for the thermal conductivity (k)

Using the formula for thermal resistance, we can solve for the thermal conductivity (k): $$ R = \dfrac{d}{kA} \Rightarrow k = \dfrac{d}{RA} $$ Now, plug in the given values and solve for \(k\): $$ k = \dfrac{0.2~\mathrm{m}}{(0.025^{\circ} \mathrm{C} / \mathrm{W}) \times 10~\mathrm{m}^2} = 0.8~\mathrm{W}/\mathrm{m} \cdot \mathrm{K} $$ The correct answer is (a) \(0.8~\mathrm{W}/\mathrm{m} \cdot \mathrm{K}\).