Question

Consider a 3-m-high, 6-m-wide, and \(0.25\)-m-thick brick wall whose thermal conductivity is \(k=0.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). On a certain day, the temperatures of the inner and the outer surfaces of the wall are measured to be \(14^{\circ} \mathrm{C}\) and \(5^{\circ} \mathrm{C}\), respectively. Determine the rate of heat loss through the wall on that day.

Short answer

Answer: The rate of heat loss through the wall is 259.2 W.

Step-by-step solution

Understand Fourier's Law of heat conduction

Fourier's law states that the rate of heat transfer through a material is proportional to the negative of the temperature gradient and the material's thermal conductivity. Mathematically, it is represented as: \(q = -kA\frac{\Delta T}{\Delta x} \) where, \(q\) = rate of heat transfer (W) \(k\) = thermal conductivity (W/mK) \(A\) = the area through which heat is transferred (m²) \(\Delta T\) = temperature difference between the two surfaces (K) \(\Delta x\) = thickness of the material (m)

Calculate the area of the wall

We are given the dimensions of the wall: 3 m height and 6 m width. To find the area through which heat is transferred, we multiply the height and width of the wall: \(A = 3\,\text{m} \times 6\,\text{m} = 18\,\text{m}^{2}\)

Calculate the temperature difference

We are given the temperatures of the inner and outer surfaces of the wall as \(14^{\circ} \mathrm{C}\) and \(5^{\circ} \mathrm{C}\), respectively. The temperature difference between the two surfaces is: \(\Delta T = T_{\text{inner}} - T_{\text{outer}} = 14^{\circ} \mathrm{C} - 5^{\circ} \mathrm{C} = 9\,\mathrm{K}\) Note that the temperature difference in Kelvin (K) is the same as the difference in Celsius.

Use the formula to find the rate of heat loss

Now, we have all the necessary values to calculate the rate of heat loss, also known as the rate of heat transfer, using Fourier's Law: \(q = -kA\frac{\Delta T}{\Delta x} \) Plug in the given values into the equation: \(q = -0.8\,\mathrm{W/mK} \times 18\,\text{m}^{2} \,\times \dfrac{9\,\mathrm{K}}{0.25\,\text{m}}\) Solve the expression: \(q = -259.2\,\text{W}\) Since the rate of heat loss is negative, it means that the heat is being transferred from the inner surface to the outer surface, which is expected as the inner surface has a higher temperature. The magnitude of the rate of heat loss, which is what the exercise asks for, is: \(q_{\text{loss}} = 259.2\, \mathrm{W}\) Therefore, the rate of heat loss through the wall on that day is 259.2 W.