Question

The inner and outer surfaces of a \(4-\mathrm{m} \times 7-\mathrm{m}\) brick wall of thickness \(30 \mathrm{~cm}\) and thermal conductivity $0.69 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\( are maintained at temperatures of \)20^{\circ} \mathrm{C}\( and \)5^{\circ} \mathrm{C}$, respectively. Determine the rate of heat transfer through the wall in W.

Short answer

Answer: The rate of heat transfer through the brick wall is 920 W.

Step-by-step solution

Identify the given parameters

We are given the following values: - Length of the wall (L): \(4 \mathrm{~m}\) - Height of the wall (H): \(7 \mathrm{~m}\) - Thickness of the wall (d): \(0.3 \mathrm{~m}\) (converting \(30 \mathrm{~cm}\) to meters) - Thermal conductivity (k): \(0.69 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) - Inner surface temperature (T1): \(20^{\circ} \mathrm{C}\) - Outer surface temperature (T2): \(5^{\circ} \mathrm{C}\)

Calculate the area of the wall

The area (A) of the wall can be calculated using the formula: \(A = L \times H\). \(A = 4 \mathrm{~m} \times 7 \mathrm{~m} = 28 \mathrm{~m^{2}}\)

Calculate the temperature difference

The temperature difference (\(\Delta T\)) between the inner and outer surfaces can be calculated by subtracting the outer surface temperature (T2) from the inner surface temperature (T1). \(\Delta T = T1 - T2 = 20^{\circ} \mathrm{C} - 5^{\circ} \mathrm{C} = 15^{\circ} \mathrm{C}\)

Calculate the rate of heat transfer through the wall

Now we can calculate the rate of heat transfer (Q) using the formula: \(Q = k * A * \frac{\Delta T}{d}\) \(Q = 0.69 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \times 28 \mathrm{~m^{2}} \times \frac{15^{\circ} \mathrm{C}}{0.3 \mathrm{~m}}\) \(Q = 920 \mathrm{~W}\) So, the rate of heat transfer through the brick wall is \(920 \mathrm{~W}\).