# Heat Flow Through a Wall: Step-by-Step Let's solve for the **heat flow (Q, in watts)** through a wall using the data provided: - **Area, \(A\)**: \(3\,\text{m}^2\) - **Thickness, \(d\)**: \(.15\,\text{m}\) - **Thermal conductivity, \(k\)**: \(.6\,\text{W}/(\text{m}\cdot\text{K})\) - **Temperature difference, \(\Delta T\)**: \(25^\circ\text{C}\) (same as 25 K for differences) --- ## 1. Formula for Heat Flow For **steady-state conduction** through a flat wall, the formula is: \[ Q = \frac{kA\Delta T}{d} \] Where: - \(Q\) = heat flow (W) - \(k\) = thermal conductivity (W/m·K) - \(A\) = area (m²) - \(\Delta T\) = temperature difference (K or °C) - \(d\) = thickness (m) --- ## 2. Plug in the Values Substitute the given values into the formula: \[ Q = \frac{.6 \times 3 \times 25}{.15} \] --- ## 3. Calculate the Numerator \[ .6 \times 3 = 1.8 \] \[ 1.8 \times 25 = 45 \] --- ## 4. Divide by the Thickness \[ Q = \frac{45}{.15} \] \[ Q = 300 \] --- ## 5. Final Answer \[ \boxed{Q = 300\,\text{W}} \] --- ## **Summary Table** | Parameter | Value | |------------------------|--------------------| | Area (\(A\)) | \(3\,\text{m}^2\) | | Thickness (\(d\)) | \(.15\,\text{m}\) | | Conductivity (\(k\)) | \(.6\,\text{W}/\text{m}\cdot\text{K}\) | | Temp. difference (\(\Delta T\)) | \(25\,\text{K}\) | | **Heat flow (\(Q\))** | **\(300\,\text{W}\)** | --- **Conclusion:** The heat flow through the wall is **300 watts**.