Let's solve the problem step by step. --- ## **Given Data** - **Thermal conductivity,** \( k = 0.72~\text{W/(m·K)} \) - **Wall dimensions:** \( 4~\text{m} \) long × \( 3~\text{m} \) wide - **Thickness of wall,** \( L = 25~\text{cm} = 0.25~\text{m} \) - **Heat loss rate,** \( Q = 500~\text{W} \) - **Inner surface temperature,** \( T_{\text{inner}} = 22^\circ\text{C} \) We are to find the **temperature at the midplane** of the wall. --- ## **Step 1: Find the Area of the Wall** \[ A = \text{length} \times \text{width} = 4~\text{m} \times 3~\text{m} = 12~\text{m}^2 \] --- ## **Step 2: Heat Transfer Equation for Conduction** The rate of steady-state heat flow through a slab is: \[ Q = \frac{kA (T_{\text{inner}} - T_{\text{outer}})}{L} \] But we need the temperature at the **midplane**, i.e., at \( x = L/2 \). --- ## **Step 3: Temperature Drop Across Half the Thickness** Since it's steady-state conduction through a plane wall, the temperature distribution is **linear**. Let \( T_{1/2} \) be the temperature at the midpoint. The temperature drop from the inner surface to the midpoint is: \[ Q = \frac{kA (T_{\text{inner}} - T_{1/2})}{L/2} \] Rearrange to solve for \( T_{1/2} \): \[ T_{\text{inner}} - T_{1/2} = \frac{Q (L/2)}{kA} \] \[ T_{1/2} = T_{\text{inner}} - \frac{Q (L/2)}{kA} \] --- ## **Step 4: Substitute Values** Plug in the numbers: \[ T_{1/2} = 22 - \frac{500 \times (0.25/2)}{0.72 \times 12} \] Calculate step by step: ### **a) Area** \[ A = 12~\text{m}^2 \] ### **b) Half the Thickness** \[ L/2 = 0.25/2 = 0.125~\text{m} \] ### **c) Denominator** \[ kA = 0.72 \times 12 = 8.64 \] ### **d) Numerator** \[ Q \cdot (L/2) = 500 \times 0.125 = 62.5 \] ### **e) The Whole Fraction** \[ \frac{62.5}{8.64} \approx 7.23 \] ### **f) Final Temperature** \[ T_{1/2} = 22 - 7.23 = 14.77^\circ\text{C} \] --- ## **Step 5: Compare with Options** Closest option: - (d) \( 14.8^\circ\text{C} \) --- ## **Final Answer** \[ \boxed{14.8^\circ\text{C}} \] --- **Explanation:** The temperature drops linearly through the wall. By calculating the drop over half the thickness, we determined the temperature at the wall's midplane using the principles of steady-state heat conduction.