A brick wall $$ (k=0.9 \mathrm{~W} / \mathrm{m} . \mathrm{K}) $$ of thickness 0.18 m separates the warm air in a room from the cold ambient air. On a particular winter day, the outside air temperature is $$ -5^{\circ} \mathrm{C} $$ and the room needs to be maintained at $$ 27^{\circ} \mathrm{C} $$. The heat transfer coefficient associated with outside air is $$ 20 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K} $$. Neglecting the convective resistance of the air inside the room, the heat loss, in $$ \mathrm{W} / \mathrm{m}^{2} $$, is . . (round off to the nearest integer).

Question

UpStudy AI · Accepted Answer

To calculate the heat loss per unit area through the brick wall, we'll consider both conduction through the wall and convection on the outside.

1. **Conduction Through the Wall:**
   - **Thermal Conductivity ($$k$$)**: 0.9 W/m·K
   - **Thickness ($$L$$)**: 0.18 m
   - **Temperature Difference ($$\Delta T$$)**: $$27^\circ \text{C} - (-5^\circ \text{C}) = 32 \text{ K}$$
   
   The thermal resistance for conduction ($$R_{\text{cond}}$$) is:
   $$
   R_{\text{cond}} = \frac{L}{k} = \frac{0.18}{0.9} = 0.2 \, \text{m}^2\text{K/W}
   $$

2. **Convection on the Outside:**
   - **Heat Transfer Coefficient ($$h$$)**: 20 W/m²·K
   
   The thermal resistance for convection ($$R_{\text{conv}}$$) is:
   $$
   R_{\text{conv}} = \frac{1}{h} = \frac{1}{20} = 0.05 \, \text{m}^2\text{K/W}
   $$

3. **Total Thermal Resistance:**
   $$
   R_{\text{total}} = R_{\text{cond}} + R_{\text{conv}} = 0.2 + 0.05 = 0.25 \, \text{m}^2\text{K/W}
   $$

4. **Heat Loss Per Unit Area ($$q$$):**
   $$
   q = \frac{\Delta T}{R_{\text{total}}} = \frac{32}{0.25} = 128 \, \text{W/m}^2
   $$

**Final Answer:**  
The heat loss is **128 W/m²**.
The heat loss is **128 W/m²**.

Upstudy AI Solution

Answer

Solution

Extra Insights

Related Questions

Latest Physics Questions