Let's break down the problem step by step: ### **Given Data** - **Thermal conductivity (\( k \))**: \( 4.8 \times 10^{-4} \) kW/m·°C = \( 0.00048 \) kW/m·°C = \( 0.48 \) W/m·°C - **Wall thickness (\( d \))**: \( 20.5 \) cm = \( 0.205 \) m - **Inside temperature (\( T_{\text{in}} \))**: \( 24.0 \)°C - **Outside temperature (\( T_{\text{out}} \))**: \( 0.0 \)°C - **Heat of combustion**: \( 9,300 \) kcal/m³ - **House dimensions**: \( 10.0 \) m (length) × \( 8.0 \) m (width) × \( 5.0 \) m (height to roof top) - **Roof angle**: \( 37.0^\circ \) - **Time period**: 1 day = 24 hours = 86400 seconds --- ### **Step 1: Find the Total Surface Area for Heat Loss** **Assume**: All four walls and the roof lose heat (ignore the floor). #### **Walls** - The house has two walls of \( 10.0 \) m × \( 5.0 \) m and two of \( 8.0 \) m × \( 5.0 \) m. - Total wall area: \[ A_{\text{walls}} = 2 \times (10.0 \times 5.0) + 2 \times (8.0 \times 5.0) = 2 \times 50 + 2 \times 40 = 100 + 80 = 180\, \text{m}^2 \] #### **Roof** - The roof is a rectangle, but sloped. - Each side of the roof has a horizontal projection of \( 10.0 \) m (length) × \( 8.0/2 = 4.0 \) m (half the width, since it's a gable roof). - The actual length of the sloped side is \( \frac{4.0}{\cos 37^\circ} \). So, area of one side: \[ A_{\text{roof, one side}} = 10.0 \times \left( \frac{4.0}{\cos 37^\circ} \right) \] Total roof area: \[ A_{\text{roof}} = 2 \times 10.0 \times \frac{4.0}{\cos 37^\circ} \] \[ \cos 37^\circ \approx 0.7986 \] \[ A_{\text{roof}} = 2 \times 10.0 \times \frac{4.0}{0.7986} = 2 \times 10.0 \times 5.009 = 2 \times 50.09 = 100.18\, \text{m}^2 \] #### **Total Area** \[ A_{\text{total}} = 180 + 100.18 = 280.18\, \text{m}^2 \] --- ### **Step 2: Calculate Heat Loss Rate (\( Q \))** \[ \dot{Q} = \frac{k \cdot A_{\text{total}} \cdot \Delta T}{d} \] Where: - \( k = 0.48 \) W/m·°C - \( \Delta T = 24.0 - 0.0 = 24.0 \)°C - \( d = 0.205 \) m - \( A_{\text{total}} = 280.18 \) m² \[ \dot{Q} = \frac{0.48 \times 280.18 \times 24.0}{0.205} \] Calculate numerator: \[ 0.48 \times 280.18 = 134.4864 \] \[ 134.4864 \times 24.0 = 3,227.6736 \] Now divide by \( 0.205 \): \[ \dot{Q} = \frac{3,227.6736}{0.205} = 15,746.7\, \text{W} = 15.75\, \text{kW} \] --- ### **Step 3: Total Energy Needed Per Day (\( Q_{\text{day}} \))** \[ Q_{\text{day}} = \dot{Q} \times \text{Time} \] \[ Q_{\text{day}} = 15,746.7\, \text{W} \times 86,400\, \text{s} \] \[ Q_{\text{day}} = 1,360,477,680\, \text{J} \] --- ### **Step 4: Convert to kcal** \[ 1\, \text{kcal} = 4,184\, \text{J} \] \[ Q_{\text{day}} = \frac{1,360,477,680}{4,184} = 325,309\, \text{kcal} \] --- ### **Step 5: Find Volume of Gas Needed** - **Heat of combustion**: \( 9,300 \) kcal/m³ \[ \text{Volume} = \frac{Q_{\text{day}}}{\text{Heat of combustion}} \] \[ \text{Volume} = \frac{325,309}{9,300} = 34.97\, \text{m}^3 \] --- ### **Final Answer** \[ \boxed{35.0\ \text{m}^3/\text{day}} \] --- ### **Explanation of Previous Error** The calculation above gives approximately 35.0 m³/day, which matches your previous answer of 34.9 quite closely (the difference likely arises from more careful area calculation or more significant figure accuracy). Try entering **35.0** (rounded to one decimal) as your answer. **If minor differences persist, double-check the area calculation or rounding at each step.**