A wall with an area of 10ft² and 4 inches thick has a thermal conductivity of 0.12BTU/ h ft°F . What is the heat loss if the temperature inside is 900°F and the outside temperature is 80° F. Show step-by-step solution 

To calculate the heat loss through the wall, we can use the formula for heat transfer through a solid surface, which is given by:

\[
Q = \frac{k \cdot A \cdot (T_{\text{inside}} - T_{\text{outside}})}{d}
\]

Where:
- \( Q \) is the heat transfer rate (BTU/h)
- \( k \) is the thermal conductivity (BTU/h ft°F)
- \( A \) is the area of the wall (ft²)
- \( T_{\text{inside}} \) is the inside temperature (°F)
- \( T_{\text{outside}} \) is the outside temperature (°F)
- \( d \) is the thickness of the wall (ft)

### Given Data:
- Area \( A = 10 \, \text{ft}^2 \)
- Thickness \( d = 4 \, \text{inches} = \frac{4}{12} \, \text{ft} = \frac{1}{3} \, \text{ft} \)
- Thermal conductivity \( k = 0.12 \, \text{BTU/h ft°F} \)
- Inside temperature \( T_{\text{inside}} = 900 \, \text{°F} \)
- Outside temperature \( T_{\text{outside}} = 80 \, \text{°F} \)

### Step-by-Step Solution:

1. **Temperature Difference**:  
   The temperature difference between the inside and the outside is:
   \[
   \Delta T = T_{\text{inside}} - T_{\text{outside}} = 900 - 80 = 820 \, \text{°F}
   \]

2. **Calculate Heat Loss**:  
   Using the formula for heat transfer, substitute the given values:
   \[
   Q = \frac{0.12 \cdot 10 \cdot 820}{\frac{1}{3}}
   \]

3. **Simplify**:  
   First, simplify the division by \( \frac{1}{3} \):
   \[
   Q = 0.12 \cdot 10 \cdot 820 \cdot 3
   \]

4. **Final Calculation**:  
   Perform the multiplications:
   \[
   Q = 0.12 \cdot 10 \cdot 2460 = 2952 \, \text{BTU/h}
   \]

Thus, the heat loss through the wall is:

\[
Q = 2952 \, \text{BTU/h}
\]

### Conclusion:
The heat loss through the wall is **2952 BTU/h**.

Would you like more details on this process or have any further questions?

---

### Related Questions:
1. What factors affect the rate of heat transfer through a material?
2. How does increasing wall thickness impact heat loss?
3. What would be the heat loss if the thermal conductivity was 0.08 BTU/h ft°F?
4. How does reducing the temperature difference between inside and outside affect heat loss?
5. What are common materials with low thermal conductivity values used in insulation?

### Tip:
The thicker the wall, the less heat is lost, because the heat transfer rate is inversely proportional to the wall's thickness.

A wall with an area of 10ft² and 4 inches thick has a thermal conductivity of 0.12 BTU/h ft°F. What is the heat loss if the temperature inside is 900°F and the outside temperature is 80°F? Show step-by-step solution.

Learn how to calculate the heat loss through a wall with an area of 10ft², 4 inches thick, and a thermal conductivity of 0.12 BTU/h ft°F. Using the heat transfer formula, we solve the problem step-by-step, accounting for an inside temperature of 900°F and an outside temperature of 80°F. Suitable for advanced high school or college-level students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation