To solve this problem, we use the formula for steady-state heat conduction through a flat wall:

$Q = \frac{k \cdot A \cdot (T_1 - T_2)}{L}$

Where:

• $Q$ is the heat loss (in BTU/h),
• $k$ is the thermal conductivity (in BTU/h·ft·°F),
• $A$ is the area of the wall (in ft²),
• $T_1$ and $T_2$ are the temperatures inside and outside the furnace (in °F),
• $L$ is the thickness of the wall (in feet).

Given:

• $k = 0.14 \, \text{BTU/h·ft·°F}$,
• $A = 12 \, \text{ft²}$,
• $T_1 = 1100^\circ \text{F}$ (inside temperature),
• $T_2 = 102^\circ \text{F}$ (outside temperature),
• $L = 6 \, \text{in} = 0.5 \, \text{ft}$.

Plugging in the values:

$Q = \frac{0.14 \cdot 12 \cdot (1100 - 102)}{0.5}$

Now, calculate step by step:

1. $1100 - 102 = 998 \, \text{°F}$,
2. $0.14 \cdot 12 = 1.68$,
3. $1.68 \cdot 998 = 1676.64 \, \text{BTU/h}$,
4. $\frac{1676.64}{0.5} = 3353.28 \, \text{BTU/h}$.

Final Answer:

The heat loss is approximately $3353 \, \text{BTU/h}$.

Would you like a more detailed explanation of any part of the calculation?

Here are 5 related questions to expand on this:

1. How does increasing the wall thickness affect heat loss in a furnace?
2. What happens to the heat loss if the outside temperature increases?
3. Can we apply the same formula for heat loss if the material has varying thermal conductivity?
4. How does thermal insulation impact the heat transfer in a furnace?
5. What units are used in heat conduction calculations in the metric system?

Tip: Always ensure the units are consistent when applying heat transfer formulas!