A cabin has a 0.159 m thick wooden floor [k = 0.14] `W//(m.""^(@)C )]` with an area of `13.4m^(2)`. A roaring fire keeps the interior of the cabin at a comfortable `18.0^(@)C` while the air temperature in the crawl space below the cabin is `- 20.6^(@)C`. What is the rate of heat conduction through the wooden floor ?

Options

A. 31 J `//`s
B. 459 J `//` s
C. 138 J `//` s
D. 245 J `//`s

Answer

To find the rate of heat conduction through the wooden floor of the cabin, we can use Fourier's law of heat conduction, which is given by the formula:

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

Where:
- \( Q \) is the rate of heat transfer (in Watts, or Joules per second),
- \( k \) is the thermal conductivity of the material (in \( W/(m \cdot °C) \)),
- \( A \) is the area through which heat is being conducted (in \( m^2 \)),
- \( T_1 \) and \( T_2 \) are the temperatures on either side of the material (in °C),
- \( L \) is the thickness of the material (in meters).

### Step-by-Step Solution:

1. **Identify the given values:**
- Thickness of the wooden floor, \( L = 0.159 \, m \)
- Thermal conductivity, \( k = 0.14 \, W/(m \cdot °C) \)
- Area, \( A = 13.4 \, m^2 \)
- Temperature inside the cabin, \( T_1 = 18.0 \, °C \)
- Temperature in the crawl space, \( T_2 = -20.6 \, °C \)

2. **Calculate the temperature difference:**
\[ \Delta T = T_1 - T_2 = 18.0 \, °C - (-20.6 \, °C) = 18.0 + 20.6 = 38.6 \, °C \]

3. **Substitute the values into the formula:**
\[ Q = \frac{k \cdot A \cdot \Delta T}{L} \]
\[ Q = \frac{0.14 \, W/(m \cdot °C) \cdot 13.4 \, m^2 \cdot 38.6 \, °C}{0.159 \, m} \]

4. **Calculate the numerator:**
\[ \text{Numerator} = 0.14 \cdot 13.4 \cdot 38.6 \]
\[ = 0.14 \cdot 517.64 \approx 72.464 \, W \]

5. **Calculate the rate of heat conduction:**
\[ Q = \frac{72.464}{0.159} \approx 455.4 \, W \]

Thus, the rate of heat conduction through the wooden floor is approximately **455 Joules per second**.