The human body transports heat from the interior tissues, at temperature `37.0^(@)C`, to the skin surface, at temperature `27.0^(@)C`, at a rate of 100 W. If the skin area is `1.5 m^(2)` and its thickness is 3.0 mm, what is the effective thermal conductivity, k, of skin?

Options

A. `4.0 xx 10^(-2) W//(mk)`
B. `2.0 xx 10^(-2) W//(mk)`
C. `2.0 xx 10^(-3) W//(k)`
D. `3.0 xx 10^(-2) W//(mk)`

Explanation

To find the effective thermal conductivity \( k \) of the skin, we can use the formula for heat transfer through a material:

\[ Q = \frac{k \cdot A \cdot \Delta T}{L} \]

Where:
- \( Q \) is the heat transfer rate (in Watts),
- \( k \) is the thermal conductivity (in W/m·K),
- \( A \) is the area through which heat is being transferred (in m²),
- \( \Delta T \) is the temperature difference across the material (in K),
- \( L \) is the thickness of the material (in m).

### Step-by-step Solution:

1. **Identify the given values:**
- Heat transfer rate, \( Q = 100 \, \text{W} \)
- Area, \( A = 1.5 \, \text{m}^2 \)
- Temperature difference, \( \Delta T = 37.0^\circ C - 27.0^\circ C = 10 \, \text{K} \)
- Thickness, \( L = 3.0 \, \text{mm} = 3.0 \times 10^{-3} \, \text{m} \)

2. **Rearrange the formula to solve for \( k \):**
\[ k = \frac{Q \cdot L}{A \cdot \Delta T} \]

3. **Substitute the values into the equation:**
\[ k = \frac{100 \, \text{W} \cdot (3.0 \times 10^{-3} \, \text{m})}{1.5 \, \text{m}^2 \cdot 10 \, \text{K}} \]

4. **Calculate the numerator:**
\[ 100 \, \text{W} \cdot 3.0 \times 10^{-3} \, \text{m} = 0.3 \, \text{W·m} \]

5. **Calculate the denominator:**
\[ 1.5 \, \text{m}^2 \cdot 10 \, \text{K} = 15 \, \text{m}^2 \cdot \text{K} \]

6. **Now, calculate \( k \):**
\[ k = \frac{0.3 \, \text{W·m}}{15 \, \text{m}^2 \cdot \text{K}} = 0.02 \, \text{W/(m·K)} \]

7. **Express in scientific notation:**
\[ k = 2.0 \times 10^{-2} \, \text{W/(m·K)} \]

### Final Answer:
The effective thermal conductivity \( k \) of the skin is \( 2.0 \times 10^{-2} \, \text{W/(m·K)} \).