Show that the SI units of thermal conductivity are `W//m-K`.

Explanation

To show that the SI units of thermal conductivity (k) are \( \text{W/m·K} \), we will start with Fourier's law of heat conduction.

### Step-by-Step Solution:

1. **Understand Fourier's Law**:
According to Fourier's law, the rate of heat transfer (Q) through a material is proportional to the negative temperature gradient. The formula is given by:
\[ Q = -k \cdot A \cdot \frac{dT}{dx} \]
where:
- \( Q \) is the heat transfer rate (in Watts, W),
- \( k \) is the thermal conductivity (in unknown units),
- \( A \) is the cross-sectional area through which heat is being transferred (in square meters, m²),
- \( \frac{dT}{dx} \) is the temperature gradient (in Kelvin per meter, K/m).

2. **Rearranging the Formula**:
From the equation, we can express thermal conductivity (k) as:
\[ k = \frac{Q}{A \cdot \frac{dT}{dx}} \]

3. **Identify the Units**:
- The unit of heat transfer rate \( Q \) is Watts (W).
- The unit of area \( A \) is square meters (m²).
- The unit of temperature gradient \( \frac{dT}{dx} \) is Kelvin per meter (K/m).

4. **Substituting the Units**:
Now, substituting these units into the equation for k:
\[ k = \frac{\text{W}}{\text{m}^2 \cdot \left(\frac{\text{K}}{\text{m}}\right)} \]

5. **Simplifying the Units**:
When we simplify the expression:
\[ k = \frac{\text{W}}{\text{m}^2} \cdot \frac{\text{m}}{\text{K}} = \frac{\text{W}}{\text{m} \cdot \text{K}} \]

6. **Final Result**:
Thus, we find that the SI unit of thermal conductivity \( k \) is:
\[ k = \frac{\text{W}}{\text{m} \cdot \text{K}} \]

### Conclusion:
The SI units of thermal conductivity are \( \text{W/m·K} \).