Unit of Thermal Conductivity Explained

The question asks for the standard unit of thermal conductivity. Thermal conductivity (\(k\)) quantifies a material's ability to conduct heat.

Deriving the Unit

We can determine the unit using Fourier's Law of Heat Conduction:

\(q = -k \cdot A \cdot \frac{dT}{dx}\)

Where:

• \(q\) is the rate of heat transfer (Power), measured in Watts (W) or Joules per second (\(\text{J} \cdot \text{s}^{-1}\)).
• \(A\) is the cross-sectional area, measured in square meters (\(\text{m}^2\)).
• \(\frac{dT}{dx}\) is the temperature gradient, measured in Kelvin per meter (\(\text{K} \cdot \text{m}^{-1}\)).

Rearranging the formula to solve for \(k\):

\(k = -\frac{q \cdot dx}{A \cdot dT}\)

Calculating Thermal Conductivity Unit

Substituting the units into the rearranged formula:

\(k \text{ unit} = \frac{(\text{J} \cdot \text{s}^{-1}) \cdot \text{m}}{\text{m}^2 \cdot \text{K}}\)

Simplifying the expression:

\(k \text{ unit} = \frac{\text{J} \cdot \text{s}^{-1}}{\text{m} \cdot \text{K}}\)

\(k \text{ unit} = \text{J} \cdot \text{s}^{-1} \cdot \text{m}^{-1} \cdot \text{K}^{-1}\)

Conclusion

The unit of thermal conductivity is derived as \(\text{J}\cdot\text{s}^{-1}\cdot\text{m}^{-1}\cdot\text{K}^{-1}\). This corresponds to Option 4.