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.
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