The unit of thermal conductivity is :

Options

A. ` Js^(-1) K`
B. `Js^(-1) m^(2)k`
C. `Jm^(-1)K`
D. `Js^(-1) m^(-1) K^(-1)`

Explanation

To find the unit of thermal conductivity, we start with the formula that relates heat transfer to thermal conductivity. The formula is:

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

Where:
- \( q \) is the heat transferred (in joules, J)
- \( k \) is the thermal conductivity (what we want to find)
- \( A \) is the area through which heat is being transferred (in square meters, m²)
- \( \Delta T \) is the temperature difference (in Kelvin, K)
- \( L \) is the length of the material (in meters, m)

We can rearrange this formula to solve for \( k \):

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

Now, we need to determine the units of each component in this equation:

1. The unit of heat \( q \) is joules (J).
2. The unit of length \( L \) is meters (m).
3. The unit of area \( A \) is square meters (m²).
4. The unit of temperature difference \( \Delta T \) is Kelvin (K).

Substituting these units into the equation for \( k \):

\[ k = \frac{J \cdot m}{m^2 \cdot K} \]

Now, simplifying this expression:

- The meter in the numerator and one of the meters in the denominator will cancel out:

\[ k = \frac{J}{m \cdot K} \]

Next, we need to consider the time factor. The heat transfer occurs over a period of time, which is typically measured in seconds (s). Therefore, we can express the thermal conductivity in terms of time:

\[ k = \frac{J}{m \cdot K \cdot s} \]

Thus, the unit of thermal conductivity can be expressed as:

\[ k = \frac{J}{s \cdot m \cdot K} \]

In conclusion, the unit of thermal conductivity is:

**Joule per second per meter per Kelvin (J/s·m·K)**.