Question

The dimensional formula for thermal conductivity is (here $\mathrm{K}$ denotes the temperature):

• A. $ML{T}^{-3}{K}^{-1}$
• B. $ML{T}^{-2}{K}^{-2}$
• C. $ML{T}^{-2}K$
• D. $ML{T}^{-3}K$

Answer 1

The correct option is A

$ML{T}^{-3}{K}^{-1}$

Explanation for the correct option:

Step 1: Thermal Conductivity:

The ability of a material to conduct or transfer heat from a hot body to a cold body is called Thermal Conductivity.

Step 2: Fourier's law of Thermal Conduction:

According to Fourier's law of thermal conduction, the rate at which the heat is transferred through a material is directly proportional to the negative value of the temperature gradient and the Area of the material through which the heat flows and is inversely proportional to the distance between the two isothermal planes, i.e., $q\alpha \frac{-∆T·A·t}{d}$

Where, $q$ is the Heat transferred through a material (in $WattorJ·K^{-1}$)

$A$ is the Area of the surface (in $m^2$)

$t$ is the time at which the heat flows (in $sec$)

$d$ is the distance between the two isothermal planes (in $m$)

$∆T=T_2-T_1$ is the Temperature Gradient (in $K$)

Step 3: Formula of the Thermal Conductivity:

$$\begin{gathered} \Rightarrow q=\kappa ·\frac{-\left(T_2-T_1\right)·A·t}{d} \\ \Rightarrow q=\kappa ·\frac{\left(T_1-T_2\right)·A·t}{d} \\ \Rightarrow \kappa =\frac{q·d}{\left(T_1-T_2\right)·A·t} \end{gathered}$$

Step 4: Dimensional formulas for the Thermal conductivity:

For Heat, $q=\left[M{L}^2{T}^{-2}\right]$

For Area, $A=\left[L^2\right]$

For the Distance between the two isothermal planes, $d=\left[L\right]$

For Temperature Gradient, $∆T=\left[K\right]$

For Time, $t=\left[T\right]$

$$\begin{gathered} \Rightarrow {\kappa }_{thermal}=\frac{\left[M{L}^2{T}^{-2}\right]·\left[L\right]}{\left[L^2\right]·\left[K\right]·\left[T\right]} \\ \Rightarrow {\kappa }_{thermal}=\left[M{L}^{3-2}{T}^{-2-1}\right]·\left[K^{-1}\right] \\ \Rightarrow {\kappa }_{thermal}=M{L}^1{T}^{-3}{K}^{-1} \end{gathered}$$

Hence, option A is correct. The dimensional formula of the Thermal Conductivity is ${\kappa }_{thermal}=M{L}^1{T}^{-3}{K}^{-1}$.