Understanding Thermal Diffusivity Dimensions in M-L-t-T System

The question asks for the dimension of thermal diffusivity in the M-L-t-T system. This system uses Mass (M), Length (L), time (t), and Temperature (T) as fundamental dimensions.

Thermal diffusivity ($\alpha$) is a material property that indicates how quickly heat diffuses through a material. It is defined by the relationship:

$$\alpha = \frac{k}{\rho c_p}$$

Where:

• $k$ is thermal conductivity
• $\rho$ is density
• $c_p$ is specific heat capacity

To find the dimension of thermal diffusivity, we first need to determine the dimensions of thermal conductivity, density, and specific heat capacity in the M-L-t-T system.

Dimensions of Constituent Properties

1. Dimension of Density ($\rho$)

Density is defined as mass per unit volume.

$$\rho = \frac{\text{Mass}}{\text{Volume}}$$

The dimension of mass is $[M]$, and the dimension of volume (length cubed) is $[L^3]$.

So, the dimension of density is:

$$[\rho] = \frac{[M]}{[L^3]} = [M L^{-3}]$$

2. Dimension of Specific Heat Capacity ($c_p$)

Specific heat capacity is defined as the amount of energy required to raise the temperature of a unit mass by one degree.

$$\text{Energy} = \text{Mass} \times \text{Specific Heat Capacity} \times \text{Change in Temperature}$$

$$\Delta Q = m c_p \Delta T$$

Rearranging for $c_p$: $c_p = \frac{\Delta Q}{m \Delta T}$.

The dimension of energy (which is equivalent to work) is $[M L^2 t^{-2}]$. The dimension of mass is $[M]$, and the dimension of temperature change is $[T]$.

So, the dimension of specific heat capacity is:

$$[c_p] = \frac{[M L^2 t^{-2}]}{[M][T]} = \frac{[L^2 t^{-2}]}{[T]} = [L^2 t^{-2} T^{-1}]$$

3. Dimension of Thermal Conductivity ($k$)

Thermal conductivity relates the heat flow rate through a material to the temperature gradient and the area.

$$\text{Heat Flow Rate} = -k \times \text{Area} \times \text{Temperature Gradient}$$

The heat flow rate is energy per unit time, so its dimension is $[\text{Energy}/ \text{time}] = [M L^2 t^{-2} / t] = [M L^2 t^{-3}]$.

The dimension of area is $[L^2]$.

The dimension of temperature gradient ($\frac{\Delta T}{\Delta x}$) is $[\frac{\text{Temperature}}{\text{Length}}] = [T L^{-1}]$.

Rearranging the formula for $k$: $k = \frac{\text{Heat Flow Rate}}{\text{Area} \times \text{Temperature Gradient}}$.

So, the dimension of thermal conductivity is:

$$[k] = \frac{[M L^2 t^{-3}]}{[L^2][T L^{-1}]} = \frac{[M L^2 t^{-3}]}{[L^2 T L^{-1}]} = \frac{[M L^2 t^{-3}]}{[L T]} = [M L^{2-1} t^{-3} T^{-1}] = [M L t^{-3} T^{-1}]$$

Calculating the Dimension of Thermal Diffusivity ($\alpha$)

Now we substitute the dimensions of $k$, $\rho$, and $c_p$ into the formula for $\alpha = \frac{k}{\rho c_p}$.

$$[\alpha] = \frac{[k]}{[\rho][c_p]}$$

$$[\alpha] = \frac{[M L t^{-3} T^{-1}]}{[M L^{-3}][L^2 t^{-2} T^{-1}]}$$

First, combine the dimensions in the denominator:

$$[\rho][c_p] = [M L^{-3}][L^2 t^{-2} T^{-1}] = [M^{1} L^{-3+2} t^{-2} T^{-1}] = [M L^{-1} t^{-2} T^{-1}]$$

Now, divide the dimension of $k$ by the combined dimension:

$$[\alpha] = \frac{[M L t^{-3} T^{-1}]}{[M L^{-1} t^{-2} T^{-1}]}$$

$$[\alpha] = [M^{1-1} L^{1 - (-1)} t^{-3 - (-2)} T^{-1 - (-1)}]$$

$$[\alpha] = [M^0 L^{1+1} t^{-3+2} T^{-1+1}]$

$$[\alpha] = [M^0 L^2 t^{-1} T^0]$

Since any dimension raised to the power of zero is 1 (in dimensional analysis), $M^0=1$ and $T^0=1$.

Therefore, the dimension of thermal diffusivity in the M-L-t-T system is:

$$[\alpha] = [L^2 t^{-1}]$$

Comparing with Options

Let's compare the calculated dimension $[L^2 t^{-1}]$ with the given options:

• Option 1: $M L^2 t^{-1}$
• Option 2: $L^2 t^{-1}$
• Option 3: $M^2 L^2 T^{-1}$
• Option 4: $L T^{-2}$

The calculated dimension $[L^2 t^{-1}]$ matches Option 2.

Revision Table: Dimensions of Related Quantities

Additional Information on Thermal Diffusivity

Thermal diffusivity is a measure of how quickly temperature changes propagate through a material. It's different from thermal conductivity, which measures how much heat flows through a material under a given temperature gradient. A material with high thermal diffusivity will heat up or cool down quickly because heat energy diffuses rapidly through it. A material with low thermal diffusivity will take longer to heat up or cool down.

Consider the units of thermal diffusivity: $[L^2 t^{-1}]$. This dimension is similar to kinematic viscosity in fluid mechanics, which is also $[L^2 t^{-1}]$. Both represent a form of diffusion (heat diffusion and momentum diffusion, respectively).

In practical terms, thermal diffusivity is important in applications like designing heat exchangers, predicting cooling rates of materials, and understanding thermal insulation properties.