Question

If 'k' is the thermal conductivity, '$ \rho $' is the mass density and 'c' is the specific heat then the thermal diffusivity of substance is

• A. \( \frac{kc}{\rho} \)
• B. \( \frac{k\rho}{c} \)
• C. \( \frac{k}{\rho c} \)
• D. \( \frac{\rho c}{k} \)

Hint:

Thermal diffusivity \( \alpha \) is a key property in transient heat conduction. It represents how quickly temperature changes propagate through a material and is calculated as the ratio of thermal conductivity \( k \) to volumetric heat capacity (\( \rho c \)).

Answer 1

The Correct Option is C

Step 1: Define Thermal Diffusivity.
Thermal diffusivity, denoted by \( \alpha \), is a material property that measures the rate at which heat is conducted through a material relative to the rate at which it stores thermal energy. It quantifies how quickly temperature changes propagate through a material.
Step 2: Recall the formula for Thermal Diffusivity.
Thermal diffusivity is defined as the ratio of thermal conductivity to the volumetric heat capacity of the material.
Volumetric heat capacity is the product of mass density and specific heat. The formula is: \[ \alpha = \frac{k}{\rho c_p} \] where:
\( k \) = thermal conductivity (W/(m K) or W/(m \(^\circ\)C))
\( \rho \) = mass density (kg/m\( ^3 \))
\( c_p \) (or \( c \)) = specific heat capacity at constant pressure (J/(kg K) or J/(kg \(^\circ\)C))
Step 3: Verify the units (optional but good practice).
Units of \( \alpha \):
\[ \frac{\text{W/(m K)}}{\text{(kg/m}^3\text{)(J/(kg K))}} = \frac{\text{W}}{\text{m K}} \times \frac{\text{m}^3 \text{ kg K}}{\text{kg J}} = \frac{\text{W m}^2}{\text{J}} \] Since \( \text{J} = \text{W s} \), \[ = \frac{\text{W m}^2}{\text{W s}} = \frac{\text{m}^2}{\text{s}} \] The units of thermal diffusivity are indeed \( \text{m}^2/\text{s} \), which confirms the formula. Given the symbols 'k' for thermal conductivity, '\( \rho \)' for mass density, and 'c' for specific heat, the thermal diffusivity is \( \frac{k}{\rho c} \). The final answer is $\boxed{\text{3}}$.