Question

If $d$ is the density and $S$ is the specific heat, then thermal conductivity $K$ and thermometric conductivity $h$ are related as:
$\begin{align}
  & (A)h=\dfrac{k}{S} \\
 & (B)h=\dfrac{dK}{S} \\
 & (C)h=\dfrac{K}{dS} \\
 & (D)h=KSd \\
\end{align}$

Answer 1

Hint: To find the required relation, between the density, specific heat, the thermal conductivity of a substance, we will first understand what are thermal conductivity and thermometric conductivity. Then, we will use the respective formulas of conduction and convection to get our required relation.

Complete answer:
Thermal conductivity is a physical measure of how much heat has been transferred from one body to another. According to thermodynamics, this heat flow always takes place from a higher potential to a lower potential, that is, from warmer object to colder object.
Now, thermometric conductivity is the control over time rate of temperature change as heat transfers through a body or an object. It is basically the measurement of the rate at which an object with non-linear (or, non-uniform) temperature reaches a stable state of equilibrium.
Now, we will derive the thermometric conductivity (h) in terms of thermal conductivity (K).
Let us consider a conducting rod of length (L), area of cross section (A), volume (V), and having a mass (M).
Now, let the two ends of the rod be at a temperature difference of $(\vartriangle T)$.
Then, by the property of conduction, the heat flow through the rod will be given as:
$\Rightarrow Q=\dfrac{KA(\vartriangle T)}{L}$ [Let this expression be equation number (1)]
This heat transfer can also be given using the specific heat of the rod as:
$\Rightarrow Q=mS(\vartriangle T)$ [Let this expression be equation number (2)]
Now, using the criteria of convection for thermometric conductivity, this heat can be written as:
$\Rightarrow Q=hA(\vartriangle T)$ [Let this expression be equation number (3)]
On equating equation number (1), (2) and (3). We get:
$\Rightarrow \dfrac{KA(\vartriangle T)}{L}=mS(\vartriangle T)=hA(\vartriangle T)$
On further simplifying the above equation, we get the thermometric conductivity as:
$\begin{align}
  & \Rightarrow K=hdS \\
 & \therefore h=\dfrac{K}{dS} \\
\end{align}$
Hence, the thermometric conductivity comes out to be $\dfrac{K}{dS}$ .

Hence, option (C) is the correct option.

Note:
This relation between thermal conductivity and thermometric conductivity is universal in nature, that is, it is independent of the nature of the object. Also, the more common term used in place of thermometric conductivity is thermal diffusivity of an object.