Question

The thermal conductivity of a material in the CGS system is $0.4$. In steady state the rate of flow of heat $10\text{ cal}{{s}^{-1}}$, then the thermal gradient will be
$\begin{align}
  & \text{A}\text{. }{{15}^{\circ }}Cc{{m}^{-1}} \\
 & \text{B}\text{. }{{25}^{\circ }}Cc{{m}^{-1}} \\
 & \text{C}\text{. }{{50}^{\circ }}Cc{{m}^{-1}} \\
 & \text{D}\text{. }{{75}^{\circ }}Cc{{m}^{-1}} \\
\end{align}$

Answer 1

Hint: Thermal conductivity refers to the ability of a given material to conduct or transfer heat. The thermal gradient is the ratio of the temperature difference and the distance between two points, equivalently, the thermal gradient is the change in temperature over a given length. For calculating the thermal gradient over a conduction process, we need to know the values of thermal conductivity and the rate of heat flow.

Formula used:
The thermal gradient is given as,
\[\dfrac{\Delta T}{x}=\dfrac{1}{KA}\left( \dfrac{\Delta Q}{\Delta t} \right)\]
Where,
$\dfrac{\Delta Q}{\Delta t}$ is the rate of heat flow
$KA$ is the thermal conductivity of the material

Complete step-by-step solution:
The thermal conductivity of a material is defined as a measure of its ability to conduct heat. Fourier’s law of thermal conduction also called the law of heat conduction states that the rate at which heat is conducted or transferred through a material is proportional to the temperature gradient is negative and is also proportional to the area through which the heat energy flows.
The differential form of Fourier’s law of thermal conduction can be expressed through the following equation:
$Q=-K\cdot \nabla T$
Where,
$Q$ denotes the thermal flux or heat flux
$K$ refers to the thermal conductivity of the material
$\nabla T$ refers to the temperature gradient
The thermal conductivity of a material is expressed using the following formula:
$K=\dfrac{QL}{A\Delta T}$
Where,
$K$ is the thermal conductivity
$Q$ is the amount of heat transferred by the material
$L$ is the distance between two materials
$A$ is the area of a surface of the material
$\Delta T$ is the difference in temperature
A temperature gradient is a physical quantity, which is used to describe in which direction and at what rate the temperature varies the most rapidly around a particular location. The temperature gradient is a dimensional quantity and is expressed in the units of degrees per unit length. The SI unit of the temperature gradient is kelvin per meter.
Re-writing the differential form for law of heat conduction,
We get.
$\dfrac{\Delta Q}{\Delta t}=\dfrac{KA\Delta T}{\Delta x}$
Thermal gradient is given as,
\[\begin{align}
  & \dfrac{\Delta T}{x}=\dfrac{\dfrac{\Delta Q}{\Delta t}}{KA} \\
 & \dfrac{\Delta T}{x}=\dfrac{1}{KA}\left( \dfrac{\Delta Q}{\Delta t} \right) \\
\end{align}\]
Where,
$\dfrac{\Delta Q}{\Delta t}$ is the rate of heat flow
$KA$ is the thermal conductivity of the material
Putting values,
$\begin{align}
  & \dfrac{\Delta Q}{\Delta t}=10\text{ cal}{{s}^{-1}} \\
 & KA=0.4 \\
\end{align}$
We get,
$\dfrac{\Delta T}{x}=\dfrac{10}{0.4}={{25}^{\circ }}Cc{{m}^{-1}}$
The thermal gradient of the material in steady state is ${{25}^{\circ }}Cc{{m}^{-1}}$
Hence, the correct option is B.

Note: The thermal conductivity of a material is a measure of its ability to conduct or transfer the heat energy. In a heat conduction process, the value of the thermal gradient is calculated in order to know the variation of heat flow with the distance, increasing or decreasing. For a particular material, the rate at which heat is conducted or transferred through the material is proportional to the negative of the temperature gradient.