Question

The heat is flowing through a rod of length 50 cm and area of cross-section \[5c{m^2}\]. Its ends are respectively at \[{25^0}C\] and \[{125^0}C\]. The coefficient of thermal conductivity of the material of the rod is \[0.092\,Kcal/s{m^0}C\]. Find the temperature gradient in the rod.
A. \[{2^0}C/cm\]
B. \[{2^0}C/m\]
C. \[{20^0}C/cm\]
D. \[{20^0}C/m\]

Answer 1

Hint: A temperature gradient is defined as the change in temperature over a specified distance between two isothermal surfaces or two points in a material. Using this term we are going to find the solution to this question.

Formula Used:
To find the temperature gradient the formula is,
\[\dfrac{{dT}}{{dx}} = \dfrac{{\Delta T}}{{\Delta x}}\]
Where, \[\Delta T\] is temperature difference between two ends of the rod and \[\Delta x\] is the length of the rod.

Complete step by step solution:
Consider a rod where the heat is flowing through it having length 50cm and the area of cross-section \[5c{m^2}\]. The temperature at its ends is \[{25^0}C\] and\[{125^0}C\]. They have given the coefficient of thermal conductivity of the material of the rod is \[0.092Kcal/s{m^0}C\]. We need to find the temperature gradient in the rod.

In order to find the temperature gradient, we have,
\[\dfrac{{dT}}{{dx}} = \dfrac{{\Delta T}}{{\Delta x}}\]
Given, \[\Delta x = 50cm\]
\[\Delta T = {T_2} - {T_1}\]
\[ \Rightarrow \Delta T = 125 - 25\]
Substitute the value of the difference in temperature and length of the rod in the above equation we get,
\[\dfrac{{dT}}{{dx}} = \dfrac{{\left( {125 - 25} \right)}}{{50}}\]
\[\Rightarrow \dfrac{{dT}}{{dx}} = \dfrac{{100}}{{50}}\]
\[\therefore \dfrac{{dT}}{{dx}} = {2^0}C/cm\]
Therefore, the temperature gradient in the rod is \[{2^0}C/cm\].

Hence, option A is the correct answer.

Note:The temperature gradients in the atmosphere are important in the topic atmospheric sciences. Here, we need to find the temperature gradient of a rod in which the temperature of the ends of the rod and the length of the rod are given.