Question

A long rod has one end at $ {0^ \circ }C $ and other end at a high temperature. The coefficient of thermal conductivity varies with distance from the low temperature end as $ k = {k_0}\left( {1 + ax} \right) $ , where $ {{\text{k}}_{\text{0}}}{\text{ = 1}}{{\text{0}}^{\text{2}}}{\text{ SI unit}} $ and $ a = 1{m^{ - 1}} $ . At what distance from the first end the temperature will be $ {100^ \circ }C $ ?
The area of cross-section is $ 1c{m^2} $ and rate of heat conduction is $ 1W $ .
(A) $ 2.7m $
(B) $ 1.7m $
(C) $ 3m $
(D) $ 1.5m $

Answer 1

Hint: This question requires application of Fourier’s Law of Heat Conduction. Provide the mathematical expression of the law and then substitute the values provided in the question to solve the problem.

Formula Used: The formulae used in the solution are given here.
 $ Q = - kA\dfrac{{dT}}{{dx}} $ where $ Q $ is the local heat flux density in, $ k $ is the conductivity of the material, $ A $ is the area and $ \dfrac{{dT}}{{dx}} $ is the temperature gradient.

Complete step by step answer:
Fourier’s law of heat conduction states that the negative gradient of temperature and the time rate of heat transfer is proportional to the area at right angles of that gradient through which the heat flows. The derivation of Fourier’s law was explained with the help of an experiment which explained the Rate of heat transfer through a plane layer is proportional to the temperature gradient across the layer and heat transfer area.
Using Fourier’s law of heat conduction,
 $ Q = - kA\dfrac{{dT}}{{dx}} $ where $ Q $ is the local heat flux density in, $ k $ is the conductivity of the material, $ A $ is the area and $ \dfrac{{dT}}{{dx}} $ is the temperature gradient.
It has been given that a long rod has one end at $ {0^ \circ }C $ and the other end at a high temperature. The coefficient of thermal conductivity varies with distance from the low temperature end as $ k = {k_0}\left( {1 + ax} \right) $ , where $ {{\text{k}}_{\text{0}}}{\text{ = 1}}{{\text{0}}^{\text{2}}}{\text{ SI unit}} $ and $ a = 1{m^{ - 1}} $ . The area of cross-section is $ 1c{m^2} $ and rate of heat conduction is $ 1W $ .
Thus, substituting $ k = {k_0}\left( {1 + ax} \right) $ in Fourier’s law, we get,
 $ Q = - {k_0}\left( {1 + ax} \right)A\dfrac{{dT}}{{dx}} $ .
We also have been given that, $ Q = 1W $ , $ dT = {100^ \circ }C $ , $ {{\text{k}}_{\text{0}}}{\text{ = 1}}{{\text{0}}^{\text{2}}}{\text{ SI unit}} $ and $ a = 1{m^{ - 1}} $ . Substituting all these values, and integrating from 0 to $ L $ , where $ L $ is the distance from the first end,
 $ \int\limits_0^L {\dfrac{{dx}}{{\left( {1 + ax} \right)}}} = - 100 \times \left( {1 \times {{10}^{ - 4}}} \right) \times \int\limits_0^{100} {dT} $
 $ \Rightarrow \ln \left( {1 + L} \right) = 1 $
Solving the equation above, we get,
 $ 1 + L = 2.73 $ .
Thus, the distance from the first end $ L = 2.73 - 1 = 1.73m $
Hence the correct answer is Option B.

Note:
Fourier’s law is the other name of the law of heat conduction. Newton’s law of cooling and Ohm’s law are a discrete and electrical analog of Fourier’s law.