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Question: A long rod has one end at \( {0^ \circ }C \) and other end at a high temperature. The coefficient of...

A long rod has one end at 0C{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=k0(1+ax)k = {k_0}\left( {1 + ax} \right) , where k0 = 102 SI unit{{\text{k}}_{\text{0}}}{\text{ = 1}}{{\text{0}}^{\text{2}}}{\text{ SI unit}} and a=1m1a = 1{m^{ - 1}} . At what distance from the first end the temperature will be 100C{100^ \circ }C ?
The area of cross-section is 1cm21c{m^2} and rate of heat conduction is 1W1W .
(A) 2.7m2.7m
(B) 1.7m1.7m
(C) 3m3m
(D) 1.5m1.5m

Explanation

Solution

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=kAdTdxQ = - kA\dfrac{{dT}}{{dx}} where QQ is the local heat flux density in, kk is the conductivity of the material, AA is the area and dTdx\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=kAdTdxQ = - kA\dfrac{{dT}}{{dx}} where QQ is the local heat flux density in, kk is the conductivity of the material, AA is the area and dTdx\dfrac{{dT}}{{dx}} is the temperature gradient.
It has been given that a long rod has one end at 0C{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=k0(1+ax)k = {k_0}\left( {1 + ax} \right) , where k0 = 102 SI unit{{\text{k}}_{\text{0}}}{\text{ = 1}}{{\text{0}}^{\text{2}}}{\text{ SI unit}} and a=1m1a = 1{m^{ - 1}} . The area of cross-section is 1cm21c{m^2} and rate of heat conduction is 1W1W .
Thus, substituting k=k0(1+ax)k = {k_0}\left( {1 + ax} \right) in Fourier’s law, we get,
Q=k0(1+ax)AdTdxQ = - {k_0}\left( {1 + ax} \right)A\dfrac{{dT}}{{dx}} .
We also have been given that, Q=1WQ = 1W , dT=100CdT = {100^ \circ }C , k0 = 102 SI unit{{\text{k}}_{\text{0}}}{\text{ = 1}}{{\text{0}}^{\text{2}}}{\text{ SI unit}} and a=1m1a = 1{m^{ - 1}} . Substituting all these values, and integrating from 0 to LL , where LL is the distance from the first end,
0Ldx(1+ax)=100×(1×104)×0100dT\int\limits_0^L {\dfrac{{dx}}{{\left( {1 + ax} \right)}}} = - 100 \times \left( {1 \times {{10}^{ - 4}}} \right) \times \int\limits_0^{100} {dT}
ln(1+L)=1\Rightarrow \ln \left( {1 + L} \right) = 1
Solving the equation above, we get,
1+L=2.731 + L = 2.73 .
Thus, the distance from the first end L=2.731=1.73mL = 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.