Question

A heat flux of $4000{\rm{ J/s}}$ is to be passed through a copper rod of length $10{\rm{ cm}}$ and area of cross-section $100\;{\rm{sq cm}}$. The thermal conductivity of copper is $400{\rm{ W/mC}}$. The two ends of this rod must be kept at a temperature difference of…

Answer 1

Hint: In this question the heat is being transferred through a copper rod by means of the conduction process. For a material having length $L$, cross sectional area $A$, thermal conductivity of the material being $K$ and the temperature difference between two ends of the material being $\Delta T$, the heat transferred through the material by conduction $\left( {\Delta Q} \right)$ is given by the “Fourier Equation”. According to this formula-
Heat transferred by conduction, $\Delta Q = KA\dfrac{{\Delta T}}{L}$

Complete step by step answer:
Given:
The heat flux through the copper rod $\Delta Q = 4000{\rm{ J/s}}$
The length of the copper rod $\begin{array}{l}L = 0.1{\rm{ m}}\end{array}$
The cross-sectional area of the copper rod $\begin{array}{l}A = 0.01{\rm{
}}{{\rm{m}}^2}\end{array}$
And, the thermal conductivity of the material $K = 400{\rm{ W/mC}}$
Now, according to the Fourier equation for the conduction process, the heat flux through the copper rod is given by-
$\Delta Q = KA\dfrac{{\Delta T}}{L}$
Substituting the values into the formula we get,
$\begin{array}{l}
\Rightarrow 4000 = 400 \times 0.01 \times \dfrac{{\Delta T}}{{0.1}}\\
\Rightarrow \Delta T = \dfrac{{4000 \times 0.1}}{{400 \times 0.01}}
\end{array}$
Solving this we get,
$\Rightarrow \Delta T = 100{\rm{ }}^\circ {\rm{C}}$
Therefore, the two ends of this rod must be kept at a temperature difference of $100{\rm{ }}^\circ {\rm{C}}$.

Note: In this type of conduction process, the value of heat flux through the conductive material at any point of the material along its length is always constant and since the value of thermal conductivity of the material is assumed to be constant at any point of material, the temperature at any point of the material along its length can be easily calculated.