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Question: A heat flux of \(4000{\rm{ J/s}}\) is to be passed through a copper rod of length \(10{\rm{ cm}}\) a...

A heat flux of 4000J/s4000{\rm{ J/s}} is to be passed through a copper rod of length 10cm10{\rm{ cm}} and area of cross-section 100  sqcm100\;{\rm{sq cm}}. The thermal conductivity of copper is 400W/mC400{\rm{ W/mC}}. The two ends of this rod must be kept at a temperature difference of…

Explanation

Solution

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

Complete step by step answer:
Given:
The heat flux through the copper rod ΔQ=4000J/s\Delta Q = 4000{\rm{ J/s}}
The length of the copper rod L=0.1m\begin{array}{l}L = 0.1{\rm{ m}}\end{array}
The cross-sectional area of the copper rod A=0.01m2\begin{array}{l}A = 0.01{\rm{ }}{{\rm{m}}^2}\end{array}
And, the thermal conductivity of the material K=400W/mCK = 400{\rm{ W/mC}}
Now, according to the Fourier equation for the conduction process, the heat flux through the copper rod is given by-
ΔQ=KAΔTL\Delta Q = KA\dfrac{{\Delta T}}{L}
Substituting the values into the formula we get,
⇒4000=400×0.01×ΔT0.1 ⇒ΔT=4000×0.1400×0.01\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,
⇒ΔT=100∘C\Rightarrow \Delta T = 100{\rm{ }}^\circ {\rm{C}}
Therefore, the two ends of this rod must be kept at a temperature difference of 100∘C100{\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.