A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts a amount of heat Q in time t. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod, when placed in thermal contact with the two reservoirs in time t ?
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Answer:
Explanation:
In steady state the amount of heat flowing from one face to the other face in time t is given by -
Q = KA(θ1−θ2)t/l
where K is coefficient of thermal conductivity of material of rod
= Q/t ∝ A/l ∝ r²/l
Since the metallic rod is melted and the material is formed into a rod of half the radius, thus V1 = V2
= πr²1l1 = πr²2l2
Therefore,
= Q1/Q2 = r²1/l1 × l2/r²2=
= r²1l/1 × 4l1/(r1/2)²
= Q1 = 16Q2
Thus, the amount of heat conducted by the new rod, when placed in thermal contact with the two reservoirs in time t is 16.
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this is the relation between them
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