A cylindrical metallic rod in thermal contact | vedclass

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Question

(A) The rate of heat conduction is given by the formula: $\frac{Q}{t} = \frac{k A (T_1 - T_2)}{l}$.Since the volume of the material remains constant d

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(A) The rate of heat conduction is given by the formula: $\frac{Q}{t} = \frac{k A (T_1 - T_2)}{l}$.Since the volume of the material remains constant during melting and reshaping,we have $A_1 l_1 = A_2 l_2$.Given that the new length $l_2 = 4 l_1$,we substitute this into the volume equation: $A_1 l_1 = A_2 (4 l_1)$,which gives $A_2 = \frac{A_1}{4}$.For the new rod,the heat conducted in time $t$ is $Q_2 = \frac{k A_2 (T_1 - T_2) t}{l_2}$.Substituting $A_2 = \frac{A_1}{4}$ and $l_2 = 4 l_1$ into the equation for $Q_2$:$Q_2 = \frac{k (A_1 / 4) (T_1 - T_2) t}{4 l_1} = \frac{1}{16} \frac{k A_1 (T_1 - T_2) t}{l_1}$.Since $Q_1 = \frac{k A_1 (T_1 - T_2) t}{l_1}$,we get $Q_2 = \frac{1}{16} Q_1$.Therefore,$\frac{Q_1}{Q_2} = 16$.

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