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A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an 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?
A
$\frac{Q}{2}$
B
$\frac{Q}{4}$
C
$\frac{Q}{16}$
D
2Q
Detailed Solution
Heat conducted in time t: $Q = \frac{KA(T_1 - T_2)t}{L}$, so for the same material, temperatures and time, $Q \propto \frac{A}{L}$
New radius = $\frac{r}{2}$, so new area $A' = \pi\left(\frac{r}{2}\right)^2 = \frac{A}{4}$
The volume of metal is unchanged: $A'L' = AL$, so $L' = \frac{AL}{A/4} = 4L$
$\frac{Q'}{Q} = \frac{A'}{A}\times\frac{L}{L'} = \frac{1}{4}\times\frac{1}{4} = \frac{1}{16}$
$Q' = \frac{Q}{16}$
New radius = $\frac{r}{2}$, so new area $A' = \pi\left(\frac{r}{2}\right)^2 = \frac{A}{4}$
The volume of metal is unchanged: $A'L' = AL$, so $L' = \frac{AL}{A/4} = 4L$
$\frac{Q'}{Q} = \frac{A'}{A}\times\frac{L}{L'} = \frac{1}{4}\times\frac{1}{4} = \frac{1}{16}$
$Q' = \frac{Q}{16}$