First slide

Heat transfer

Question

Two identical conducting rods are first connected independently to two vessels, one containing water at 100^oC and the other containing ice at 0^oC. In the second case, the rods are joined end to end and connected to the same vessels. Let q₁ and q₂g/s be the rate of melting of ice in two cases, respectively. The ratio of q₁/q₂ is

Moderate

A

$\frac{1}{2}$
B

$\frac{2}{1}$
C

$\frac{4}{1}$
D

$\frac{1}{4}$

Solution

Initially the rods are placed in vessels as shown below.
$\frac{Q}{t} = \frac{(θ_{1} - θ_{2})}{R} \Rightarrow {(\frac{Q}{t})}_{1} = \frac{mL}{t} = q_{1} L = \frac{(100 - 0)}{\frac{R}{2}}$ …..(i)
Finally when rods are joined end to end as shown,
$\Rightarrow {(\frac{Q}{t})}_{2} = \frac{mL}{t} = q_{2} L = \frac{(100 - 0)}{2 R}$ ……..(ii)
From equations (i) and (ii), $\frac{q_{1}}{q_{2}} = \frac{4}{1}$
IN series combination thermal resistance is four times to that in parallel combination of identical conducting rods .so in parallel combination heat flow is 4 times and hence rate of melting of ice will be 4 times to that in parallel combination.

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