Question

Two identical square rods of metal are welded end to end as shown in figure (a). Assume that 10 cal of heat flows through the rods in 2 min. Now the rods are welded as shown in figure, (b). The time it would take for 10 cal to flow through the rods now, is

• A. 0.75 min
• B. 0.5 min
• C. 1.5 min
• D. 1 min

Answer 1

The correct option is B 0.5 min

Given: Two identical square rods of metal are welded end to end as shown in figure (a). Assume that 10 cal of heat flows through the rods in 2 min. Now the rods are welded as shown in figure, (b).

To find the time it would take for 10 cal to flow through the rods now

Solution:

As the temperature difference in this case $\Delta T=T_{hot}–T_{cold}$ is same for both the case, thus the thermal conductivity will be same for both the cases.

The rate H at which heat is transferred through the slab is given by

$H=\frac{kA\Delta T}{\Delta x}.....\left(i\right)$,

where $k$ is the proportionality constant and is called thermal conductivity of the material

$A$ is the area of the slab

$\Delta x$ is the thickness of the slab

$\Delta T$ is the temperature difference.

To obtain the rate of heat transfer $H_o$ for the square rod which are connected in series (configuration (a)), substitute $2L$ for $\Delta x$ in the equation (i), we get

$H_o=\frac{kA\Delta T}{2L}$

To obtain the rate of heat transfer $(H_o{)}^{\prime }$ for the square rod which are connected in parallel (configuration (b)), substitute 2A for A in the equation(i), we get

$H_o=\frac{k\left(2A\right)\Delta T}{L}$

So,

$\frac{(H_o{)}^{\prime }}{H_o}=\frac{\frac{k\left(2A\right)\Delta T}{L}}{\frac{kA\Delta T}{2L}}=\mathrm{4....}\left(ii\right)$

Again also rate of heat transfer H is defined as,

$H=\frac{W}{t}......\left(iii\right)$

Here $W$ is the energy and $t$ is the time.

To obtain rate of heat transfer $H_o$ for the square rod which are connected in series (configuration (a)), substitute $10cal$ for $W$ and $2min$ for $t$ in the equation(iii), we get

$H_o=\frac{10}{2}=5cal/min......\left(iv\right)$

To obtain The rate of heat transfer $(H_o{)}^{\prime }$ for the square rod which are connected in parallel (configuration (b)), substitute eqn (iv) in eqn (ii), we get

$(H_o{)}^{\prime }=4\times 5=20cal/min$

The time $t^{\prime }$ for $10cal$ to flow through the rods if they were welded in the figure parallel configuration will be,

$t^{\prime }=\frac{W}{(H_o{)}^{\prime }}⟹t^{\prime }=\frac{10}{20}⟹t^{\prime }=0.5min$

is the required time it would take for 10 cal to flow through the rods now.