Question

Three metal rods made of copper, aluminium and brass, each $20cm$ long and $4cm$ in diameter, are maintained at $100^\circ $ and $0^\circ $ respectively. Assume that the thermal conductivity of copper is twice that of aluminium and four times that of brass. The approximately equilibrium temperatures of copper-aluminium and aluminium-brass junctions are respectively.

(A) $68^\circ $ and $75^\circ C$
(B) $75^\circ $ and $68^\circ C$
(C) $57^\circ $ and $86^\circ C$
(D) $86^\circ $ and $57^\circ C$

Answer 1

Hint Using the formula for the heat current, we can equate the heat current in the three rods since they are connected in series. Then we can find the value of the equilibrium temperatures of the junctions from the equations.

Formula Used: In the solution we will be using the following formula,
$H = \dfrac{{\Delta TKA}}{L}$
where $H$ is the heat current flowing, $\Delta T$ is the difference in temperatures,
$K$ is the thermal conductivity of the material, $A$ is the area of cross-section,
and $L$ is the length of the wire.

Complete step by step answer
According to the picture that is given in the question, we see that the three rods are connected in series with each other. So the heat current that is flowing through the three rods will be equal to one another. Now according to the question the length and the area of cross-section of all the three rods are the same. So we take them as$L$ and $A$ for all the rods. The thermal conductivity of copper is twice that of aluminium and four times that of brass. So we take the thermal conductivity of aluminium as $K$. So the thermal conductivity of copper is $2K$ and that of brass is $\dfrac{K}{2}$
Therefore we can use the formula for the heat current through each of the rods as,
${H_{Cu}} = \dfrac{{\left( {100 - {T_1}} \right)2KA}}{L}$
${H_{Al}} = \dfrac{{\left( {{T_1} - {T_2}} \right)KA}}{L}$ and
${H_{Brass}} = \dfrac{{\left( {{T_2} - 0} \right)KA}}{{2L}}$
On equating the three we have,
$\dfrac{{\left( {100 - {T_1}} \right)2KA}}{L} = \dfrac{{\left( {{T_1} - {T_2}} \right)KA}}{L} = \dfrac{{\left( {{T_2} - 0} \right)KA}}{{2L}}$
Now from ${H_{Al}} = {H_{Brass}}$
$\dfrac{{\left( {{T_1} - {T_2}} \right)KA}}{L} = \dfrac{{\left( {{T_2} - 0} \right)KA}}{{2L}}$
On cancelling the common terms on both the sides we have,
${T_1} - {T_2} = \dfrac{{{T_2}}}{2}$
On taking the similar terms to one side we have,
${T_1} = {T_2} + \dfrac{{{T_2}}}{2}$
On adding we have
${T_1} = \dfrac{{3{T_2}}}{2}$
Now from ${H_{Cu}} = {H_{Al}}$ we get,
$\dfrac{{\left( {100 - {T_1}} \right)2KA}}{L} = \dfrac{{\left( {{T_1} - {T_2}} \right)KA}}{L}$
Again on cancelling the common terms we have,
$2\left( {100 - {T_1}} \right) = {T_1} - {T_2}$
On opening the brackets and taking common terms to one side we get,
$200 = 2{T_1} + {T_1} - {T_2}$
Hence we have,
$200 = 3{T_1} - {T_2}$
Now on substituting ${T_1} = \dfrac{{3{T_2}}}{2}$ we get,
\[200 = 3 \times \dfrac{{3{T_2}}}{2} - {T_2}\]
On subtracting we get,
\[200 = \dfrac{{\left( {9 - 2} \right){T_2}}}{2}\]
Hence we get the temperature as,
\[{T_2} = \dfrac{{200 \times 2}}{7}\]
On calculating this gives us,
\[{T_2} = 57.14^\circ C\]
This is approximately equal to \[{T_2} \simeq 57^\circ C\]
Substituting this value in ${T_1} = \dfrac{{3{T_2}}}{2}$ we get,
${T_1} = \dfrac{{3 \times 57}}{2}$
Hence, ${T_1} = 85.5^\circ C$
This is approximately equal to,
${T_1} \simeq 86^\circ C$
So the temperatures are $86^\circ $ and $57^\circ C$. So the correct option is D.

Note
The heat current is described as the rate of exchange of kinetic energies between two molecules. It can also be described as the rate of transfer of heat with respect to the time. It can be written in the form of $H = \dfrac{{dQ}}{{dt}}$, where $Q$ is the heat and $t$ is the time.