Question

ABCDE is a regular pentagon of uniform wire. The rate of heat entering at A and leaving at C is equal \[{T_B}\] ​and \[{T_D}\] ​are the temperature of B and D. Find the temperature \[{T_C}\]​:

A)$\dfrac{{3{T_B} + 2{T_D}}}{5}$
B)$3{T_D} - 2{T_B}$
C)$2{T_D} - 2{T_B}$
D) Can have any value

Answer 1

Hint: The heat flow in the pentagon will behave similarly to the current flow in wires with a similar configuration. The heat flow will split into the two branches similar to the splitting of current in parallel resistors.

Complete step by step answer:
We’ve been given that ABCDE is a regular pentagon. Let us denote the heat conductivity of each wire with $K$. Then the net conductivity for the flow of heat from the branch ABC (branch-1) will be $2K$ since wires are in series.
Similarly, the conductivity in the branch AEDC (branch-2) will be $3K$.
Hence the heat flow in the two branches will be in the inverse ratio of the conductivity of the two branches which will be
\[\dfrac{{{Q_1}}}{{{Q_2}}} = \dfrac{{3K}}{{2K}} = \dfrac{3}{2}\]
Let us consider the wires BC and DC. In both the wires, the heat flow will depend on the temperature at point C. For wire BC, the heat flow will be proportional to
${Q_1} \propto ({T_C} - {T_B})$
And the heat flow in the DC wire will be
${Q_2} \propto ({T_C} - {T_D})$
Then taking the ratio of the two above equations, we get
$\dfrac{{{Q_1}}}{{{Q_2}}} = \dfrac{{{T_C} - {T_B}}}{{{T_C} - {T_D}}}$
Since we know \[\dfrac{{{Q_1}}}{{{Q_2}}} = \dfrac{3}{2}\], we can write
$\dfrac{3}{2} = \dfrac{{{T_C} - {T_B}}}{{{T_C} - {T_D}}}$
Cross multiplying the denominators, it can be simplified to
$3{T_C} - 3{T_D} = 2{T_C} - 2{T_B}$
Which gives us
${T_C} = 3{T_D} - 2{T_B}$
Hence, the correct choice is option (B).

Note: We should know that heat flow in a series of wires is constant. However, the neat heat flow will split into the inverse of the conductivity of the cables when two branches are connected in parallel. We don’t need to know the numerical value of all the temperatures at all the different wires but only need to use the laws of heat conduction.