Question

A metal rod of length 2 m has cross-sectional area 2A and A as shown in the figure. The ends are maintained at temperatures $100^\circ C$ and $70^\circ C$. The temperature at middle point C is
A. $90^\circ C$
B. $30^\circ C$
C. $45^\circ C$
D. $60^\circ C$

Answer 1

Use the formula for heat transfer through the slab of thickness L. In the series, the rate of heat transfer is constant.
Therefore, for every point on the rod, the rate of heat transfer is the same.

Formula used:
The rate of heat transfer is,
$\dfrac{Q}{t} = \dfrac{{KA\left( {{T_2} - {T_1}} \right)}}{L}$
Here, K is the thermal conductivity, A is the area and L is the thickness of the slab.

Complete step by step answer:
We assume the temperature at middle point C is $T$. Also, ${T_2} = 100^\circ C$ and ${T_1} = 70^\circ C$.
The heat energy transferred through the slab of area A whose two surfaces are maintained at temperatures ${T_1}$ and ${T_2}$ is,
$\dfrac{Q}{t} = \dfrac{{KA\left( {{T_2} - {T_1}} \right)}}{L}$
Here, K is the thermal conductivity, A is the area and L is the thickness of the slab.
We know that the rate of heat flow in the series rod is constant. Therefore, we can write,
$K\left( {2A} \right)\dfrac{{\left( {{T_2} - T} \right)}}{L} = KA\dfrac{{\left( {T - {T_1}} \right)}}{L}$
$\Rightarrow 2\left( {{T_2} - T} \right) = \left( {T - {T_1}} \right)$
Substitute ${T_2} = 100^\circ C$ and ${T_1} = 70^\circ C$ in the above equation.
$\Rightarrow 2\left( {100^\circ C - T} \right) = \left( {T - 70^\circ C} \right)$
$\Rightarrow 200^\circ C - 2T = T - 70^\circ C$
$\Rightarrow T = \dfrac{{270^\circ C}}{3}$
$\therefore T = 90^\circ C$

So, the correct answer is “Option A”.

Note:
To use the formula, $\dfrac{Q}{t} = \dfrac{{KA\left( {{T_2} - {T_1}} \right)}}{L}$, the temperature is in Kelvin, but since it is the difference in the temperatures, the difference will be the same for both Kelvin and Celsius temperatures. Therefore, we don’t need to convert the temperature in Kelvin in this question.