Question

Four identical rods $\text{AB, CD. CF}$ and $\text{DR}$ are joined as shown in figure. The length, cross-sectional area and thermal conductivity of each rod are $l,A$ and $K$ respectively. The ends $\text{A, E}$ and $\text{F}$ are maintained at temperatures $T_1={70}^{\circ }\text{C},T_2={30}^{\circ }\text{C}$ and $T_3={40}^{\circ }\text{C}$ respectively. Assuming no loss of heat to the atmosphere, find the temperature at $\text{B}$.

• A. ${45}^{\circ }\text{C}$
• B. ${50}^{\circ }\text{C}$
• C. ${55}^{\circ }\text{C}$
• D. ${60}^{\circ }\text{C}$

Answer 1

The correct option is B ${50}^{\circ }\text{C}$


Let the temperature at point $\text{B}$ be $T$.

Rate of heat flow in Rod $\text{FCB}$ due to conduction

$\frac{\Delta Q_1}{\Delta t}=\frac{KA(T-40)}{l+\frac{l}{2}}=\frac{2KA(T-40)}{3l}[\frac{\Delta Q}{\Delta t}=\frac{KA\Delta T}{l}]$

Rate of heat flow in the rod $\text{EDB}$ is

$\frac{\Delta Q_2}{\Delta t}=\frac{KA(T-30)}{l+\frac{l}{2}}=\frac{2KA(T-30)}{3l}$

Similarly, the rate of heat flow in tod $\text{AB}$ will be

$\frac{\Delta Q_3}{\Delta t}=\frac{KA(70-T)}{l}$

In steady state, heat coming is equal to heat outgoing at any point. So, at point $\text{B}$

Heat in flow = Heat out flow

$\frac{KA(70-T)}{l}=\frac{2KA(T-40)}{3l}+\frac{2KA(T-30)}{3l}$

$\Rightarrow 3(70-T)=2T-80+2T-60$

$\therefore T=50{}^{\circ }\text{C}$

Hence, option $\left(\text{B}\right)$ is the correct answer.

Why this question ? Concept : In steady state, net heat flow at any point of the system is zero. It means heat incoming is equal to heat outgoing.