Question

For the shown figure below, calculate $\left|\begin{array}{c}{T}_1-T_2\end{array}\right|$. The surface area and length is the same for all bars.

• A. $42.86{}^{\circ }\text{C}$
• B. $10{}^{\circ }\text{C}$
• C. $28.54{}^{\circ }\text{C}$
• D. $20{}^{\circ }\text{C}$

Answer 1

The correct option is D $20{}^{\circ }\text{C}$

Let $A$ be area of cross-section and $l$ be length of each bar.

The rate of flow of heat in series will be the same,

Rate of heat flow is given by,

$\frac{dQ}{dt}=kA\frac{dT}{dx}$

At the junction $\text{A}$,

$\Rightarrow \frac{(200-T_1)\left(kA\right)}{l}=\frac{(T_1-T_2)\left(2kA\right)}{l}$

$\Rightarrow T_1-200=2T_2-2T_1$

$\Rightarrow 3T_1-2T_2=200......\left(1\right)$

At the junction $\text{B}$

$\Rightarrow \frac{(T_1-T_2)\left(2kA\right)}{l}=\frac{(T_2-100)kA}{l}$

$\Rightarrow 2(T_2-T_1)=100-T_2$

$\Rightarrow 3T_2-2T_1=100......\left(2\right)$

Solving (1) & (2) we get,

$T_1=160{}^{\circ }\text{C}$

$T_2=140{}^{\circ }\text{C}$

$|T_1-T_2|=|160-140|=20{}^{\circ }\text{C}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{D}\right)$ is the correct answer.