Question

Three identical thermal conductors are connected as shown in figure. Considering no heat loss due to radiation, temperature at the junction will be

• A. $40^{\circ }C$
• B. $60^{\circ }C$
• C. $50^{\circ }C$
• D. $35^{\circ }C$

Answer 1

Answer: (C)

Let the temperature of junction be $\theta .$ In equilibrium, rate of flow of heat through rod 1 is equal to sum of rate of flow of heat through rods 2 and $3,ie$,
${\left(\frac{dQ}{dt}\right)}_1={\left(\frac{dQ}{dt-}\right)}_2={\left(\frac{dQ}{dt}\right)}_{-3}$
$\therefore \frac{KA(\theta -20)}{l}=\frac{KA(60-\theta )}{l}+\frac{KA(70-\theta )}{l}$
$\Rightarrow \theta -20=130-2\theta$
or $3\theta =150$
$\therefore \theta =50^{\circ }C$