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 $3, i e$,
$\left(\frac{d Q}{d t}\right)_{1}=\left(\frac{d Q}{d t-}\right)_{2}=\left(\frac{d Q}{d t}\right)_{-3}$
$\therefore \frac{K A(\theta-20)}{l}=\frac{K A(60-\theta)}{l}+\frac{K A(70-\theta)}{l}$
$\Rightarrow \theta-20=130-2 \theta$
or $3 \theta=150$
$\therefore \theta=50^{\circ} C $