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. $40$
• B. $60$
• C. $50$
• D. $35$

Answer 1

Answer: (C)

$50$

Let the temperature of junction be $T$.

Also let the thermal conductivity, area and length of conductor be $K,A$ and $L$ respectively.

Using Junction law, $\frac{d{q}_1}{dt}=\frac{d{q}_2}{dt}+\frac{d{q}_3}{dt}$

$\therefore$ $\frac{KA(20-T)}{L}=$ $\frac{KA(T-60)}{L}+$ $\frac{KA(T-70)}{L}$

Or $20-T=T-60+T-70$

$\therefore$ $T=\frac{150}{3}={50}^{o}C$