Question

Find the temperature of junction if rods are identical and are placed symmetrically

(a) 95^0 C

(b)  90^0 C

(c) 85^0 C

(d) 80^0 c

Solution

Assume temperature at junction, T

We will calculate the heat flow from each branch. We know total out going heat will be equal to total incoming heat.

Rate of heat flow (DeltaQ)/(Deltat)= - (kxxAxxDeltaT)/l

Since both the rods are identical, therefore k,A and l will be same for all.

-(kxxAxx(100-T))/l+ (-(kxxAxx(100-T))/l)= (-(kxxAxx(T-70))/l)+(-(kxxAxx(T-50))/l)

=> 100-T+100-T= T-70+T-50

=> 4T = 100+100+50+70= 320

=> T = 80^0 C

Therefore, answer (d) is correct.