Three rods of identical cross-sectional area and made from the same metal form the sides of an isosceles triangle ABC right angled at B. The points A and B are maintained at temperatures T and √2T, respectively, in the steady state. Assuming that only heat conduction takes place, temperature of point C is
As TB > TA, heat flows from B to A through both paths BA and BCA.
Rate of heat flow in BC = Rate of heat flow in CA
KA(√2T−Tc)l = KA(Tc−T)√2l
Solving this, we get Tc = 3T√2+1