at equilibrium ,
QB→C = QC→A
kA(√2T - Tc)l = kA(Tc-T)√2l
Tc = 3*T√2+1
Three rods of identical cross-sectional area and made from the same metal form the sides of an isosceles triangle ABC, right angeled at B. The points A and B are maintained at temperatures T and (√2)T respectively. In steady state, temperature of point C is Tc. Assuming that only heat conduction takes place, Tc/T is
I am getting 1/(√2 + 1) but ans given is 3/(√2+1)
at equilibrium ,
QB→C = QC→A
kA(√2T - Tc)l = kA(Tc-T)√2l
Tc = 3*T√2+1
no.
it is very clear that Ta <Tc<Tb
hence heat will flow from B→C→A and B→A
heat will flow from C to A and not the other way.