A rod CD of thermal resistance `5.0KW^(-1)` is joined at the middle of an identical rod AB as shown in figure. The ends A, B and D are maintained at `100^(@)C` , `0^(@)C` and `25^(@)C` respectively. Find the heat current in CD.

Explanation

Given, `AC=CB=(CD)/2`
As resistance of the rod is proportional to the length,
so `R_(AC)=R_(CB)=(R_(CD))/2=5/2=2.5K//W`
Let `H_(1)` is the heat current in CD and `H_(2)` in CB, then
`H_(1)=(T_(C)-T_(D))/R_(CD)=(T_(C)-25)/5andH_(2)=(T_(C)-T_(B))/R_(CB)=(T_(C)-0)/2.5`
If H is the heat current through AC, then
`H=(T_(A)-T_(C))/R_(AC)=(100-T_(C))/2.5`
We know that `H=H_(1)+H_(2)`
`therefore(100-T_(C))/2.5=(T_(C)-25)/5+T_(C)/2.5orT_(C)=45^(@)C`
From (i) `H_(1)=(45-25)/5=4.0W`