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

The thermal resistance of AC is equal to that of CB and is equal to `2.5KW_(-1)` . Suppose, the temperature at C is (theta). The heat current through AG, CB and CD are
``(DeltaQ_(1))/(Deltat)=(100^(@)C-theta)/(2.5KW^(-1)` . `(DeltaQ_(2))/(Deltat)=(theta-0^(@)C)/(2.5KW^(-1)` . and `(DeltaQ_(3))/(Deltat)=(theta-25^(@)C)/(5.0KW^(-1)` . We also have
`(DeltaQ_(1))/(Deltat)=(DeltaQ_(2))/(Deltat)+(DeltaQ_(3))/(Deltat)` . or, ``(100^(@)C-theta)/(2.5)=(theta-0^(@)C)/(2.5)+(theta-25^(@)C)/(5)` . or, `225^(@)C=5theta` . or, `theta=45^(@)C` . Thus, `(DeltaQ_(3))/(Deltat)=(45^(@)C-45^(@)C)/(5.0KW^(-1))+(20K)/(5.0KW^(-1))` . `=4.0W` .