An iron bar (`L_(1) = 0.1 m, A_(1) = 0.02 m^(2) , K_(1) = 79 Wm^(-1) K^(-1)`) and a brass bar `(L_(2)=0.1 m , A_(2) = 0.02 m^(2), K_(2) = 109 Wm^(-1)K^(-1)`) are soldered end to end as shown in fig. the free ends of iron bar and brass bar are maintained at 373 K and 273 K respectively. Obtain expressions for and hence compute (i) the temperature of the junction of the two bars, (ii) the equivalent thermal conductivity of the compound bar and (iii) the heat current through the compound bar.
.

Explanation

Given, Length `L_(1) = L_(2) = L=0.1`m
Area, `A_(1) = A_(2) = A = 0.02m^(2)`
Thermal conductivity, `K_(1)`= 79 `Wm^(-1)K^(-1)`,`K_(2) = 109 Wm^(-1)K^(-1)`
Temperature , `T_(1)`=373 K and `T_(2)=273K`
At steady state, heat transferred from any section of thermal conductor is same i.e. `H_(1) =H_(2)=H`
`rArr ((K_(1)A_(1)(T_(1)-T_(0)))/L_(1)=((K_(2)A_(2)(T_(0)-T_(2)))/L_(2)).......(ii)
For `A_(1)=A_(2)=A` and `L_(1) = L_(2) = L`, Eq. (i) becomes `K_(1)(T_(1)-T_(0)) = K_(2)(T_(0)-T_(2)) rArrT_(0)=(K_(1)T_(1)+K_(2)T_(2))/(K_(1)+K_(2))`
Therefore, heat current through other bar
`H= (K_(1)A(T_(1)-T_(0)))/L=(K_(2)A(T_(0)-T_(2)))/L = A(T_(1)-T_(2))/(L(1/K_(1)+1/K_(2)))`
Now, heat current through composite bar of length `L_(1)+L_(2) = 2L` and equivalent thermal conductivity `K^(')`, can be given by
`H^(') = (K^(')A(T_(1)-T_(2)))/(2L) = H rArrK=(2K_(1)K_(2))/(K_(1)+K_(2))`
(i) So, the temperature of the junction of two bars is
``T_(0)= (K_(1)T_(1) + K_(2)K_(2))/(K_(1)+K_(2))`
`=(79Wm^(-1)xx373 K + 109Wm^(-1)K^(-1)xx 273 K)/(79Wm^(-1)K^(-1) + 109 Wm^(-1)K^(-1))`
`91.6 Wm^(-1)K^(-1)`
iii) Heat current through the composite bar
`H^(') = H = (K^(')A(T_(1)-T_(2))/2L`
`(91.6 Wm^(-1)K^(-1) xx 0.002m^(3) xx (373 -273)K)/(2 xx 0.1)m