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.
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Explanation

Given, `L_(1) = L_(2)= L = 0.1 m, A_(1) = A_(2)= A= 0.02 m^(2) K_(1) = 79 W m^(–1) K^(–1), K_(2) = 109 W m^(–1) K^(–1), T_(1) = 373 K, and T_(2) = 273 K.`
Under steady state condition, the heat current (H1) through iron bar is equal to the heat current (H2) through brass bar.
So, `H=H_(1)=H_(2)`
`(K_(1)A_(1)T_(1)-T_(0))/(L_(1))=(K_(2)A_(2)T_(0)-T_(2))/(L_(2))`
For `A_(1) = A_(2) = A" and "L_(1) = L_(2) = L`, this equation leads to `K_(1)(T_(1)-T_(0))=K_(2)(T_(0)-T_(2))`
Thus the junction temperature `T_(0)` of the two bars is

`T_(0)=((K_(1)T_(1)+K_(2)T_(2))/((K_(1)+K_(2))))`
Using this equation, the heat current H through either bar is
`H=(K_(1)A(T_(1)-T_(0)))/(L)=(K_(2)A(T_(0)-T_(2)))/(L)`
`((K_(1)K_(2))/(K_(1)+K_(2)))(A(T_(1)T_(2)))/(L)=(A(T_(1)-T_(2)))/(L((1)/(K_(1))+(1)/(K_(2))))`
Using these equations, the heat current H′ through the compound bar of length `L_(1) + L_(2) = 2L` and the equivalent thermal conductivity K′, of the compound bar are given by
`H'=(K'A(T_(1)-T_(2)))/(2L)=H`
`K'=(2K_(1)K_(2))/(K_(1)+K_(2))`
(i) `T_(0)=((K_(1)T_(1)+K_(2)T_(2)))/((K_(1)+K_(2)))`
` ((79Wm^(-1)K^(-1))(373K)+(109 W m^(-1)K^(-1))(273K))/(79W m^(-1)K^(-1)+109 W m^(-1) K_(-1))`
=315 K
(ii) `K'=(2K_(1)K_(2))/(K_(1)+K_(2))`
`=(2xx(79WM^(-1)K^(-1))xx(109W m^(-1)K^(-1)))/(79W m^(-1)K^(-1)+109W m^(-1)K^(-1)) `
` =91.6 W m^(-1)K^(-1)`
(iii) `H'=H=(K'A(T_(1)-T_(2)))/(2L)`
`=((91.6W m^(-1) K^(-1))xx(0.02m^(2))xx(373K-273K))/(2xx(0.1m))`
916.1 W