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 juction 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

Let `T_(0)` be the temperature of the junction of the two bars and A be the area of cross-section of each bar.
(i) Under steady state condition, the heat conducted per second through iron bar is equal to the heat conducted per second through brass bar,
i.e., `H = (K_(1)A_(1)(T_(1)-T_(0)))/(L_1) = (K_(2)A_(2)(T_(0)-T_(2)))/(L_2)`
As, `A_(1) = A_(2) = A and L_(1)=L_(2)=L`, so
`H = (K_(1)A(T_(1)-T_(0)))/(L) = (K_(2)A(T_(0)-T_(2)))/(L)` ...(i)
or `K_(1) (T_(1)-T_(0) = K_(2)(T_(0)-T_(2))`
or `K_(1)T_(1)-K_(1)T_(0)=K_(2)T_(0)-K_(2)T_(2)`
or `T_(0)= (K_(1)T_(1)+K_(2)T_(2))/(K_(1)+K_(2))` ...(ii)
`:. T_(0) = (79xx 373+109 xx 273)/(79+109) = 315 K`.
(ii) Putting the value of `T_(0)` from (ii) in (i) we get
` H = (K_(1)A)/(L) (T_(1)-(K_(1)T_(1)+K_(2)T_(2))/(K_(1)+K_(2)))`
`=((K_(1)K_(2)))/((K_(1)+K_(2)))(T_(1)-T_(2))`...(iii)
If K is the equivalent thermal conductivity of the two bars, then
`H = (KA(T_(1)-T_(2)))/(2L)` ...(iv)
From (iii) and (iv) ,
`K/2 = (K_(1)K_(2))/(K_(1)+K_(2)) or K = (2K_(1)K_(2))/(K_(1)+K_(2))`
`:. K =(2xx79xx109)/(79+109) = 91.6 W m^(-1)K^(-1)`
(iii) `H = (KA(T_(1)-T_(2)))/(2L)`
`=(91.6 xx 90.02) xx (373-273)/(2 xx 0.1) = 916.1W`.