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 = 0.1m`
`A_1 = A_2 = A = 0.02 m^2`
` T_1 = 373 K, T_2 = 273K`
(i) In steady state , the rate of flow of heat through both the bars will be same .
` therefore (K_1(T_1 - T_0))/(L_1) = (K_2 A_2 (T_0 - T_2))/(L_2)`
` rArr (K_1A (T_1 - T_0))/(L_1) = (K_2A(T_0 - T_2) )/(L)`
` rArr K_1/K_2 = (T_0 - T_2)/(T_1 - T_0)`
` therefore T_0 = (K_1 T_1 + K_2T_2)/(K_1 + K_2) = (79 xx 373 xx 109 xx 273)/(79+109) `
` =(29467 + 29757)/(188) = 315.02K`
(ii) Heat current through any of the bars will be
`H = (K_1A (T_1 - T_0))/(L) = (K_1A)/(L) (T_1 - (K_1T-1 - K_2T_2)/(K_1 + K_2) )`
` = (K_1K_2)/(K_1 + K_2) ( (T_1 -T_2)/(L) )`
Heat current H. through the compound bar of length 2L and thermal conductivity K. = H
` therefore (K.A (T_1 -T_2) )/(2L) =(K_1 K_2 A)/(K_1 + K_2) = (2 xx 79 xx 109)/(79 + 109) `
` = 91.6 m^(-1) K^(-1)`
(iii) ` H. = (K.A (T_1 -T_2))/(2L)`
` = (91.6 xx 0.02 xx (373 - 273) )/(2 xx 0.1)`
` = 916.1 W`