An iron bar (L1 = 0.1 m, A1= 0.02 m2
, K1 = 80 W m?1 K?1) and a brass bar (L2 = 0.1 m, A2 = 0.02
, K2 = 120 W m?1K?1) are soldered end to end. The free ends of the iron bar and brass bar are
maintained at 400K and 300 K respectively. What is the the temperature of the junction of the two

Let emperature of the junction be T0. As both the bars are connected in series so heat current through them is equal .Heat current through iron rod =KironAiron(T1-T)L1 =KironAiron(400-T)0.1=80×0.02(400-T)0.1Heat current through brass rod =KbrassAbrass(T-T2)L2=KbrassAbrass(T-300)0.1=120×0.02×(T-300)0.1So80×0.02(400-T)0.1 =120×0.02×(T-300)0.180(400-T) =120(T-300)(400-T)(T-300)=12080=1.5400- T=1.5 T-450950= 2.5 TT =380 K

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