The outer faces of a rectangular slab made of equal thickness of iron and brass are maintained at $100^{\circ} C$ and $0^{\circ} C$ respectively. The temperature at the interface is ........... $^{\circ} C$ (Thermal conductivity of iron and brass are $0.2$ and $0.3$ respectively.)
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(b)
Rate of flow of heat energy through iron=rate of flow of heat energy through brass
$KA ( dT ) / l = KA ( dT ) / l$
As Area and length are same, therfore,
$0.2(100- T )=0.3( T - O )$
$2 O -0.2 T =0.3\,T$
$T =20 / 0.5$
$T =40^{\circ} C$
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