10-2.Transmission of Heat
easy

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.)

A

$100$

B

$40$

C

$50$

D

$70$

Solution

(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,

$0.2(100- T )=0.3( T – O )$

$2 O -0.2 T =0.3\,T$

$T =20 / 0.5$

$T =40^{\circ} C$

Standard 11
Physics

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