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Q.

The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity k and 2 K and thickness x and 4 x respectively are T2 and T1 (T2 > T1). The rate of heat transfer through the slab, in a steady-state is AT2−T1Kx f with f equal to:

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a

1

b

12

c

23

d

13

answer is D.

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Detailed Solution

For slabs in series, we have :Req=R1+R2 i.e., 5xKcq. A=4x2KA+xKAKeq=5K3Now, in steady state, rate of heat transfer through the slab=Keq⋅AT2−T15x=AT2−T1KxfPutting the value of Keq. we get f=13
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