JEE Main · 2004 · 24 Apr 2004 · Offline · Heat & Thermodynamics

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The temperature of two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively are

T_{2}

and

T_{1}

(T_{2} > T_{1})

. The rate of heat transfer through the slab, in a steady state is

(\frac{A ( T _{2} - T _{1} ) K}{x}) f

, with

f

equal to

(A)1
(B) $1/2$
(C) $2/3$
(D) $1/3$ ✓ Correct

Answer: (D)

Solution

In the steady state the same heat flows through both slabs. Across the first,

\frac{d Q}{d t} = \frac{K A ( T _{2} - T )}{x}

, and across the second,

\frac{d Q}{d t} = \frac{2 K A ( T - T _{1} )}{4 x} = \frac{K A ( T - T _{1} )}{2 x}

. Equating the two gives

2 (T_{2} - T) = T - T_{1}

, so the interface settles at

T = \frac{2 T _{2} + T _{1}}{3}

. Feeding this back,

\frac{d Q}{d t} = \frac{K A}{x} \cdot \frac{T _{2} - T _{1}}{3}

, so the rate of flow is one-third of

\frac{K A ( T _{2} - T _{1} )}{x}

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option

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JEE Main · 2004 · 24 Apr 2004 · Offline · Heat & Thermodynamics

Solution

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