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Heat and Thermodynamics question

2004 · Shift 0 · Q175

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Heat and Thermodynamics
2004 · Shift 0 · Q175

Heat and Thermodynamics question

2004 · Shift 0 · Q175

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1

The temperature of the 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}\left( {{T_2} \gt {T_1}} \right).

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

\left( {{{A\left( {{T_2} - {T_1}} \right)K} \over x}} \right)f,

with

f

equal to

AIEEE 2004 Physics - Heat and Thermodynamics Question 406 English

A
${2 \over 3}$
B
${1 \over 2}$
C
$1$
D
${1 \over 3}$

View written solutionFree

Correct answer: D

Use thermal resistance for each layer

For steady-state heat conduction through slabs in series,

\dot Q = \frac{A\,(T_2-T_1)}{R_1+R_2}

where thermal resistance of a slab is

R = \frac{L}{kA}

Here the two layers are:

Layer 1: conductivity $K$ , thickness $x$
Layer 2: conductivity $2K$ , thickness $4x$

So,

R_1 = \frac{x}{KA}

and

R_2 = \frac{4x}{(2K)A} = \frac{2x}{KA}

Find total resistance

R_{\text{total}} = R_1 + R_2 = \frac{x}{KA} + \frac{2x}{KA} = \frac{3x}{KA}

Compute heat current

\dot Q = \frac{A(T_2-T_1)}{\frac{3x}{KA}} = \frac{AK(T_2-T_1)}{3x}

This is given in the form

\left(\frac{A(T_2-T_1)K}{x}\right)f

Therefore,

f = \frac{1}{3}

Check options

A: $\frac{2}{3}$ ❌
B: $\frac{1}{2}$ ❌
C: $1$ ❌
D: $\frac{1}{3}$ ✅

Hence the correct answer is D.

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