Heat and Thermodynamics 3 Question 1

1. Two materials having coefficients of thermal conductivity " 3K ’ and ’ K ’ and thickness ’ d ’ and ’ 3d ’ respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are ’ θ2 ’ and ’ θ1 ’ respectively, (θ2>θ1). The temperature at the interface is

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(a) θ2+θ12

(b) θ13+2θ23

(c) θ16+5θ26

(d) θ110+9θ210

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Solution:

  1. Let interface temperature in steady state conduction is θ, then assuming no heat loss through sides;

( Rate of heat  flow through  first slab )=( Rate of heat  flow through  second slab )(3K)A(θ2θ)d=KA(θθ1)3d9(θ2θ)=θθ19θ2+θ1=10θθ=910θ2+110θ1



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