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The figure shows the face and interface temperature of a composite slab containing of four layers of two materials having identical thickness. Under steady state condition, find the value of temperature $θ$ .

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$5 ° C$

$10 ° C$

$- 15 ° C$

$15 ° C$

answer is A.

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Detailed Solution

$\frac{∆ Q}{∆ t} = s a m e S o \frac{K A (20 - 10)}{l} = \frac{2 K A (10 - θ)}{l} \Rightarrow 2 θ = 10 \Rightarrow θ = 5 ° C$

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