Question

Figure shows in cross section a wall consisting of four layers with thermal conductivities $K_{1}=0.06\, W /$ $mK ; K_{3}=0.04\, W / mK$ and $K_{4}=0.10\, W / mK$. The layer thicknesses are $L_{1}=1.5\, cm , L_{3}=2.8\, cm$ and $L_{4}=3.5\, cm$. The temperature of interfaces is as shown in figure. Energy transfer through the wall is in steady state. The temperature of the interface between layers 2 and 3 is

• A. $11^{\circ} C$
• B. $8^{\circ} C$
• C. $7.2^{\circ} C$
• D. $5.4^{\circ} C$

Answer 1

Answer: (A)

$\left.\frac{\Delta Q}{\Delta t}\right|_{\text {layer } 2}=\left.\frac{\Delta Q}{\Delta t}\right|_{\text {layer } 4}$
$\Rightarrow \frac{0.06 \times A \times 5}{1.5 \times 10^{-2}}$
$=\frac{0.04 \times A \times \Delta T}{2.8 \times 10^{-2}}$
$\Delta T=14^{\circ} C$
$T_{3}=(3+14)^{\circ} C =11^{\circ} C$