Figure shows in cross section a wall consisting of four layers with thermal conductivities K1=0.06 W / mK ; K3=0.04 W / mK and K4=0.10 W / mK. The layer thicknesses are L1=1.5 cm , L3=2.8 cm and L4=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

Question

Figure shows in cross section a wall consisting of four layers with thermal conductivities K1=0.06 W / mK ; K3=0.04 W / mK and K4=0.10 W / mK. The layer thicknesses are L1=1.5 cm , L3=2.8 cm and L4=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° C (B) 8° C (C) 7.2° C (D) 5.4° C

Tardigrade Admin · Accepted Answer

.(Δ Q/Δ t)| text layer 2=.(Δ Q/Δ t)| text layer 4 ⇒ (0.06 × A × 5/1.5 × 10-2) =(0.04 × A × Δ T/2.8 × 10-2) Δ T=14° C T3=(3+14)° C =11° C

$1 1^{\circ} C$

$8^{\circ} C$

$7. 2^{\circ} C$

$5. 4^{\circ} C$

$\frac{Δ Q}{Δ t} ∣ ∣_{layer 2} = \frac{Δ Q}{Δ t} ∣ ∣_{layer 4}$
$\Rightarrow \frac{0.06 \times A \times 5}{1.5 \times 1 0 ^{- 2}}$
$= \frac{0.04 \times A \times Δ T}{2.8 \times 1 0 ^{- 2}}$
$Δ T = 1 4^{\circ} C$
$T_{3} = (3 + 14)^{\circ} C = 1 1^{\circ} C$

11∘C

8∘C

7.2∘C

5.4∘C

ΔtΔQ​∣∣​layer 2​=ΔtΔQ​∣∣​layer 4​ ⇒1.5×10−20.06×A×5​ =2.8×10−20.04×A×ΔT​ ΔT=14∘C T3​=(3+14)∘C=11∘C

$1 1^{\circ} C$

$8^{\circ} C$

$7. 2^{\circ} C$

$5. 4^{\circ} C$

$\frac{Δ Q}{Δ t} ∣ ∣_{layer 2} = \frac{Δ Q}{Δ t} ∣ ∣_{layer 4}$
$\Rightarrow \frac{0.06 \times A \times 5}{1.5 \times 1 0 ^{- 2}}$
$= \frac{0.04 \times A \times Δ T}{2.8 \times 1 0 ^{- 2}}$
$Δ T = 1 4^{\circ} C$
$T_{3} = (3 + 14)^{\circ} C = 1 1^{\circ} C$