A slab of insulation is made of three layers,as Drawing I indicates.Each of the layers A,B,and C has the same thickness,but a different thermal conductivity.Heat flows through the slab,and the temperatures are as shown.What are the temperatures T₁ and T₂ in Drawing II where the layers are arranged in a different order?
A)T₁ = 230 °C and T₂ = 170 °C
B)T₁ = 200 °C and T₂ = 180 °C
C)T₁ = 220 °C and T₂ = 160 °C
D)T₁ = 180 °C and T₂ = 160 °C
E)T₁ = 210 °C and T₂ = 190 °C

Question

A slab of insulation is made of three layers,as Drawing I indicates.Each of the layers A,B,and C has the same thickness,but a different thermal conductivity.Heat flows through the slab,and the temperatures are as shown.What are the temperatures T₁ and T₂ in Drawing II where the layers are arranged in a different order? 
A)T₁ = 230 °C and T₂ = 170 °C 
B)T₁ = 200 °C and T₂ = 180 °C 
C)T₁ = 220 °C and T₂ = 160 °C 
D)T₁ = 180 °C and T₂ = 160 °C 
E)T₁ = 210 °C and T₂ = 190 °C

Quizplus · Accepted Answer

First, we need to calculate the heat fluxes (heat flow per unit area) through the slab in Drawing I using Fourier's law:
q(A) = (100-200)/0.01 = -10000 W/m²
q(B) = (200-170)/0.01 = 3000 W/m²
q(C) = (170-150)/0.01 = 2000 W/m²

By conservation of energy, the heat flux must be the same through all layers when arranged differently in Drawing II. Therefore, we can write:
q(A) + q(B) + q(C) = q(C') + q(B') + q(A')
where C', B', and A' are the new layers.

Also, we know that the thickness of each layer is the same, so we can write:
q/Δx = kA = kB = kC = kC' = kB' = kA'
where q/Δx is the heat flux, and k is the thermal conductivity of each layer.

Solving these equations, we find:
q = -2000 W/m²
kA' = kA = 0.2 W/m·K
kB' = kB = 0.1 W/m·K
kC' = kC = 0.15 W/m·K

Now we can use the heat flux and thermal conductivity to calculate the temperatures T1 and T2 in Drawing II:
q = (T1-T2)/Δx
T1 = qΔx + T2

For the left side of the slab:
q = -2000 W/m²
Δx = 0.03 m (thickness of each layer)
T2 = 200 °C (temperature at the left interface)
T1 = -2000*0.03 + 200 = 140 °C

For the right side of the slab:
q = -2000 W/m²
Δx = 0.03 m
T2 = 150 °C (temperature at the right interface)
T1 = -2000*0.03 + 150 = 90 °C

Therefore, the answer is C) T1 = 220 °C and T2 = 160 °C.

A Slab of Insulation Is Made of Three Layers,as Drawing