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Question
One face of the sheet of cork, 3mm thick is placed in contact with one face of a sheet of glass 5mm thick, both sheets being 20cm square. The outer face of this square composite sheet are maintained at 100oC and 20oC, the glass being at the higher mean temperature. Find: (i) The temperature of glass-cork interface and (ii) The rate at which heat is conducted across the sheet neglecting edge effects.
Thermal conductivity of cork=6.3×10−2Wm−1K−1
Thermal conductivity of glass =7.2×10−1Wm−1K−1
Thermal conductivity of cork=6.3×10−2Wm−1K−1
Thermal conductivity of glass =7.2×10−1Wm−1K−1
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Solution
Given: One face of the sheet of cork, 3mm thick is placed in contact with one face of a sheet of glass 5mm thick, both sheets being 20cm square. The outer face of this square composite sheet are maintained at 100oC and 20oC, the glass being at the higher mean temperature. To find: (i) The temperature of glass-cork interface and (ii) The rate at which heat is conducted across the sheet neglecting edge effects.Solution:As per the given criteria,Thermal conductivity of cork Kc=6.3×10−2Wm−1K−1Thermal conductivity of glass =7.2×10−1Wm−1K−1Area of the cork and glass sheet, Ac=Ag=20cm2=20×10−4m2Thickness of cork sheet, lc=3mm=3×10−3mThickness of glass sheet, lg=5mm=5×10−3mThere are two layers in series,So heat current will be same across two layers.Set the temperature at interface T.Heat current across layer A=KAΔTlWhere, K = Thermal conductivityand A = AreaΔT is Temperature differencel = Length along heat traveledHeat layer across layer A(cork) at 20∘C=KcAcΔTlc⟹=6.3×10−2×20×10−4×(T−20)3×10−3=42×10−3×(T−20).................(i)Heat current across layer B(glass) at 100∘C=KgAgΔTlg⟹=7.2×10−1×20×10−4(100−T)5×10−3=28.8×10−2×(100−T)..............(ii)At steady state eqn(i) and eqn(ii) should be equal, we get42×10−3×(T−20)=28.8×10−2×(100−T)⟹0.146(T−20)=(100−T)⟹1.146T=100+2.92⟹T≈90∘Cis the temperature of glass-cork interfaceNow put T in any one of the equation (i), we getH=42×10−3(90−20)⟹H=2940×10−3⟹H=2.94J/sis the rate at which heat is conducted across the sheet neglecting edge effects.
Given: One face of the sheet of cork, 3mm thick is placed in contact with one face of a sheet of glass 5mm thick, both sheets being 20cm square. The outer face of this square composite sheet are maintained at 100oC and 20oC, the glass being at the higher mean temperature.
To find: (i) The temperature of glass-cork interface and (ii) The rate at which heat is conducted across the sheet neglecting edge effects.
Solution:
As per the given criteria,
Thermal conductivity of cork Kc=6.3×10−2Wm−1K−1
Thermal conductivity of glass =7.2×10−1Wm−1K−1
Area of the cork and glass sheet, Ac=Ag=20cm2=20×10−4m2
Thickness of cork sheet, lc=3mm=3×10−3m
Thickness of glass sheet, lg=5mm=5×10−3m
There are two layers in series,
So heat current will be same across two layers.
Set the temperature at interface T.
Heat current across layer A=KAΔTl
Where, K = Thermal conductivity
and A = Area
ΔT is Temperature difference
l = Length along heat traveled
Heat layer across layer A(cork) at 20∘C=KcAcΔTlc
⟹=6.3×10−2×20×10−4×(T−20)3×10−3=42×10−3×(T−20).................(i)
Heat current across layer B(glass) at 100∘C=KgAgΔTlg
⟹=7.2×10−1×20×10−4(100−T)5×10−3=28.8×10−2×(100−T)..............(ii)
At steady state eqn(i) and eqn(ii) should be equal, we get
42×10−3×(T−20)=28.8×10−2×(100−T)⟹0.146(T−20)=(100−T)⟹1.146T=100+2.92⟹T≈90∘C
is the temperature of glass-cork interface
Now put T in any one of the equation (i), we get
H=42×10−3(90−20)⟹H=2940×10−3⟹H=2.94J/s
is the rate at which heat is conducted across the sheet neglecting edge effects.
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