Solveeit Logo
Login

Question

Question: One face of the sheet of cork, 3mm thick is placed in contact with one face of a sheet of glass 5mm ...

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 20 cm square. The outer face of this square composite sheet is maintained at 100∘C{{100}^{\circ }}Cand 20∘C{{20}^{\circ }}C, 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−16.3\times {{10}^{-2}}W{{m}^{-1}}{{K}^{-1}}
Thermal conductivity of glass: 7.2×10−2Wm−1K−17.2\times {{10}^{-2}}W{{m}^{-1}}{{K}^{-1}}

Explanation

Solution

Assume that temperature of glass-cork interface is T∘C{{T}^{\circ }}C. As we know that, heat current across a layer is given by: Q=KAlΔTQ=\dfrac{KA}{l}\Delta T, where Q is heat current, k is thermal conductivity, A is the area, l is the length and ΔT\Delta T is the temperature difference. Find the heat current for both the sheets in terms of T and then equate both the equations to get the value of the temperature of the glass-cork interface. Later, substitute the value of T in any of the equations to get the rate of heat conducted.

Complete step by step answer:
We have the following setup of glass and cork sheet:

We have:
Thermal conductivity of cork: 6.3×10−2Wm−1K−16.3\times {{10}^{-2}}W{{m}^{-1}}{{K}^{-1}}
Thermal conductivity of glass: 7.2×10−2Wm−1K−17.2\times {{10}^{-2}}W{{m}^{-1}}{{K}^{-1}}
Length of cork: 3 mm =3×10−3m=3\times {{10}^{-3}}m
Length of glass: 5 mm =5×10−3m=5\times {{10}^{-3}}m
Area of glass = Area of cork = 20cm2=20×10−4m220c{{m}^{2}}=20\times {{10}^{-4}}{{m}^{2}}
So, by using the formula for heat rate: Q=KAlΔTQ=\dfrac{KA}{l}\Delta T, we get:
For cork sheet:
Qc=6.3×10−2×20×10−43×10−3(100−T)......(1){{Q}_{c}}=\dfrac{6.3\times {{10}^{-2}}\times 20\times {{10}^{-4}}}{3\times {{10}^{-3}}}\left( 100-T \right)......(1)
For glass sheet:
Qg=7.2×10−2×20×10−45×10−3(T−20)......(2){{Q}_{g}}=\dfrac{7.2\times {{10}^{-2}}\times 20\times {{10}^{-4}}}{5\times {{10}^{-3}}}\left( T-20 \right)......(2)
Since, heat released by the cork sheet is equal to heat gained by glass sheet, we get:

& {{Q}_{c}}={{Q}_{g}} \\\ & \dfrac{6.3\times {{10}^{-2}}\times 20\times {{10}^{-4}}}{3\times {{10}^{-3}}}\left( 100-T \right)=\dfrac{7.2\times {{10}^{-2}}\times 20\times {{10}^{-4}}}{5\times {{10}^{-3}}}\left( T-20 \right) \\\ & 42\times {{10}^{-3}}\left( 100-T \right)={{28.810}^{-2}}\left( T-20 \right) \\\ & 1.146T=100+2.92 \\\ & T\approx {{90}^{\circ }}C \end{aligned}$$ Hence, the temperature of glass-cork interface is $$T\approx {{90}^{\circ }}C$$ Now, put the value of T in any of the equation (say equation (1)), we get: $$\begin{aligned} & Q=\dfrac{6.3\times {{10}^{-2}}\times 20\times {{10}^{-4}}}{3\times {{10}^{-3}}}\left( 100-90 \right)......(1) \\\ & =2940\times {{10}^{-3}} \\\ & =2.94J{{s}^{-1}} \end{aligned}$$ **Note:** Thermal conductivity refers to the ability of a given material to conduct/transfer heat. It is generally denoted by the symbol 'k'. The reciprocal of this quantity is known as thermal resistivity.