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 20 cm square. The outer face of this square composite sheet is maintained at ${{100}^{\circ }}C$and ${{20}^{\circ }}C$, 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: $6.3\times {{10}^{-2}}W{{m}^{-1}}{{K}^{-1}}$
Thermal conductivity of glass: $7.2\times {{10}^{-2}}W{{m}^{-1}}{{K}^{-1}}$

Answer 1

Hint: Assume that temperature of glass-cork interface is ${{T}^{\circ }}C$. As we know that, heat current across a layer is given by: $Q=\dfrac{KA}{l}\Delta T$, where Q is heat current, k is thermal conductivity, A is the area, l is the length and $\Delta T$ 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: $6.3\times {{10}^{-2}}W{{m}^{-1}}{{K}^{-1}}$
Thermal conductivity of glass: $7.2\times {{10}^{-2}}W{{m}^{-1}}{{K}^{-1}}$
Length of cork: 3 mm $=3\times {{10}^{-3}}m$
Length of glass: 5 mm $=5\times {{10}^{-3}}m$
Area of glass = Area of cork = $20c{{m}^{2}}=20\times {{10}^{-4}}{{m}^{2}}$
So, by using the formula for heat rate: $Q=\dfrac{KA}{l}\Delta T$, we get:
For cork sheet:
\[{{Q}_{c}}=\dfrac{6.3\times {{10}^{-2}}\times 20\times {{10}^{-4}}}{3\times {{10}^{-3}}}\left( 100-T \right)......(1)\]
For glass sheet:
${{Q}_{g}}=\dfrac{7.2\times {{10}^{-2}}\times 20\times {{10}^{-4}}}{5\times {{10}^{-3}}}\left( T-20 \right)......(2)$
Since, heat released by the cork sheet is equal to heat gained by glass sheet, we get:
\[\begin{align}
  & {{Q}_{c}}={{Q}_{g}} \\
 & \dfrac{6.3\times {{10}^{-2}}\times 20\times {{10}^{-4}}}{3\times {{10}^{-3}}}\left( 100-T \right)=\dfrac{7.2\times {{10}^{-2}}\times 20\times {{10}^{-4}}}{5\times {{10}^{-3}}}\left( T-20 \right) \\
 & 42\times {{10}^{-3}}\left( 100-T \right)={{28.810}^{-2}}\left( T-20 \right) \\
 & 1.146T=100+2.92 \\
 & T\approx {{90}^{\circ }}C
\end{align}\]
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: \[\begin{align}
  & Q=\dfrac{6.3\times {{10}^{-2}}\times 20\times {{10}^{-4}}}{3\times {{10}^{-3}}}\left( 100-90 \right)......(1) \\
 & =2940\times {{10}^{-3}} \\
 & =2.94J{{s}^{-1}}
\end{align}\]

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.