Question

A double-pane window used for insulating a room thermally from outside consists of two glass sheets each of area 1\[{{m}^{2}}\]and thickness 0.01m separated by 0.05m thick stagnant air space. In the steady state, the room-glass interface and the glass-outdoor interface are at constant temperatures of 270 \[{}^\circ C\] and 0\[{}^\circ C\] respectively. The thermal conductivity of glass is 0.8 and of air 0.08W/mK. The rate of flow of heat through the window pane is nearly equal to
A. 1000J/s
B. 1224 J/s
C. 3000J/s
D. 4000J/s

Answer 1

Hint: In steady state cases the flow of heat takes place at a constant pace. If a system is thermally isolated then it is not possible for heat to move from outside into the system and vice versa.

Complete step by step answer:
Given a double-pane window consisting of two glass sheets each of area 1\[{{m}^{2}}\]and thickness 0.01m separated by 0.05m thick stagnant air space. The steady state temperatures are 270 \[{}^\circ C\]and 0\[{}^\circ C\]. The thermal conductivity of glass is 0.8 and of air 0.08W/mK. We need to find the rate of flow of heat through the window pane.
Area of each glass sheet, A=1\[{{m}^{2}}\]
Thermal conductivity of air, \[{{k}_{a}}=0.08W/mK\]
​Thermal conductivity of glass, \[{{k}_{g}}=0.8W/mK\]
Temperature at room-glass interface, \[{{T}_{1}}=270{}^\circ C+273=543K\]
Temperature at glass-outdoor interface, \[{{T}_{2}}=0{}^\circ C+273=273K\]
Thickness of the glass sheet, \[{{d}_{g}}=0.01m\]
Thickness of the air space, \[{{d}_{a}}=0.05m\]
The equation of heat flow is given as: \[\dfrac{d{{Q}_{1}}}{dt}=\dfrac{{{k}_{g}}A({{T}_{1}}-{{T}_{2}})}{{{d}_{g}}}\]
substituting value, we get,
\[\dfrac{d{{Q}_{1}}}{dt}=\dfrac{0.8\times 1\times (543-{{T}_{2}})}{0.01}\]-----------(1)
Similarly, at second interface \[\dfrac{d{{Q}_{2}}}{dt}=\dfrac{0.08\times 1\times ({{T}_{2}}-{{T}_{3}})}{{{d}_{g}}}\]--------(2)
Similarly, at third interface \[\dfrac{d{{Q}_{3}}}{dt}=\dfrac{{{k}_{g}}A({{T}_{3}}-{{T}_{4)}}}{{{d}_{g}}}\]
Substituting the values, \[\dfrac{d{{Q}_{3}}}{dt}=\dfrac{0.8\times 1\times ({{T}_{3}}-273)}{0.01}\]---------(3)
But in steady state \[\dfrac{d{{Q}_{1}}}{dt}=\dfrac{d{{Q}_{2}}}{dt}=\dfrac{d{{Q}_{3}}}{dt}\]
Therefore, equating (1) and (2) we get,
\[\begin{align}
  & \dfrac{0.8\times 1\times (543-{{T}_{2}})}{0.01}=\dfrac{0.08\times 1\times ({{T}_{2}}-{{T}_{3}})}{0.05} \\
 & \Rightarrow 0.05\{0.8\times (543-{{T}_{2}})\}=0.01\{0.08\times ({{T}_{2}}-{{T}_{3}})\} \\
 & \Rightarrow 0.04(543-{{T}_{2}})=0.008({{T}_{2}}-{{T}_{3}}) \\
 & \Rightarrow 543-{{T}_{2}}=0.02{{T}_{2}}-0.02{{T}_{3}} \\
 & \Rightarrow 1.02{{T}_{2}}-0.02{{T}_{3}}=543------(4) \\
\end{align}\]
Now equating (2) and (3) we get
\[\begin{align}
  & \Rightarrow \dfrac{0.8\times 1\times ({{T}_{2}}-{{T}_{3}})}{0.05}=\dfrac{0.8\times 1\times ({{T}_{3}}-273)}{0.01} \\
 & \Rightarrow 0.01\{0.08\times ({{T}_{2}}-{{T}_{3}})\}=0.05\{0.8\times ({{T}_{3}}-273)\} \\
 & \Rightarrow 0.0008({{T}_{2}}-{{T}_{3}})=0.04({{T}_{3}}-273) \\
 & \Rightarrow 0.02{{T}_{2}}-0.02{{T}_{3}}={{T}_{3}}-273 \\
 & \Rightarrow 0.02{{T}_{2}}-1.02{{T}_{3}}=273------(5) \\
 & \\
\end{align}\]
Multiply (4) with 0.02 and (5) with 1.02 and by subtracting them, we get
\[\begin{align}
  & \Rightarrow 0.0204{{T}_{2}}-0.0004{{T}_{3}}=10.86 \\
 & \Rightarrow 1.04{{T}_{3}}=-267.6 \\
 & \Rightarrow {{T}_{3}}=-257.3K \\
\end{align}\]
Substituting the value of \[T\] in (5), we get
\[\begin{align}
  & \Rightarrow 0.02{{T}_{2}}-1.02(-257.3)=273 \\
 & \Rightarrow 0.02{{T}_{2}}=273-262.446 \\
 & \Rightarrow {{T}_{2}}=527.7K \\
\end{align}\]
Thus, rate of flow of the heat through the window pane is given by
\[\begin{align}
  & \Rightarrow \dfrac{d{{Q}_{1}}}{dt}=\dfrac{0.8\times 1\times (543-{{T}_{2}})}{0.01} \\
 & \Rightarrow \dfrac{d{{Q}_{1}}}{dt}=\dfrac{0.8\times 1\times (543-527.7)}{0.01} \\
 & \Rightarrow \dfrac{d{{Q}_{1}}}{dt}=1224W \\
\end{align}\]
Also 1W= 1J/s
So \[\dfrac{d{{Q}_{1}}}{dt}=1224J/s\]

So, the correct answer is “Option B”.

Note:
While solving such kinds of problems we have to keep in mind that all temperature readings have to be put in Kelvin. But if there is change in temperature then there is no issue of converting the given unit into Kelvin. Because the difference always remains the same in any units.