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A concrete wall has the dimensions 3 m × 5.5 m and a depth \((d)\) of 0.25 metres.
Calculate the heat energy that passes through the wall if the outside temperature is 5°C and the temperature inside is 26°C across a 5 minute period. \((k= 2.25\ \text{W m}^{-1}\ K^{-1}) \) (3 marks)
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\(935\ 550\ \text{J}\)
| \(\dfrac{Q}{t}\) | \(=\dfrac{kA(T_{hot}-T_{cold})}{d}\) | |
| \(Q\) | \(=\dfrac{kA(T_{hot}-T_{cold})}{d} \times t\) | |
| \(=\dfrac{2.25 \times 3 \times 5.5 \times (26-5)}{0.25} \times (5 \times 60)\) | ||
| \(=935\ 550\ \text{J}\) |
A glass panel has thickness \(d\) mm and a temperature difference between its two sides of \(\Delta T\).
If the surface area of a glass panel was increased from \(A\) m² to \(4A\) m², what two changes could be made to the glass panel to ensure the rate of change of heat energy through the glass remains the same. (3 marks)
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Thermal conductivity equation \(\ \Rightarrow \dfrac{Q}{t}=\dfrac{kA\Delta T}{d}\).
Given \(A\) increases to \(4A\):
→ \(\dfrac{Q}{t} \) remains the same if the distance through which the heat travels through the glass increases from \(d\) to \(4d\).
→ \(\dfrac{Q}{t} \) remains the same if the temperature difference between the two sides of the glass panel decreases from \(\Delta T\) to \(\dfrac{\Delta T}{4}\).
Thermal conductivity equation \(\ \Rightarrow \dfrac{Q}{t}=\dfrac{kA\Delta T}{d}\).
Given \(A\) increases to \(4A\):
→ \(\dfrac{Q}{t} \) remains the same if the distance through which the heat travels through the glass increases from \(d\) to \(4d\).
→ \(\dfrac{Q}{t} \) remains the same if the temperature difference between the two sides of the glass panel decreases from \(\Delta T\) to \(\dfrac{\Delta T}{4}\).