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Sagot :
The distance between the glass to have the given heat loss is 2.54 m.
The given parameters:
- dimension of the window, = 60 cm by 90 cm
- temperature, T = 24 K
- heat lost, Q = 4.09 W
- thermal conductivity of glass, k = 0.8 W/mK
The area of the glass window is calculated as follows;
[tex]A = 0.6 \times 0.9\\\\A = 0.54 \ m^2[/tex]
The distance between the glass is calculated as follows;
[tex]Q = \frac{KA \Delta T}{\Delta x} \\\\\Delta x = \frac{kA \Delta T}{Q} \\\\\Delta x = \frac{0.8 \times 0.54 \times 24 }{4.09} \\\\\Delta x = 2.54 \ m[/tex]
Thus, the distance between the glass to have the given heat loss is 2.54 m.
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