Complete Question
The schematics is shown on the first uploaded image
Answer:
For single pane is [tex]Q =1.96 *10^{4}} \ W [/tex]
For double pane is [tex]Q_2 = 120 \ W [/tex]
Explanation:
Generally the heat loss through a single pane is mathematically represented by Fourier’s law as
[tex]Q = k_g * H^2 * \frac{ T_1 - T_2 }{L}[/tex]
=> [tex]Q = 1.4 * 2^2 * \frac{ 15 -(-20) }{ 5 * \frac{1 \ m}{1000 \ mm}}[/tex]
=> [tex]Q =1.96 *10^{4}} \ W [/tex]
Generally the heat loss through a double pane is mathematically represented by Fourier’s law as
[tex]Q_2 = k_a * H^2 * \frac{ T_1 - T_2 }{L}[/tex]
=> [tex]Q_2 = 0.024 * 2^2 * \frac{ 10 -(-15) }{ 10 * \frac{1 \ m}{1000 \ mm}}[/tex]
=> [tex]Q_2 = 120 \ W [/tex]
