Answer:
[tex]\frac{\dot Q}{A} = 151.33 W/m^2[/tex]
[tex]T_c = 25.153°C[/tex]
Explanation:
Given data:
one side wall temperature 200°C
other side wall temperature 10°C
h = 100 W/m^2 °C
k = 2.6W/m °C
wall thickness L = 30 cm
we know that heat flux is given as
[tex]\frac{\dot Q}{A} = \frac{ T_A - T\infty}{\frac{L}{K} + \frac{1}{h}}[/tex]
[tex]\frac{\dot Q}{A} = \frac{ 20- 10} {\frac{0.30}{2.6} + \frac{1}{100}}[/tex]
[tex]\frac{\dot Q}{A} = 151.33 W/m^2[/tex]
[tex]1515.33 W/m^2 = \frac{ T_A - T_c}{\frac{L}{K}}[/tex]
solving for temperature for cold surface is given as
[tex]T_c = -1515.33 \times \frac{0.3}{2.6} + 200[/tex]
[tex]T_c = 25.153°C[/tex]
