Answer:
Q=116.37 W
Explanation:
Given that
Emissivity ,ε= 0.200
Surface temperature ,T₁ = 10⁰ C = 283 K
Surrounding T₂ = - 15⁰ C = 258 K
Area ,A= 1.6 m²
The net heat transfer given as
[tex]Q=\sigma A \varepsilon (T_1^4-T_2^2)[/tex]
σ = 5.67 x 10⁻⁸
The temperature should be in Kelvin.
Now by putting the values
[tex]Q=5.67\times 10^{-8}\times 1.6\times 0.2\times (283^4-258^2)\ W[/tex]
Q=116.37 W
Therefore answer will be 116.37 W.
