The outer surface of a grill hood is at 87 °c and the emissivity is 0.93. the heat transfer coefficient for convection between the hood and surroundings at 27 °c is 10 w/m2 /k. determine the net rate of heat transfer between the grill hood and the surroundings by convection and radiation, in kw per m2 of surface area.

The net flux, or rate of heat transfer per area, for convection and radiation are the following:

Convection:
Q/A = hΔT
Radiation:
Q/A = [εσ(T₁⁴ - T₂⁴)]/{1 - [A₁/A₂(1-ε)²]}
where
h is the heat transfer coefficient: 10 W/m²·K or 0.01 kW/m²·K
ε is the emissivity: 0.93
σ is the Stephen-Boltzmann constant: 5.67×10⁻¹¹ kW/m²·K⁴
A₁ = A₂ because they have the same surface area
T is absolute temperature in Kelvin: K = °C + 273

Thus,

Q/A = hΔT + [εσ(T₁⁴ - T₂⁴)]/{1 - [A₁/A₂(1-ε)²]}
Q/A = (0.01 kW/m²·K)(360 K - 300 K) + [0.93*5.67×10⁻¹¹ kW/m²·K⁴*(360⁴ - 300⁴)]/{1 - [1*(1-0.93)²]}
Q/A =  0.6 + 0.461
Q/A = 1.061 kW/m²

Ithink the answer is c but then again i'm probably wrong