Answer:
a)  ≈ 11.7 watts
b)  ≈ 21.4 watts
Explanation:
Given data :
Oven door dimensions : 0.6m  by 0.7m
average surface temperature = 32°c
Temperature of ambient air = 22°c
emissivity of door = 0.4
surrounding temperature = 22°c
a) Calculate heat transfer by free convection
q = 3.34 ( 0.5 * 0.7 ) (32-22 ) ≈ 11.7 watts
attached below is the detailed solution
b) calculate heat transfer by radiation
q = ∈.As б ( Ts^4 - Tsun^4 )
  = 1( 0.5*0.7 ) * 5.67 * 10^-8 [( 273+32)^4 - ( 273+22)^4 ]
  ≈ 21.4 watts
