1. The problem statement, all variables and given/known data A wooden box is painted completely with black paint. The interior dimensions are 18”x24”x3.5”. The bottom and sides are made with 9 pieces of 2x4 douglas-fir-larch dimension wood lumber. The top of the box is open and is covered with 0.118” thick clear acrylic. The acrylic is sealed to the box and only allows light to enter the box. The air surrounding the box has negligible velocity. The box and the ambient air temperature are 47 degrees F. At time, t = 0, t a 500 W light bulb that is suspended 6 feet above the box is turned on. Thermal equilibrium of the air and box occurs 10 minutes later. What is the temperature of the air inside the box at this time? Black Paint ε = 0.80 500 W light ε = 0.95 T light = 3000 K CWood 1700 J/kg*K CAir 1009 J/kg*K ρ air = 1.225 kg/m3 ρ wood = 530 kg/m3 σ = 5.670373E -8 J/S*m2*K4 2. Relevant equations E = A ε σ T4 Q = m c ΔT 3. The attempt at a solution To heat the Air: Vair = 18x24*3.5 in3 = 0.02477 m3 ma = ρ airVair = 0.0304 kg Qair = ma cair ΔT To heat the Wood: Vwood = 3.5x1.5x 24 in3 = 0.0021 m3 mw = ρ woodVwood = 1.094 kg Qwood = mw cwood ΔT To heat the wood/air: Qtotal = ma cair ΔT + mw cwood ΔT Energy from 500 W light: Distribute the 500 W over a sphere with a 6 ft (1.8288 m) radius, A = 4π r2 = 42.028 m2 500 W/ 42.028 m2 = 11.9 w/m2 (at the box from the light) The box is 18x24 in 2 (0.2787 m2) so it receives 3.317 W on the surface at that distance. 3.317 W applied for 10 minutes (600 seconds) = 1990 J applied to the air and box. 1990 J = ma cair ΔT + mw cwood ΔT solve for ΔT = 1.05 K ?