RadiationContents:• Basic Concept• Example• Whiteboards

Radiation

TOC

Hot objects radiate photons (black body radiation)Heating up a poker

Demo - Show red hot paperclipDemo – Start silver/black demoBonfire/barn fire/stovesThe sunFrostPreheat the oven/drying tennis balls

Radiation

4ATet

Q

ΔQ/t = Heat flow rate in J/s or Wattse = emissivity (0 < e < 1) (Temp?)

σ = 5.67x10-8 Wm-2K-4 (Stefan Boltzmann constant)

A = Area (m2)T = High temperature (K)

Net heat transfer

(emitted) - (absorbed)T1 = objectT2 = surroundings

e = e?

42

41 ATeATe

t

Q

)( 42

41 TTAe

Q/t = (0.75)(1.72 m2)(5.67E-8 Wm-2K-4)((273.15+33 K)4-(273.15+18 K)4) = 116.97 W

120 W

Example: A person has a surface area of 1.72 m2 and a skin temperature of 33oC. They are standing in a room whose walls are at 18 oC, and their skin has an emissivity of 0.75. What is the net rate of radiative heat flow out of their body?

Demo – check temperatures on radiation demo

Silver lined blankets/bags reflect radiation back to you

Radiation1-3

Q/t = (0.92)(0.0096 m2)(5.67E-8 Wm-2K-4)((273.15 + 20.50)4-(273.15+20.00 K)4) = 0.025296… W

0.025 W, 25 mW

A power converter is a cube 4.0 cm on a side, and is made of black plastic with an emissivity of 0.92. It maintains a temperature of 20.50 oC in a room whose walls are at 20.00 oC. What is its net rate of radiative heat transfer?

34 = e(4(0.03 m)2)(5.67E-8 Wm-2K-4)((273.15 + 250.0)4-(273.15+19.0 K)4)

0.78

A 3.0 cm radius (sperical) light bulb radiates 34 Watts of energy. If it has a surface temperature of 250.0 oC in a room whose walls are at 19.0 oC, what is its emissivity?

1000 = (0.95)(6.00 m2)(5.67E-8 Wm-2K-4)(T4-(273.15+18.00 K)4), T = 318.417T = 45 oC

318 K, 45 oC

If you put a 1000. Watt hair dryer into a 1.00 m cube box with an emissivity of 0.95, it would heat up the box’s surface until energy flowed out (assume radiatively) as fast as it flowed in. What would be the surface temperature of the box in Celsius if the room it was in had walls at 18.0 oC?