Radiation
Contents:
•Basic Concept
•Example
•Whiteboards
Radiation
Hot objects radiate photons (black body radiation)
Heating up a poker
Demo - Show red hot paperclip
Demo – Start silver/black demo
Bonfire/barn fire/stoves
The sun
Frost
Preheat the oven/drying tennis balls
Radiation
Q
4
 eAT
t
ΔQ/
e
σ
A
T
t
= Heat flow rate in J/s or Watts
= emissivity (0 < e < 1) (Temp?)
= 5.67x10-8 Wm-2K-4 (Stefan Boltzmann constant)
= Area (m2)
= High temperature (K)
Net heat transfer
Q
4
4
4
4
 eAT1  eAT2  eA(T1  T2 )
t
(emitted) - (absorbed)
T1 = object
T2 = surroundings
e = e?
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?
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
Silver lined blankets/bags reflect radiation back to you
Demo – check temperatures
on radiation demo
Radiation
1-3
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?
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 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?
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
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?
1000 = (0.95)(6.00 m2)(5.67E-8 Wm-2K-4)(T4-(273.15+18.00 K)4), T = 318.417
T = 45 oC
318 K, 45 oC