Question

Radiation of Energy 

The rate of heat transfer by emitted radiation is determined by the Stefan-Boltzmann law of radiation: = aeAT4 where o 5.67x10-8 J/s - m2 K is the Stefan-Boltzmann constant, A is the surface area of the object, and T is its absolute temperature in kelvin. The symbol e stands for the emissivity of the object, which is a measure of how well it radiates An ideal jet-black (or black body) radiator has e 1,whereas a perfect reflector has e 0. Real objects fall between these two values. Take, for example, tungsten light bulb filaments which have an e of about 0.5, and carbon black (a material used in printer toner), which has the (greatest known) emissivity of about 0.99 If an object is surrounded by an environment with uniform temperature T the object absorbs energy from the encironment. The rate of heat transfer by absorbed radiation is: geATSar Absorb Radinte Inciddent radiant energy Emitted Reflected Retained Absorbed Black Black Incident iant eneroy Reflected Emittod Retaineed Absorbmd Silver coated Silver oated A person with a surface area of 1.20 m2, and a skin temperature of 27 °C, is in a room that is at a temperature of 16.6 °C. The emissivity of the skin is 0.895. The Stefan-Boltzmann constant is 5.67 x 108 W/(m2-K4). 

(a) How much power (power is the energy per second) is radiated by the person? 

(b) How much power is absorbed by the person from the surroundings?

(c) What is his NET Power loss? Report as a Positive value.

Answer 1

@) power radiated by the penson Ppersona re At's 5.67 x10°x 0. 895 X 1.2x13003 492.48 watt 'penson = 6.08 X 10 X 8 loooooo00o = b) Pronounding = 6.08810X (2896) = 6.08 870.33 = 427.6 watt I e) Perea(7-74) P = 492.48- 427.6 = 64.88 Wate net