Question

A sphere of radius 0.348 m, temperature 47.1°C, and emissivity 0.704 is located in an environment of temperature 71.6°C. At what rate does the sphere (a) emit and (b) absorb thermal radiation? (c) What is the sphere's net rate of energy exchange?

Answer 1

given :

R = radius = 0.348 m

T = 47.1oC = 320.1 K

Ts = 71.6oC = 344.6 K

e = emissivity = 0.704

a = coefficien of absorption = e = 0.704

a) Rate of heat emission = [ = Stefan's constant, A = surface area = ]

= 0.704 x 5.67 x 10-8 x 4 x x 0.3482 x 320.14

= 637.77 J/s [answer]

b) Rate of Heat radiation absorbed =

= 0.704 x 5.67 x 10-8 x 4 x x 0.3482 x 344.64

= 856.61 J/s [answer]

c) Net Rate of Energy Exchange = Rate of Heat radiation absorbed - Rate of heat emission [absorption rate is greater than emission rate]

= 856.61 - 637.77 = 218.84 J/s [answer]