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Grade: 12
        A metal ball of surface area 200 sq. cm and temp 527 degree celcius is surrounded by a vesses at 27 degree  celcius . If the emmisivity of the metal is 0.4, then the rate of loss of heat from the ball is 
10 years ago

Answers : (1)

AskIITians Expert Hari Shankar IITD
17 Points
							

By the stefan boltzmann law,

Rate of heat loss from the ball will be

E = seA(T4 -Ts4)

 

where s =Stefan-Boltzmann constant = 5.67 e-8 units (SI units, I dont remember the units exactly, but you should always right the correct units)

e = emissivity = 0.4

A = 200 cm2 = 200 * 10-6 m2 = 2e-4 m2

T = Temp. of ball = 527 C = 527+273 K = 800 K

Ts = Temp of surroundings = 27 C = 300 K

 

Putting in all the values,

E = (5.67e-8)(0.4)(2e-4)(8004-3004) = 1.85 J/s

 

 

10 years ago
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