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

Question

AskIITians Expert Hari Shankar IITD · Answer

By the stefan boltzmann law,

Rate of heat loss from the ball will be

E = seA(T⁴ -T_s4)

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

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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

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

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