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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
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
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
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