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Physics, Astronomy & Cosmology / Re: How many gamma rays per second would 15 million degree celsius object emit?
« on: 07/05/2019 00:21:58 »
To calculate the power radiated by a black body with a known temperature, emissivity and surface area, you can use the following equation:
P = AεσT4, where
"P" is the resulting power in watts
"A" is the surface area of the black body in square meters
"ε" is the emissivity of the material that the black body is composed of (a number between 0 and 1)
"σ" is the Stefan-Boltzmann constant, which is 5.670373 x 10-8 W m-2 K-4, and
"T" is the temperature of the black body in kelvins
Let's consider a sphere of anodized aluminum (which has an emissivity of 0.9) that has a radius of 0.05 meters and a temperature of 900 kelvins. Using the equation A = 4πr2, we determine that a sphere with a radius of 0.05 meters has a surface area of 0.031 square meters:
P = AεσT4
P = (0.031)(0.9)(5.670373 x 10-8)(9004)
P = (0.031)(0.9)(5.670373 x 10-8)(656,100,000,000)
P = (0.0279)((5.670373 x 10-8)(656,100,000,000)
P = (0.000000001582034067)(656,100,000,000)
P = 1,037.97 watts
If you want to know the peak wavelength of the radiation emitted by a black body, you can use this equation:
λmax = b/T, where
"λmax" is the peak wavelength in meters
"b" is Wien's displacement constant, which is 2.8977729 x 10−3 m K, and
"T" is the temperature in kelvins
So to find the peak wavelength that the aluminum sphere emits at, we enter the values into the equation:
λmax = b/T
λmax = (2.8977729 x 10−3)/(900)
λmax = 3.2197477 x 10-6 meters (or 3.2197477 micrometers, which is in the infrared)
P = AεσT4, where
"P" is the resulting power in watts
"A" is the surface area of the black body in square meters
"ε" is the emissivity of the material that the black body is composed of (a number between 0 and 1)
"σ" is the Stefan-Boltzmann constant, which is 5.670373 x 10-8 W m-2 K-4, and
"T" is the temperature of the black body in kelvins
Let's consider a sphere of anodized aluminum (which has an emissivity of 0.9) that has a radius of 0.05 meters and a temperature of 900 kelvins. Using the equation A = 4πr2, we determine that a sphere with a radius of 0.05 meters has a surface area of 0.031 square meters:
P = AεσT4
P = (0.031)(0.9)(5.670373 x 10-8)(9004)
P = (0.031)(0.9)(5.670373 x 10-8)(656,100,000,000)
P = (0.0279)((5.670373 x 10-8)(656,100,000,000)
P = (0.000000001582034067)(656,100,000,000)
P = 1,037.97 watts
If you want to know the peak wavelength of the radiation emitted by a black body, you can use this equation:
λmax = b/T, where
"λmax" is the peak wavelength in meters
"b" is Wien's displacement constant, which is 2.8977729 x 10−3 m K, and
"T" is the temperature in kelvins
So to find the peak wavelength that the aluminum sphere emits at, we enter the values into the equation:
λmax = b/T
λmax = (2.8977729 x 10−3)/(900)
λmax = 3.2197477 x 10-6 meters (or 3.2197477 micrometers, which is in the infrared)
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