If you quadruple the temperature of a black body, by what factor will the total energy radiated per second per square meter increase? 1,024 64 16 4 256 submit answer

Respuesta :

In modern physics, as it was called "Stefan-Boltzmann law", the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of the black body's temperature T

as:

[tex]P\alpha T^4[/tex]

where: P is the power (total energy radiated per second per square meter) and T is the temperature of a black body.

then we can make a ratio between the state of before quadruple (with subscript 1) and after (with subscript 2) as:

[tex]\frac{P_{1} }{P_{2} } =\frac{T_{1}^4 }{T_{2}^4}[/tex]

As

[tex]T_{2}=4T_{1}[/tex]

Then

[tex]\frac{P_{1} }{P_{2} } =\frac{T_{1}^4 }{(4T_{1})^4}[/tex]

then

[tex]P_{2}=256*P_{1} [/tex]

  • The factor will the total energy radiated per second per square meter increase = 256