Question

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

Answer 1

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:

$P\alpha T^4$

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:

$\frac{P_{1} }{P_{2} } =\frac{T_{1}^4 }{T_{2}^4}$

As

$T_{2}=4T_{1}$

Then

$\frac{P_{1} }{P_{2} } =\frac{T_{1}^4 }{(4T_{1})^4}$

then

$P_{2}=256*P_{1}$

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