To solve this problem we must consider the expressions of Stefan Boltzmann's law for which the rate of change of the radiation of energy H from a surface must be

$H = Ae\sigma T^4$

Where

A = Surface area

e = Emissivity that characterizes the emitting properties of the surface

$\sigma$ = Universal constant called the Stefan-Boltzmann constant $(5.67*10^{-8}m^{-2}K^{-4})$

T = Absolute temperature

The total heat loss would be then

$Q = H_2 -H_1$

$Q =Ae\sigma T_2^4-Ae\sigma T_1^4$

$Q = Ae\sigma (T_2^4-T_1^4)$

$Q = (1.2)(1)(5.67*10^{-8})(303^4-280^4)$

$Q = 155.29J$

Therefore the net rate of heat loss from the body by radiation is 155.29J

8 0

2 years ago

What is free fall? Explain it in brief.

xxTIMURxx [149]

· free fall is any motion of a body where gravity is the only forceacting upon it. In the context of general relativity, where gravitation is reduced to a space-time curvature, a body in free fall has no force acting on it.

3 0

3 years ago