Question

find the rate of energy radiated by a man by assuming the surface area of his body 1.7m²and emissivity of his body 0.4​

Answer 1

The rate of energy radiated by the man is 3.86 x $10^{-8}$  J/s. $m^{2}$.

The amount of energy radiated by an object majorly depends on the area of its surface and its temperature. The is well explained in the Stefan-Boltzmann's law which states that:

Q(t) = Aeσ$T^{4}$

where: Q is the quantity of heat radiated, A is the surface area of the object, e is the emmisivity of the object, σ is the Stefan-Boltzmann constant and T is the temperature of the object.

To determine the rate of energy radiated by the man in the given question;

$\frac{Q(t)}{T^{4} }$ = Aeσ

But A = 1.7 m², e = 0.4 and σ = 5.67 x $10^{-8}$ J/s.

So that;

$\frac{Q(t)}{T^{4} }$ = 1.7 * 0.4 * 5.67 x $10^{-8}$

     = 3.8556 x $10^{-8}$

     = 3.86 x $10^{-8}$  J/s. $m^{2}$

Thus, the rate of energy radiated by the man is 3.86 x $10^{-8}$  J/s. $m^{2}$.

Learn more on energy radiation of objects by visiting: https://brainly.com/question/12550129