Question

In this problem you will consider the balance of thermal energy radiated and absorbed by a person.Assume that the person is wearing only a skimpy bathing suit of negligible area. As a rough approximation, the area of a human body may be considered to be that of the sides of a cylinder of length L=2.0m and circumference C=0.8m.For the Stefan-Boltzmann constant use σ=5.67×10−8W/m2/K4.Part aIf the surface temperature of the skin is taken to be Tbody=30∘C, how much thermal power Prb does the body described in the introduction radiate?Take the emissivity to be e=0.6.Express the power radiated into the room by the body numerically, rounded to the nearest 10 W.part bFind Pnet, the net power radiated by the person when in a room with temperature Troom=20∘C

Answer 1

Answer:

The thermal power emitted by the body is $P_t = 286.8 \ Wm^{-2}$

The net power radiated is  $P_{net} = 460 \ W$

Explanation:

From the question we are told that

   The length of the assumed hum$T_{room} = 20 ^oC$an body is  L =  2.0 m

   The circumference of the assumed human body is  $C = 0.8 \ m$

   The  Stefan-Boltzmann constant is  $\sigma = 5.67 * 10^{-8 } \ W\cdot m^{-2} \cdot K^{-4}.$

    The temperature of skin $T_{body} = 30^oC$

     The temperature of the room is  

    The emissivity is  e=0.6

The thermal power radiated by the body is mathematically represented as

           $P_t = e * \sigma * T_{body}^4$

substituting value

        $P_t = 0.6 * 5.67*10^{-8} * (303)^4$

        $P_t = 286.8 \ Wm^{-2}$

The net power radiated by the body is mathematically evaluated as

    $P_{net} = P_t * A$

Where A is the surface area of the body which is mathematically evaluated as

     $A = C* L$

substituting values

      $A = 0.8 * 2$

      $A = 1.6 m^2$

=>    $P_{net} = 286.8 * 1.6$

=>   $P_{net} = 460 \ W$