contestada

Following vigorous exercise, the body temperature of an 79-kg person is 41∘C. At what rate in watts must the person transfer thermal energy to reduce the body temperature to 36∘C in 26 min, assuming the body continues to produce energy at the rate of 135 W?

Respuesta :

Answer:

The person must transfer energy at the rate of 1021.21 W

Explanation:

Given:

mass of the man, m = 79 klg

time, t = 26 minute

          = 26 x 60

           =1560 sec

Power produced by body = 135 W

Change in temperature, ΔT = 41 - 36

                                              = 5°C

Let specific heat of man, [tex]c_{man}[/tex] =3500 J/kg °C

Therefore,

Heat transfer, Q = m.c.ΔT

                           = 79 x 3500 x 5

                            =1382500 J

Therefore, [tex]P_{cooling}[/tex] = Q / t

                                                     =  1382500 / 1560

                                                      = 886.21 W

Therefore, [tex]P_{requred}[/tex] = [tex]P_{cooling}[/tex] + power produced by body

                                                     = 886.21 + 135

                                                     = 1021.21 W

Thus, the person must transfer energy at the rate of 1021.21 W