Respuesta :
Answer:
(a) 79.42 %
(b) 20.58 %
(c) 0.255 kg
Solution:
As per the question:
Power generated by a person, P = 300 W
Inclination of the treadmill from the horizontal, [tex]sin\theta[/tex] = 3% = 0.03
Mass of man, m = 70 kg
Speed, v = 3 m/s
Time taken, t = 45 min = [tex]45\times 60\ s[/tex] = 2700 s
Now,
Weight of the man, W = mg = [tex]70\times 9.8 = 686\ N[/tex]
Man's weight component along the inclination of the treadmill, F = [tex]mgsin\theta = 686\times 0.03 = 20.58\ N[/tex]
In order to keep moving the power lost by the man in 1 s is:
P' = Fv = [tex]20.58\times 3 = 61.74\ W[/tex]
Therefore, the power that is utilized in raising the body temperature of the man:
P'' = P - P' = 300 - 61.74 = 238.26 W
(a) The percentage of the power that is utilized in heating up the body:
%P = [tex]\frac{P''}{P}\times 100 = \frac{238.26}{300}\times 100[/tex] = 79.42%
(b) Percentage power required to keep the man moving:
%P = [tex]\frac{P'}{P}\times 100 = \frac{61.74}{300}\times 100[/tex] = 20.58%
(c) Water evaporated by the heat equals the product of the power utilized in raising the body temperature of the man and time:
Heat required to evaporate water, Q = [tex]238.26\times 2700 = 643.302\ kJ[/tex]
At [tex]37^{\circ}C[/tex], heat required for the evaporation of 1 kg of water:
⇒ Heat required to raise the temperature of 1 kg of water at [tex]100^{\circ}C[/tex] + Heat required for conversion of 1 kg of water into steam at [tex]100^{\circ}C[/tex]
⇒ [tex]mc\Delta T + mL_{v}= 1\times 4186(100 - 37) + 1\times 2256254 = 2519972\ J[/tex]
where
[tex]L_{v}[/tex] = Latent heat of vaporization
Therefore, mass of water evaporated, [tex]m_{w} = \frac{Q}{2519972} = \frac{643302}{2519972} = 0.255\ kg[/tex]
