Aperson can generate about 300 w of power on a treadmill. if the treadmill is inclined at 3% and a 70-kg man runs at 3 m/s for 45 min, consider the following.(a) calculate the percentage of the power output that goes into heating up his body. %(b) calculate the percentage that keeps him moving on the treadmill. %(c) how much water would that heat evaporate? assume the water starts at body temperature, 37 °c.

(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, = 3% = 0.03

Mass of man, m = 70 kg

Speed, v = 3 m/s

Time taken, t = 45 min = = 2700 s

Now,

Weight of the man, W = mg =

Man's weight component along the inclination of the treadmill, F =

In order to keep moving the power lost by the man in 1 s is:

P' = Fv =

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 = = 79.42%

(b) Percentage power required to keep the man moving:

%P = = 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 =

At , heat required for the evaporation of 1 kg of water:

⇒ Heat required to raise the temperature of 1 kg of water at + Heat required for conversion of 1 kg of water into steam at

⇒

where

= Latent heat of vaporization

Therefore, mass of water evaporated,