Question

A person 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. _________kg

Answer 1

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, $sin\theta$ = 3% = 0.03

Mass of man, m = 70 kg

Speed, v = 3 m/s

Time taken, t = 45 min = $45\times 60\ s$ = 2700 s

Now,

Weight of the man, W = mg = $70\times 9.8 = 686\ N$

Man's weight component along the inclination of the treadmill, F = $mgsin\theta = 686\times 0.03 = 20.58\ N$

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

P' = Fv = $20.58\times 3 = 61.74\ W$

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 = $\frac{P''}{P}\times 100 = \frac{238.26}{300}\times 100$ = 79.42%

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

%P = $\frac{P'}{P}\times 100 = \frac{61.74}{300}\times 100$ = 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 = $238.26\times 2700 = 643.302\ kJ$

At $37^{\circ}C$, heat required for the evaporation of 1 kg of water:

⇒ Heat required to raise the temperature of 1 kg of water at $100^{\circ}C$ + Heat required for conversion of 1 kg of water into steam at $100^{\circ}C$

⇒ $mc\Delta T + mL_{v}= 1\times 4186(100 - 37) + 1\times 2256254 = 2519972\ J$

where

$L_{v}$ = Latent heat of vaporization

Therefore, mass of water evaporated, $m_{w} = \frac{Q}{2519972} = \frac{643302}{2519972} = 0.255\ kg$