   Chapter 11, Problem 38P

Chapter
Section
Textbook Problem

A 60.0-kg runner expends 3.00 × 102 W of power while running a marathon. Assuming 10.0% of the energy is delivered to the muscle tissue and mat the excess energy is removed from the body primarily by sweating, determine the volume of bodily fluid (assume it is water) lost per hour. (At 37.0°C, the latent heat of vaporization of water is 2.41 × 106 J/kg.)

To determine
The volume of fluid lost per hour.

Explanation

Section 1:

To determine : The energy given out by the runner.

Answer: The energy given out by the runner is 1.08×106J .

Explanation :

Given Info:  Runner expends 3.00×102W of power and Running time 1 hour.

Formula to calculate the energy given out by the runner is,

Qout=Pt

• Qout is the energy given out by the runner,
• P is the power expended,
• t is the running time,

Substitute 3.00×102W for P and 1 h for t to find Qout .

Qout=(3.00×102W)(J/sW)(1h)(3600s1h)=1.08×106J

Therefore, the energy given out by the runner is 1.08×106J .

Section 2:

To determine : The mass of the fluid (water) evaporated.

Answer: The mass of the fluid (water) evaporated is 0.403kg .

Explanation :

Given Info: 10 % of energy is delivered to the muscle and 90% is removed by sweating.

Formula to calculate the mass of water that will be evaporated by the excess internal energy is,

mevap=(90%)QoutLv

• mevap is the mass of water evaporated,
• Lv is the latent heat of vaporization of water.

Substitute 1.08×106J for Qexcess and 2

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