/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 8 A \(60.0\) -kg runner expends ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 34

8 A \(60.0\) -kg runner expends \(300 \mathrm{~W}\) of power while running a marathon. Assuming \(10.0 \%\) of the energy is deliyered to the muscle tissue and that 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 \(37,0^{\circ} \mathrm{C}\), the latent heat of vaporization of water is \(\left.2.41 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\right)\)

Short Answer

Expert verified
The volume of sweat lost per hour by the runner is approximately \(0.403 \mathrm{~L}\).

Step by step solution

01

Determine The Power Excess

First, find out how much power is used to evaporate sweat. We know that the runner's muscles use 10% of the power, so the other \(90\%\) is used to evaporate sweat. Thus, the power used to evaporate sweat per second is \(0.90 \times 300 \mathrm{~W} = 270 \mathrm{~W}\). This is in watts, which are joules per second, so this gives us the energy per second.
02

Find Energy Used Per Hour

Next, convert from energy per second to energy per hour. One hour is \(3600\) seconds, so the energy used per hour is \(270 \mathrm{~J/s} \times 3600 \mathrm{~s} = 972,000 \mathrm{~J}\). This is the total energy used to evaporate sweat in one hour.
03

Determine Mass of Evaporated Sweat

Now, use the latent heat of vaporization to find the mass of water that this energy can evaporate. The formula is \(L = Q/m\), where \(Q\) is the heat energy, \(m\) is the mass, and \(L\) is the latent heat of vaporization. Here, we want to find the mass, so we can rearrange to \(m = Q/L = 972,000 \mathrm{~J} / 2.41 \times 10^{6} \mathrm{~J/kg} = 0.403 kgs\). This is the mass of the sweat evaporated in one hour.
04

Convert Mass to Volume

To translate the mass of the sweat to a volume, use the fact that the density of water is \(1 \mathrm{~kg/L}\). Thus, the volume of sweat evaporated in one hour is \(0.403 \mathrm{~L}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Latent heat of vaporization
In thermodynamics, the concept of latent heat refers to the amount of heat required to change the state of a substance without altering its temperature. The latent heat of vaporization specifically refers to the heat needed to convert a liquid to a gas. For instance, when water is heated to become steam, it absorbs a certain amount of heat without increasing its temperature during this phase change.

In our example with the marathon runner, the excess energy from metabolism is expended to evaporate sweat, a prime example of using latent heat of vaporization. Here, the latent heat for water is given as \(2.41 \times 10^{6} \text{ J/kg}\) at \(37.0^{\circ} \text{C}\). This means every kilogram of water requires \(2.41 \times 10^{6}\) joules to turn into vapor. Understanding this concept provides insight into how our body cools down during vigorous activities like running.
Energy conversion
Energy conversion is the process of changing energy from one form to another. In the context of the runner, we start with metabolic energy converted into mechanical energy to power muscle movements. However, not all this energy used is efficient or purely for movement. In fact, a considerable portion is wasted or diverted into other processes, such as perspiration.

In this instance, only \(10\%\) of the energy is efficiently used for muscle contractions, while \(90\%\) is converted to heat energy used to evaporate sweat. Each joule of energy not used in muscle contractions must be dealt with by the body, mainly in thermal regulation through sweating. So, this energy conversion process leads to the evaporation of water from the skin, keeping the runner cool.
Power calculation
Power is essentially the rate at which work is being done or energy is being consumed. It's measured in watts in the SI system, where one watt equals one joule per second. Calculating power helps understand how much work is being done in a certain period.

For our runner, the power output is initially \(300\text{ W}\). However, with only \(10\%\) being actively used by muscles, \(90\%\) or \(270\text{ W}\) is utilized to evaporate sweat. Knowing the unit conversion, we multiply this by the amount of time to find the total energy expenditure in joules. Thus, over an hour, \(270\text{ J/s} \times 3600\text{ s} = 972,000\text{ J}\) of energy is spent.
  • Step by step: Convert wattage into heat energy to measure impact.
  • Calculate hourly: Multiply energy per second by number of seconds in an hour.
  • Efficiency check: Recognize only a fraction of power is used for muscle movement.

The meticulously calculated \(972,000\text{ J}\) demonstrates how a simple understanding of power and its calculus transforms into actionable knowledge about the body's regulation during tasks.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

ecp A sprinter of mass \(m\) accelerates uniformly from rest to velocity vin \(t\) seconds. (a) Write a symbolic expression for the instantaneous mechanical power \(\mathcal{P}\) required by the sprinter in terms of force \(F\) and velocity \(u\). (b) Use Newton's second law and a kinematics equation for the velocity at any time to obtain an expression for the instantaneous power in terms of \(m, a\), and \(t\) only. (c) If a \(75.0-\mathrm{kg}\) sprinter reaches a speed of \(11.0 \mathrm{~m} / \mathrm{s}\) in \(5.00 \mathrm{~s}\), calculate the sprinter's acceleration, assuming it to be constant. (d) Calculate the \(75.0\) -kg sprinter's instantaneous mechanical power as a function of time \(\ell\) and \((\mathrm{e})\) give the maximum rate at which he burns Calories during the sprint, assuming \(25 \%\) efficiency of conversion form food energy to mechanical energy.

Calculate the temperature at which a tungsten filament that has an emissivity of \(0.90\) and a surface area of \(2.5 \times 10^{-5} \mathrm{~m}^{2}\) will radiate energy at the rate of \(25 \mathrm{~W}\) in a room where the temperature is \(22^{\circ} \mathrm{C}\).

GP In the summer of 1958 in St. Petersburg, Florida, a new sidewalk was poured near the childhood home of one of the authors. No expansion joints were supplied, and by mid-July the sidewalk had been completely destroyed by thermal expansion and had to be replaced, this time with the important addition of expansion joints! This event is modeled here. A slab of concrete \(4.00 \mathrm{~cm}\) thick, \(1.00 \mathrm{~m}\) long, and \(1 \leq 00 \mathrm{~m}\) wide is poured for a sidewalk at an ambient temperature of \(25.0^{\circ} \mathrm{C}\) and allowed to set. The slab is exposed to direct sunlight and placed in a series of such slabs without proper expansion joints, so linear expansion is prevented. (a) Using the linear expansion equation (Eq. \(10.4)\), eliminate \(\Delta I\). from the equation for compressive stress and strain (Eq. 9.3). (b) Use the expression found in part (a) to eliminate \(\Delta T\) from Equation \(11.3\), obtain ing a symbolic equation for thermal energy transfer \(Q\). (c) Compute the mass of the concrete slab given that its density is \(2.40 \times 10^{5} \mathrm{~kg} / \mathrm{m}^{3}\). (d) Concrete has an ultimate compressive strength of \(2.00 \times 10^{7} \mathrm{~Pa}\), specific heat of \(880 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), and Young's modulus of \(2.1 \times 10^{10} \mathrm{~Pa}\). How much thermal energy must be transferred to the slab to reach this compressive stress? (e) What temperature change is required? (f) If the sun delivers \(1.00 \times 10^{3} \mathrm{~W}\) of power to the top surface of the slab and if half the energy, on the average, is absorbed and retained, how long does it Lake the slab to reach the point at which it is in danger of cracking due to compressive stress?

In a showdown on the streets of Laredo, the good guy drops a \(5.0-g\) silver bullet at a temperature of \(20^{\circ} \mathrm{C}\) into a \(100-\mathrm{cm}^{3}\) cup of water at \(90^{\circ} \mathrm{C}\). Simultaneously, the bad guy drops a \(5.0-\mathrm{g}\) copper bullet at the same initial temperature into an identical cup of water. Which one ends the showdown with the coolest cup of water in the West? Neglect any energy transfer into or away from the container.

Three liquids are at temperatures of \(10^{\circ} \mathrm{C}, 20^{\circ} \mathrm{C}\), and \(30^{\circ} \mathrm{C}\), respectively. Equal masses of the first two liquids are mixed, and the equilibrium temperature is \(17^{\circ} \mathrm{C}\). Equal masses of the second and third are then mixed, and the equilibrium temperature is \(28^{\circ} \mathrm{C}\). Find the equilibrium temperature when equal masses of the first and third are mixed.
See all solutions

Recommended explanations on Physics Textbooks

Astrophysics

Read Explanation

Electricity and Magnetism

Read Explanation

Kinematics Physics

Read Explanation

Fields in Physics

Read Explanation

Wave Optics

Read Explanation

Atoms and Radioactivity

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.