Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 95

It has been determined that the body can generate \(5500 \mathrm{~kJ}\) of energy during one hour of strenuous exercise. Perspiration is the body's mechanism for eliminating this heat. What mass of water would have to be evaporated through perspiration to rid the body of the heat generated during two hours of exercise? (The heat of vaporization of water is \(40.6 \mathrm{~kJ} / \mathrm{mol}\).)

Short Answer

Expert verified
The mass of water that would have to be evaporated through perspiration to rid the body of the heat generated during two hours of exercise is given by: Mass_of_water_evaporated = ( (5500 kJ/h × 2 h) / 40.6 kJ/mol ) × 18.015 g/mol Calculate the value to get the mass of water evaporated.

Step by step solution

01

Calculate the total heat generated during two hours of exercise

First, let's calculate the total heat generated during two hours of strenuous exercise. Total_heat_generated = Heat_generated_per_hour × Hours_of_exercise Total_heat_generated = 5500 kJ/h × 2 h
02

Calculate the number of moles of water evaporated

Now, we'll calculate the number of moles of water evaporated (n) using the heat of vaporization formula: Total_heat_generated = n × Heat_of_vaporization n = Total_heat_generated / Heat_of_vaporization n = (5500 kJ/h × 2 h) / 40.6 kJ/mol
03

Calculate the mass of water evaporated

Finally, we'll convert the number of moles of water evaporated into mass using the molar mass of water (18.015 g/mol): Mass_of_water_evaporated = n × Molar_mass_of_water Mass_of_water_evaporated = ( (5500 kJ/h × 2 h) / 40.6 kJ/mol ) × 18.015 g/mol Now, calculate the final value to determine the mass of water that would have to be evaporated through perspiration to rid the body of the heat generated during two hours of exercise.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Consider the substances in Table 6.1. Which substance requires the largest amount of energy to raise the temperature of \(25.0 \mathrm{~g}\) of the substance from \(15.0^{\circ} \mathrm{C}\) to \(37.0^{\circ} \mathrm{C}\) ? Calculate the energy. Which substance in Table \(6.1\) has the largest temperature change when \(550 . \mathrm{g}\) of the substance absorbs \(10.7 \mathrm{~kJ}\) of energy? Calculate the temperature change.

Given the following data $$ \begin{aligned} \mathrm{P}_{4}(s)+6 \mathrm{Cl}_{2}(g) & \longrightarrow 4 \mathrm{PCl}_{3}(g) & & \Delta H=-1225.6 \mathrm{~kJ} \\ \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) & \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) & & \Delta H=-2967.3 \mathrm{~kJ} \\ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) & \longrightarrow \mathrm{PCl}_{5}(g) & & \Delta H=-84.2 \mathrm{~kJ} \\ \mathrm{PCl}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) & \mathrm{Cl}_{3} \mathrm{PO}(g) & & \Delta H=-285.7 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the reaction $$ \mathrm{P}_{4} \mathrm{O}_{10}(s)+6 \mathrm{PCl}_{5}(g) \longrightarrow 10 \mathrm{Cl}_{3} \mathrm{PO}(g) $$

Given the following data $$ \begin{aligned} \mathrm{Ca}(s)+2 \mathrm{C}(\text { graphite }) & \longrightarrow \mathrm{CaC}_{2}(s) & \Delta H &=-62.8 \mathrm{~kJ} \\ \mathrm{Ca}(s)+\frac{1}{2} \mathrm{O}_{2}(g) & \longrightarrow \mathrm{CaO}(s) & \Delta H &=-635.5 \mathrm{~kJ} \\ \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q) & \Delta H &=-653.1 \mathrm{~kJ} \\ \mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \Delta H &=-1300 . \mathrm{kJ} \\ \mathrm{C}(\text { graphite })+\mathrm{O}_{2}(g) & \Delta H &=-393.5 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the reaction $$ \mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{C}_{2} \mathrm{H}_{2}(g) $$

The preparation of \(\mathrm{NO}_{2}(g)\) from \(\mathrm{N}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) is an endothermic reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g) \text { (unbalanced) } $$ The enthalpy change of reaction for the balanced equation (with lowest whole- number coefficients) is \(\Delta H=67.7 \mathrm{~kJ}\). If \(2.50 \times\) \(10^{2} \mathrm{~mL} \mathrm{~N}_{2}(g)\) at \(100 .{ }^{\circ} \mathrm{C}\) and \(3.50\) atm and \(4.50 \times 10^{2} \mathrm{~mL} \mathrm{O}_{2}(g)\) at \(100 .{ }^{\circ} \mathrm{C}\) and \(3.50 \mathrm{~atm}\) are mixed, what amount of heat is necessary to synthesize the maximum yield of \(\mathrm{NO}_{2}(g)\) ?

The bomb calorimeter in Exercise 108 is filled with \(987 \mathrm{~g}\) water. The initial temperature of the calorimeter contents is \(23.32^{\circ} \mathrm{C}\). A \(1.056-\mathrm{g}\) sample of benzoic acid \(\left(\Delta E_{\text {comb }}=-26.42 \mathrm{~kJ} / \mathrm{g}\right)\) is combusted in the calorimeter. What is the final temperature of the calorimeter contents?
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Making Measurements

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

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.

Sign-up for free