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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.

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Key Concepts

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

Energy Expenditure
During exercise, the body requires energy to fuel physical activities. This energy is often measured in kilojoules (kJ) or calories. In the original exercise, we see that vigorous activity can generate as much as 5500 kJ per hour. While the primary goal of exercise might be improving fitness, it also serves to illustrate how much energy the body expends to perform work.
Energy expenditure in exercise is crucial as it determines how much heat is produced. The body's natural processes, like muscle contractions and biochemical reactions, produce this heat. By knowing the energy expenditure, we can calculate how much heat needs to be managed by the body.
  • Energy measurement helps in assessing fitness levels and designing workout plans.
  • Understanding energy expenditure is key in calculating caloric needs for weight management.
Exercise Physiology
Exercise physiology spends significant focus on how the body handles demanding physical tasks. When you engage in exercise, your heart rate increases, and more blood is pumped around your body. Muscles become engaged, using energy and releasing waste products, some of which is heat.
The field looks at how systems in the body such as the cardiovascular, respiratory, and muscular systems interact to sustain effort. All these systems work together to maintain homeostasis, a state of balance despite changes in energy demands.
  • Exercise physiology helps understand how the body adapts to exercise over time.
  • It aids in improving performance and recovery strategies for athletes.

This interconnection highlights the importance of considering multiple aspects like energy supply and waste heat elimination.
Thermoregulation
Thermoregulation is the body’s ability to maintain its core internal temperature. During exercise, as energy is expended, heat is generated. The body needs to rid itself of this excess heat to prevent overheating. In the context of the exercise provided, perspiration is the key mechanism for achieving thermoregulation.
Sweat evaporates from the skin, taking heat away and cooling the body. This cooling effect is vital for maintaining optimal body function. If thermoregulation fails, it can lead to heat-related illnesses.
  • Effective heat regulation is vital for maintaining physical and mental performance during exercise.
  • The process of sweat evaporation ties directly into the heat of vaporization, an important concept in thermodynamics.
Mass Calculations
To determine how much sweat (water) is necessary to dissipate the excess heat during exercise, mass calculations become important. These calculations involve using the heat of vaporization, which is the energy needed to turn a liquid into vapor.
In our problem, we've calculated the number of moles of water needed to absorb the required heat. By converting the moles of water into grams (using the known molar mass of water, 18.015 g/mol), we ultimately find the mass of water that needs to evaporate.
  • The process involves unit conversions and ensures accurate determination of physical quantities.
  • Mass calculations are fundamental in fields such as chemistry and biology for solving practical problems.
Chemical Thermodynamics
Chemical thermodynamics is a broad area that deals with the relationships between heat and other forms of energy during chemical reactions. For the exercise, thermodynamics principles help us understand how heat is transferred from the body to the environment.
The concept of the heat of vaporization is a classic thermodynamic topic, describing the energy needed to transform liquid water into vapor. By applying formulas from thermodynamics, we can effectively calculate how the body dissipates heat.
  • The study of thermodynamics offers tools to model and predict energy changes in chemical and physical processes.
  • Understanding these principles aids in developing energy-efficient technologies and improving physiological models related to human health.

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Most popular questions from this chapter

The combustion of \(0.1584 \mathrm{~g}\) benzoic acid increases the temperature of a bomb calorimeter by \(2.54^{\circ} \mathrm{C}\). Calculate the heat capacity of this calorimeter. (The energy released by combustion of benzoic acid is \(26.42 \mathrm{~kJ} / \mathrm{g} .\) ) A \(0.2130-\mathrm{g}\) sample of vanillin \(\left(\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}_{3}\right)\) is then burned in the same calorimeter, and the temperature increases by \(3.25^{\circ} \mathrm{C}\). What is the energy of combustion per gram of vanillin? Per mole of vanillin?

At \(298 \mathrm{~K}\), the standard enthalpies of formation for \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\) and \(\mathrm{C}_{6} \mathrm{H}_{6}(l)\) are \(227 \mathrm{~kJ} / \mathrm{mol}\) and \(49 \mathrm{~kJ} / \mathrm{mol}\), respectively. a. Calculate \(\Delta H^{\circ}\) for $$ \mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow 3 \mathrm{C}_{2} \mathrm{H}_{2}(g) $$ b. Both acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) can be used as fuels. Which compound would liberate more energy per gram when combusted in air?

The equation for the fermentation of glucose to alcohol and carbon dioxide is $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g) $$ The enthalpy change for the reaction is \(-67 \mathrm{~kJ}\). Is the reaction exothermic or endothermic? Is energy, in the form of heat, absorbed or evolved as the reaction occurs?

Consider the following equations: $$ \begin{aligned} 3 \mathrm{~A}+6 \mathrm{~B} \longrightarrow & 3 \mathrm{D} & \Delta H &=-403 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{E}+2 \mathrm{~F} & \longrightarrow \mathrm{A} & \Delta H &=-105.2 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{C} & \longrightarrow \mathrm{E}+3 \mathrm{D} & \Delta H &=+64.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ Suppose the first equation is reversed and multiplied by \(\frac{1}{6}\), the second and third equations are divided by 2, and the three adjusted equations are added. What is the net reaction and what is the overall heat of this reaction?

Acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and butane \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)\) are gaseous fuels with enthalpies of combustion of \(-49.9 \mathrm{~kJ} / \mathrm{g}\) and \(-49.5 \mathrm{~kJ} / \mathrm{g}\), respectively. Compare the energy available from the combustion of a given volume of acetylene to the combustion energy from the same volume of butane at the same temperature and pressure.

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