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Chapter 7: Problem 91

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 2 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 2 hours of exercise is approximately \(4870.37\ \mathrm{g}\).

Step by step solution

01

Calculate total energy generated during 2 hours of exercise

To find the total energy generated during 2 hours of exercise, we multiply the energy generated per hour by the number of hours. Total energy = Energy per hour × Number of hours Total energy = \(5500\ \mathrm{kJ/hour} × 2\ \mathrm{hours}\) Total energy = \(11000\ \mathrm{kJ}\)
02

Calculate the number of moles of water to evaporate

We're given the heat of vaporization of water, which represents the energy needed to evaporate one mole of water. We will divide the total energy generated by the heat of vaporization to find the number of moles of water that need to evaporate. Number of moles = Total energy / Heat of vaporization per mole Number of moles = \( \frac{11000 \ \mathrm{kJ}}{40.6\ \mathrm{kJ/mol}} \) Number of moles ≈ \(270.44\ \mathrm{moles}\)
03

Convert moles to mass

We will now convert the number of moles to mass using the molar mass of water, which is \(18.015\ \mathrm{g/mol}\). Mass of water = Number of moles × Molar mass of water Mass of water = \(270.44\ \mathrm{moles} × 18.015\ \mathrm{g/mol}\) Mass of water ≈ \(4870.37\ \mathrm{g}\) The mass of water that would have to be evaporated through perspiration to rid the body of the heat generated during 2 hours of exercise is approximately \(4870.37\ \mathrm{g}\).

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Key Concepts

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

Energy During Exercise

Exercise is a vital activity for maintaining good health, and it also represents an interesting process from a physiological perspective. When we engage in physical activity, our bodies use energy to fuel our muscles. This energy is often measured in kilojoules (kJ), a unit of work or energy in the International System of Units (SI).

Different exercises burn energy at various rates, but on average, an hour of strenuous exercise can generate about 5500 kJ. It's crucial for our bodies to regulate their temperature during exercise; as the body's core temperature rises, mechanisms like perspiration help dissipate excess heat. During perspiration, sweat evaporates from the surface of the skin, which requires energy, known as the heat of vaporization. This process effectively cools the body by utilizing the thermal energy generated during the workout.

Regulating Body Temperature

The human body is quite efficient at maintaining a stable internal temperature, notwithstanding the heat generated from exercise. The thermal energy needs to be transferred to the surrounding environment to prevent overheating, and the evaporation of sweat is one of the body's primary cooling mechanisms.

Exercise Intensity

It's worth noting that the intensity of the exercise influences the amount of energy expended and, subsequently, the amount of heat that needs to be dissipated. More intense workouts will, therefore, lead to greater sweat production and a higher demand for evaporative cooling.

Molar Heat of Vaporization

The molar heat of vaporization is a crucial concept when discussing the change of state from liquid to gas. It is quantified as the amount of energy required to vaporize one mole of a substance at its boiling point under standard pressure, expressed in kilojoules per mole (kJ/mol). Water has a molar heat of vaporization value of 40.6 kJ/mol, indicating that to evaporate one mole of water, 40.6 kJ of energy is needed.

When we exercise, the body uses this energy to turn sweat, which is water in its liquid form, into vapor, thereby regulating body temperature. This energy is derived from the heat our body produces during physical activity.

Understanding the Process

Evaporative cooling through perspiration is an endothermic process, meaning it absorbs heat from the environment — in this case, the skin and the body's internal heat. The higher the molar heat of vaporization, the more effective the cooling process is because it absorbs more heat per mole of liquid evaporated.

Role in Exercise

The molar heat of vaporization directly correlates with the amount of sweat required to cool the body during exercise. The calculation of this quantity helps in estimating how much sweat needs to evaporate to eliminate a certain amount of thermal energy.

Converting Moles to Mass

The stoichiometry of chemical compounds often requires the conversion between moles and mass. Moles provide a way to quantify atoms, molecules, or ions in a consistent manner, using the Avogadro's number (approximately 6.022 x 10^23 entities per mole).

To convert moles to mass, we multiply the number of moles by the molar mass of the substance, which is expressed in grams per mole (g/mol). For water (H2O), the molar mass is 18.015 g/mol. Therefore, knowing the number of moles of water that needs to be evaporated during exercise allows us to calculate the corresponding mass of water in grams.

Application in Exercise

The conversion from moles to mass is vital in understanding the physical implications of biochemical processes, such as sweat evaporation during exercise. For instance, in our textbook problem, by calculating the mass of water to be evaporated, we can gauge the amount of sweat a person might produce to cool down after a two-hour workout session.

Practical Relevance

While these calculations might seem abstract, they have practical relevance. Sports scientists, trainers, and individuals can use these figures to estimate hydration needs post-exercise, ensuring that athletes and exercisers replenish the fluids lost through sweat.

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Most popular questions from this chapter

Calculate \(\Delta E\) for each of the following. a. \(q=-47 \mathrm{kJ}, w=+88 \mathrm{kJ}\) b. \(q=+82 \mathrm{kJ}, w=-47 \mathrm{kJ}\) c. \(q=+47 \mathrm{kJ}, w=0\) d. In which of these cases do the surroundings do work on the system?

On Easter Sunday, April \(3,1983,\) nitric acid spilled from a tank car near downtown Denver, Colorado. The spill was neutralized with sodium carbonate: $$2 \mathrm{HNO}_{3}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(s) \longrightarrow 2 \mathrm{NaNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)$$ a. Calculate \(\Delta H^{\circ}\) for this reaction. Approximately \(2.0 \times\) \(10^{4}\) gal nitric acid was spilled. Assume that the acid was an aqueous solution containing \(70.0 \%\) HNO \(_{3}\) by mass with a density of \(1.42 \mathrm{g} / \mathrm{cm}^{3} .\) What mass of sodium carbonate was required for complete neutralization of the spill, and what quantity of heat was evolved? \((\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{NaNO}_{3}(a q)=-467 \mathrm{kJ} / \mathrm{mol})\) b. According to The Denver Post for April \(4,1983,\) authorities feared that dangerous air pollution might occur during the neutralization. Considering the magnitude of \(\Delta H^{\circ}\) what was their major concern?

The best solar panels currently available are about \(15 \%\) efficient in converting sunlight to electricity. A typical home will use about \(40 .\) kWh of electricity per day \((1 \mathrm{kWh}=1\) kilowatt hour; \(1 \mathrm{kW}=1000 \mathrm{J} / \mathrm{s}\) ). Assuming 8.0 hours of useful sunlight per day, calculate the minimum solar panel surface area necessary to provide all of a typical home's electricity. (See Exercise 124 for the energy rate supplied by the sun.)

The reaction $$\mathrm{SO}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(a q) $$ is the last step in the commercial production of sulfuric acid. The enthalpy change for this reaction is \(-227 \mathrm{kJ}\). In designing a sulfuric acid plant, is it necessary to provide for heating or cooling of the reaction mixture? Explain.

One mole of \(\mathrm{H}_{2} \mathrm{O}(g)\) at 1.00 atm and \(100 .^{\circ} \mathrm{C}\) occupies a volume of 30.6 L. When 1 mole of \(\mathrm{H}_{2} \mathrm{O}(g)\) is condensed to 1 mole of \(\mathrm{H}_{2} \mathrm{O}(l)\) at 1.00 atm and \(100 .^{\circ} \mathrm{C}, 40.66 \mathrm{kJ}\) of heat is released. If the density of \(\mathrm{H}_{2} \mathrm{O}(l)\) at this temperature and pressure is \(0.996 \mathrm{g} / \mathrm{cm}^{3},\) calculate \(\Delta E\) for the condensation of 1 mole of water at 1.00 atm and \(100 .^{\circ} \mathrm{C}\).
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