Question

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

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 \(4878.53 \mathrm{~g}\) or \(4.88 \mathrm{~kg}\).

Step-by-step solution

Calculate the total energy generated in 2 hours of exercise

First, we need to calculate the total energy generated during 2 hours of exercise. We are given that the body can generate \(5500 \mathrm{~kJ}\) of energy during one hour of exercise. To find the total energy in 2 hours, we simply multiply the energy generated in one hour by 2: Total energy generated = Energy generated in one hour × Time (in hours) = \(5500 \mathrm{~kJ/hour} \times 2 \mathrm{~hours} = 11000 \mathrm{~kJ}\)

Calculate the number of moles of water evaporated

We are given the heat of vaporization of water, which is \(40.6 \mathrm{~kJ/mol}\). To find the number of moles of water needed to be evaporated, we can divide the total energy generated by the heat of vaporization of water: Number of moles = Total energy generated / Heat of vaporization of water = \(11000 \mathrm{~kJ} / 40.6 \mathrm{~kJ/mol} = 270.94 \mathrm{~mol}\)

Calculate the mass of water evaporated

Now that we have the number of moles of water that needs to be evaporated, we can convert it into mass using the molar mass of water. The molar mass of water is \(18.015 \mathrm{~g/mol}\). Mass of water evaporated = Number of moles × Molar mass of water = \(270.94 \mathrm{~mol} \times 18.015 \mathrm{~g/mol} = 4878.53 \mathrm{~g}\) Thus, 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 \(4878.53 \mathrm{~g}\) or \(4.88 \mathrm{~kg}\).

Key concepts

Understanding Heat of Vaporization

The concept of the heat of vaporization is central to the process of sweating and cooling off the body after exercise. The heat of vaporization is the amount of energy required to convert a given amount of liquid into vapor at a constant temperature. For water, this energy requirement is quite high, precisely 40.6 kJ/mol.
This means a substantial amount of energy is absorbed by the water molecules from the body's surface, facilitating the phase change from liquid to vapor.

• This energy absorption enables the body to lose heat efficiently.
• As water evaporates, it draws heat from the body, cooling it down.

The evaporation process is vital for temperature regulation and ensures that the body can maintain a stable internal environment during strenuous activities.
Hence, the heat of vaporization plays a crucial role in our body's natural cooling mechanism.

Molar Mass of Water in Vaporization

Molar mass is the mass of one mole of a given substance, typically expressed in grams per mole. For water, the molar mass is 18.015 g/mol.
This value is important when calculating how much water must evaporate to dissipate heat generated during physical activity.

• Knowing the molar mass helps in converting moles of water into grams or kilograms.
• This conversion allows us to quantify the exact mass of water required for effective cooling.

In the given exercise, after determining the number of moles of water needed to absorb the generated heat, this number is multiplied by the molar mass to find the total mass of water.
Understanding the molar mass helps clarify how chemical properties of substances, like water, relate directly to their physical behaviors.

Energy Conversion in Physical Activity

During physical activity, the body undergoes significant energy conversion processes. Energy from food and stored nutrients is converted into mechanical energy for movement and also into heat.

• Of the total energy your body generates during exertion, a portion is used for muscle contraction.
• The rest is converted into heat, which if not dissipated, would raise the body's core temperature.

This heat must be expelled efficiently to avoid overheating; hence, the body employs perspiration as a form of energy conversion, utilizing latent heat exchange to cool the body effectively.
By understanding these conversions, you appreciate how interconnected energy transformation and body regulation systems are and why managing the body's heat is crucial during exercise.

Role of Perspiration

Perspiration is an essential function of the body that aids in temperature regulation, especially during strenuous exercise. Sweat glands release water onto the skin surface, which then evaporates, pulling heat away from the body.

• Evaporation is an endothermic process, meaning it absorbs heat, cooling the body down.
• The efficiency of this cooling method reflects the physical principle of heat of vaporization.

Through sweating, the body can maintain a stable temperature even when producing large amounts of heat during intense activities.
Without perspiration, the body's core temperature could dangerously rise, making the understanding of this process vital in both scientific and athletic contexts.

Exercise Physiology and Thermoregulation

Exercise physiology examines how physical exercise impacts different body systems, particularly focusing on how the body manages heat generated during activity.

• During exercise, muscles produce a significant amount of heat; too much of it can threaten normal physiological functions.
• Thermoregulatory mechanisms such as vasodilation and increased perspiration are crucial.

The study of exercise physiology helps us understand these adaptations and processes that ensure the body maintains homeostasis during exercise.
By learning about these physiological responses, one can develop better strategies for training and recovery, optimizing performance while preventing overheating.