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Chapter 17: Problem 89

If the air temperature is the same as the temperature of your skin (about \(30^{\circ} \mathrm{C}\) ), your body cannot get rid of heat by transferring it to the air. In that case, it gets rid of the heat by evaporating water (sweat). During bicycling, a typical \(70 \mathrm{~kg}\) person's body produces energy at a rate of about \(500 \mathrm{~W}\) due to metabolism, \(80 \%\) of which is converted to heat. (a) How many kilograms of water must the person's body evaporate in an hour to get rid of this heat? The heat of vaporization of water at body temperature is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) (b) The evaporated water must, of course, be replenished, or the person will dehydrate. How many \(750 \mathrm{~mL}\) bottles of water must the bicyclist drink per hour to replenish the lost water? (Recall that the mass of a liter of water is \(1.0 \mathrm{~kg} .\) )

Short Answer

Expert verified
The person's body must evaporate approximately 0.595 kg (or 0.595 L) of water per hour to get rid of the heat produced while bicycling. This means the bicyclist must drink essentially 1 bottle of water (750 mL) per hour to avoid dehydration.

Step by step solution

01

Calculate the amount of heat produced

We know that the person's body generates energy at a rate of 500 W due to metabolism, 80% of which is converted to heat. So, the amount of heat produced per second can be calculated as follows: \(0.80 \times 500 \, \mathrm{W} = 400 \, \mathrm{W}\) Since a watt is equivalent to a joule per second, the person's body produces 400 Joules of heat per second.
02

Find the mass of water evaporated

The heat of vaporization of water is \(2.42 \times 10^6 \, \mathrm{J/kg}\), which means this is the amount of energy required to evaporate 1 kg of water. Using the formula \(q = m \cdot \Delta H\), where 'q' is the heat transferred, 'm' is the mass, and '\(\Delta H\)' is the heat of vaporization, we can solve for 'm' (mass of water evaporated). Here, 'q' is the total heat produced in one hour which is \(400 \, \mathrm{J/sec} \times 3600 \, \mathrm{sec/hr} = 1.44 \times 10^6 \, \mathrm{J/hr}\), and the heat of vaporization '\(\Delta H\)' is given as \(2.42 \times 10^6 \, \mathrm{J/kg}\). Solving the formula for 'm' gives us: \(m = \frac{q}{\Delta H} = \frac{1.44 \times 10^6 \, \mathrm{J/hr}}{2.42 \times 10^6 \, \mathrm{J/kg}} = 0.595 \, \mathrm{kg/hr}\)
03

Calculate the number of water bottles needed

Firstly, convert the mass of water evaporated to liters. Since the mass of a liter of water is 1 kg, this means 0.595 kg of water is 0.595 L. Then, find the number of 750 mL bottles equivalent to 0.595 L of water. 0.595 L is equivalent to 595 mL. Since each bottle is 750 mL, the number of bottles needed can be calculated as follows: \( \frac{595 \, \mathrm{mL}}{750 \, \mathrm{mL/bottle}} = 0.793 \, \mathrm{bottles}\) Since we can't have a part of a bottle, we round this up to 1 bottle of water.

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Key Concepts

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

Heat of Vaporization
When we talk about the heat of vaporization, we're referring to the amount of energy required to transform a certain mass of a liquid into a gas at a constant temperature. This process is a key player when our bodies utilize sweating to regulate temperature. If the air is the same temperature as our skin, direct heat transfer isn't effective. Instead, our bodies turn to sweat, which absorbs the excess heat when it evaporates from our skin.

For water, this heat of vaporization is quite high, meaning it takes a significant amount of energy to evaporate it. To be precise, at body temperature, the heat of vaporization for water is approximately 2.42 x 106 joules per kilogram (J/kg). When our skin sweats during intense activities like bicycling, our body uses this efficient thermal management system to prevent overheating by converting water into vapor, which carries heat away from our bodies.
Energy Conversion in Metabolism
Metabolism is essentially our body's engine, converting the food we eat into the energy we need to live and perform activities. This energy conversion isn't 100% efficient; some energy is always lost, mostly as heat. For example, while cycling, a significant portion鈥攁round 80%鈥攐f the metabolic energy produced is converted into heat. This heat needs to be managed or dissipated to prevent the body from overheating.

In terms of energy units, metabolism is measured in watts, with one watt being equal to one joule of energy per second. If we consider a bicyclist whose body generates energy at a rate of 500 watts, where 400 watts is effectively turned into heat every second, this accumulation of heat requires an efficient cooling mechanism, which brings us back to the role of the heat of vaporization as a cooling method.
Hydration and Dehydration
Hydration plays a pivotal role in our body's temperature regulation and overall function. When we sweat, our bodies lose water, which needs to be replenished to avoid dehydration. Dehydration can negatively impact metabolic processes and lead to overheating, as well as a decline in physical performance and cognitive function.

So, when considering the exercise scenario of a bicyclist, it's crucial to replace the water lost through sweating. Calculating the amount of water loss, and consequently the number of bottles required to stay hydrated, is dependent on understanding how much sweat (water) has evaporated to dissipate the heat produced during cycling. The replacement calculation is also straightforward - the volume of water lost to sweating is replenished by an equivalent volume of ingestion, making sure to stay ahead of dehydration.

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Most popular questions from this chapter

At very low temperatures the molar heat capacity of rock salt varies with temperature according to Debye's \(T^{3}\) law: $$ C=k \frac{T^{3}}{\theta^{3}} $$ where \(k=1940 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}\) and \(\theta=281 \mathrm{~K}\). (a) How much heat is required to raise the temperature of \(1.50 \mathrm{~mol}\) of rock salt from \(10.0 \mathrm{~K}\) to \(40.0 \mathrm{~K} ?\) (Hint: Use Eq. (17.18) in the form \(d Q=n C d T\) and integrate.) (b) What is the average molar heat capacity in this range? (c) What is the true molar heat capacity at \(40.0 \mathrm{~K} ?\)

Convert the following Kelvin temperatures to the Celsius and Fahrenheit scales: (a) the midday temperature at the surface of the moon \((400 \mathrm{~K}) ;\) (b) the temperature at the tops of the clouds in the atmosphere of Saturn \((95 \mathrm{~K}) ;\) (c) the temperature at the center of the sun \(\left(1.55 \times 10^{7} \mathrm{~K}\right)\)

\(Camels require very little }}\) water because they are able to tolerate relatively large changes in their body temperature. While humans keep their body temperatures constant to within one or two Celsius degrees, a dehydrated camel permits its body temperature to drop to \(34.0^{\circ} \mathrm{C}\) overnight and rise to \(40.0^{\circ} \mathrm{C}\) during the day. To see how effective this mechanism is for saving water, calculate how many liters of water a \(400 \mathrm{~kg}\) camel would have to drink if it attempted to keep its body temperature at a constant \(34.0^{\circ} \mathrm{C}\) by evaporation of sweat during the day ( 12 hours) instead of letting it rise to \(40.0^{\circ} \mathrm{C}\). (Note: The specific heat of a camel or other mammal is about the same as that of a typical human, \(3480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). The heat of vaporization of water at \(34^{\circ} \mathrm{C}\) is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) )

A machinist bores a hole of diameter \(1.35 \mathrm{~cm}\) in a steel plate that is at \(25.0^{\circ} \mathrm{C}\). What is the cross-sectional area of the hole (a) at \(25.0^{\circ} \mathrm{C}\) and \((\mathrm{b})\) when the temperature of the plate is increased to \(175^{\circ} \mathrm{C} ?\) Assume that the coefficient of linear expansion remains constant over this temperature range.

The rate at which radiant energy from the sun reaches the earth's upper atmosphere is about \(1.50 \mathrm{~kW} / \mathrm{m}^{2} .\) The distance from the earth to the sun is \(1.50 \times 10^{11} \mathrm{~m},\) and the radius of the sun is \(6.96 \times 10^{8} \mathrm{~m} .\) (a) What is the rate of radiation of energy per unit area from the sun's surface? (b) If the sun radiates as an ideal blackbody, what is the temperature of its surface?
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