/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 Evaporation of sweat is an impor... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 48

Evaporation of sweat is an important mechanism for temperature control in some warm-blooded animals. (a) What mass of water must evaporate from the skin of a \(70.0 \mathrm{~kg}\) man to cool his body \(1.00 \mathrm{C}^{\circ}\) ? The heat of vaporization of water at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) The specific heat of a typical human body is \(3480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) (see Exercise 17.27 ) . (b) What volume of water must the man drink to replenish the evaporated water? Compare to the volume of a soft-drink can \(\left(355 \mathrm{~cm}^{3}\right)\).

Short Answer

Expert verified
The mass of water needed to evaporate to cool a human body by one degree is around 0.1007 kg. The volume of water to replenish this evaporated water is about 100.7 cubic cm. Compared to the volume of a typical soft drink can which is 355 cubic cm, the consumed water volume is less than one-third of the can's volume.

Step by step solution

01

Calculate Energy Required to Cool the Body

To calculate the energy required to cool the body by one degree, the specific heat formula \(Q = mc\Delta T\) is used, where \(Q\) denotes heat energy, \(m\) is the body mass, \(c\) is the specific heat, and \(\Delta T\) is the change in temperature. Inputting the given values, the result is \(Q = 70 kg \times 3480 J/(kg\cdot K) \times 1 K = 243600 J\).
02

Calculate the Mass of Evaporated Water

The mass of water evaporated is calculated by equating the energy required to cool the body with the energy used for evaporation. Using the formula \(Q = mL\) where \(L\) is the latent heat or heat of vaporization, and solving for \(m\), the result is \(m = Q/L = 243600 J / (2.42 \times 10^{6} J/kg) = 0.1007 kg\).
03

Compute the Volume of Water to Replenish

To calculate the volume of water requiring replenishment, the mass of water is used, applying the property that 1 kg of water is equivalent to 1 L or 1000 cm鲁 in volume. Hence the volume is \(V = m = 0.1007 kg = 100.7 cm鲁\).
04

Compare with a Soft Drink Can's Volume

The volume of the lost water is compared to the volume of a soft drink can. The volume of the can is 355 cm鲁 which is far more than the evaporated water volume of 100.7 cm鲁. Thus, to maintain body hydration after mild evaporative cooling, drinking less than one-third of a typical soft-drink can size of water is sufficient.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Evaporative Cooling
Evaporative cooling is a natural phenomenon utilized by various organisms, including humans, to regulate body temperature. When we sweat, water molecules from the skin absorb thermal energy from the body as they enter a gaseous state, leading to a cooling effect. This energy absorption occurs because breaking the hydrogen bonds that hold water molecules together requires considerable energy, known as the heat of vaporization. For instance, during a workout, the body temperature increases, and sweat production kicks in as a biological response to avoid overheating.

In the exercise, the amount of sweat (water) needed to evaporate to cool down a 70.0 kg person by 1.00掳C demonstrates the practical application of evaporative cooling. It requires the understanding of the amount of energy needed which involves specific heat and the heat of vaporization of water at the given body temperature. Providing this knowledge prepares students to grasp the role of thermoregulation in biology and allows them to apply mathematical concepts to biological processes.
Specific Heat
Specific heat is the amount of heat per unit mass required to raise the temperature of a substance by one degree Celsius. It's a property that varies between different substances. For instance, water has a high specific heat, meaning it can absorb a lot of heat before it gets hot. This characteristic is critical in biology, as it enables water to act as a temperature buffer in organisms and environments, maintaining conditions vital for life.

When we apply the concept to the human body, we can calculate the amount of energy required to change the body's temperature. This principle is key in understanding how animals maintain their internal temperatures within a narrow range, despite external temperature fluctuations. In the provided exercise, the specific heat value for the human body is used to determine the energy needed to cool the body by 1掳C.
Heat of Vaporization
The heat of vaporization is a critical thermodynamic property that signifies the energy needed to convert a given amount of a substance from a liquid to a vapor without increasing temperature. This large energy requirement for phase change plays a vital role in biological systems, particularly in temperature regulation.

Within the context of the exercise, the heat of vaporization value for water at body temperature is a key factor to determine the mass of water that must evaporate to cool the body. Understanding this concept helps students appreciate how organisms leverage phase change as an efficient mechanism to manage heat stress.
Energy Transfer in Biological Systems
Energy transfer is fundamental in biological systems. It involves not only the metabolic processes that occur within cells but also the thermodynamic principles that govern how organisms exchange heat with their surroundings. This transfer can be through various means such as radiation, conduction, convection, and in the case of the exercise, evaporation.

Understanding how energy is transferred, especially the calculations and concepts like specific heat and heat of vaporization, is essential for students who are learning about homeostasis and other temperature-related biological phenomena. The exercise's query about replenishing the water volume that was lost through evaporative cooling prompts students to consider the balance between energy loss and intake necessary to sustain biological functions.

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

A plastic cup of negligible mass contains \(0.280 \mathrm{~kg}\) of an unknown liquid at a temperature of \(30.0^{\circ} \mathrm{C}\). A \(0.0270 \mathrm{~kg}\) mass of ice at a temperature of \(0.0^{\circ} \mathrm{C}\) is added to the liquid, and when thermal equilibrium is reached the temperature of the combined substances is \(14.0^{\circ} \mathrm{C}\). Assuming no heat is exchanged with the surroundings, what is the specific heat capacity of the unknown liquid?

Hot Air in a Physics Lecture. (a) A typical student listening attentively to a physics lecture has a heat output of \(100 \mathrm{~W}\). How much heat energy does a class of 140 physics students release into a lecture hall over the course of a 50 min lecture? (b) Assume that all the heat energy in part (a) is transferred to the \(3200 \mathrm{~m}^{3}\) of air in the room. The air has specific heat \(1020 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and density \(1.20 \mathrm{~kg} / \mathrm{m}^{3} .\) If none of the heat escapes and the air conditioning system is off, how much will the temperature of the air in the room rise during the 50 min lecture? (c) If the class is taking an exam, the heat output per student rises to \(280 \mathrm{~W}\). What is the temperature rise during \(50 \mathrm{~min}\) in this case?

On-Demand Water Heaters. Conventional hot-water heaters consist of a tank of water maintained at a fixed temperature. The hot water is to be used when needed. The drawbacks are that energy is wasted because the tank loses heat when it is not in use and that you can run out of hot water if you use too much. Some utility companies are encouraging the use of on-demand water heaters (also known as flash heaters), which consist of heating units to heat the water as you use it. No water tank is involved, so no heat is wasted. A typical household shower flow rate is \(2.5 \mathrm{gal} / \mathrm{min}(9.46 \mathrm{~L} / \mathrm{min})\) with the tap water being heated from \(50^{\circ} \mathrm{F}\left(10^{\circ} \mathrm{C}\right)\) to \(120^{\circ} \mathrm{F}\left(49^{\circ} \mathrm{C}\right)\) by the on-demand heater. What rate of heat input (either electrical or from gas) is required to operate such a unit, assuming that all the heat goes into the water?

You have probably seen people jogging in extremely hot weather. There are good reasons not to do this! When jogging strenuously, an average runner of mass \(68 \mathrm{~kg}\) and surface area \(1.85 \mathrm{~m}^{2}\) produces energy at a rate of up to \(1300 \mathrm{~W}, 80 \%\) of which is converted to heat. The jogger radiates heat but actually absorbs more from the hot air than he radiates away. At such high levels of activity, the skin's temperature can be elevated to around \(33^{\circ} \mathrm{C}\) instead of the usual \(30^{\circ} \mathrm{C}\). (Ignore conduction, which would bring even more heat into his body.) The only way for the body to get rid of this extra heat is by evaporating water (sweating). (a) How much heat per second is produced just by the act of jogging? (b) How much net heat per second does the runner gain just from radiation if the air temperature is \(40.0^{\circ} \mathrm{C}\left(104^{\circ} \mathrm{F}\right) ?\) (Remember: He radiates out, but the environment radiates back in.) (c) What is the total amount of excess heat this runner's body must get rid of per second? (d) How much water must his body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) (e) How many \(750 \mathrm{~mL}\) bottles of water must he drink after (or preferably before!) jogging for a half hour? Recall that a liter of water has a mass of \(1.0 \mathrm{~kg}\).

There is \(0.050 \mathrm{~kg}\) of an unknown liquid in a plastic container of negligible mass. The liquid has a temperature of \(90.0^{\circ} \mathrm{C}\). To measure the specific heat capacity of the unknown liquid, you add a mass \(m_{w}\) of water that has a temperature of \(0.0^{\circ} \mathrm{C}\) to the liquid and measure the final temperature \(T\) after the system has reached thermal equilibrium. You repeat this measurement for several values of \(m_{\mathrm{w}},\) with the initial temperature of the unknown liquid always equal to \(90.0^{\circ} \mathrm{C}\). The plastic container is insulated, so no heat is exchanged with the surroundings. You plot your data as \(m_{\mathrm{w}}\) versus \(T^{-1}\), the inverse of the final temperature \(T\). Your data points lie close to a straight line that has slope \(2.15 \mathrm{~kg} \cdot \mathrm{C}^{\circ} .\) What does this result give for the value of the specific heat capacity of the unknown liquid?
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Rotational Dynamics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Work Energy and Power

Read Explanation

Radiation

Read Explanation

Fields in Physics

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.