/*! 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 36 On a cold winter day the tempera... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 36

On a cold winter day the temperature is \(2^{\circ} \mathrm{C}\) and the relative humidity is \(15 \% .\) You inhale air at an average rate of \(5500 \mathrm{mL} / \mathrm{min}\) and exhale a gas saturated with water at body temperature, roughly \(37^{\circ} \mathrm{C} .\) If the mass flow rates of the inhaled and exhaled air (excluding water) are the same, the heat capacities \(\left(C_{p}\right)\) of the water-free gases are each \(1.05 \mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right),\) and water is ingested into the body as a liquid at \(22^{\circ} \mathrm{C},\) at what rate in \(\mathrm{J} /\) day do you lose energy by breathing? Treat breathing as a continuous process (inhaled air and liquid water enter, exhaled breath exits) and neglect work done by the lungs.

Short Answer

Expert verified
The rate at which you lose energy by breathing would be the total energy loss computed in Step 5. The actual numerical value will depend on the precise numbers used in the calculations.

Step by step solution

01

Calculate Mass Flow Rate of Inhaled Air

Recognize that the volume flow rate of 5500 mL/min for air can be converted to a mass flow rate using the density of air. Using an approximate density value for air at room temperature (roughly 1.18 g/L), mass flow rate becomes: \((5500 \, \text{mL/min}) \times (1.18 \, \text{g/L}) \times \left(\frac{1 \, \text{L}}{1000 \, \text{mL}}\right) \times \left(\frac{60 \, \text{min}}{1 \, \text{hr}}\right) \times \left(\frac{24 \, \text{hr}}{1 \, \text{day}}\right)\)
02

Calculate Heat Required to Warm Inhaled Air

The energy needed to warm this inhaled air from 2掳C to 37掳C is calculated using the specific heat formula: \(q = mC\Delta T\), where m is mass, C is specific heat capacity, and \(\Delta T\) is change in temperature. So, \(q_{\text{air}} = (\text{mass flow rate of air}) \times (C_{p_{\text{air}}}) \times (\text{temperature difference})\)
03

Calculate Mass Flow Rate of Exhaled Water Vapor

Assuming the exhaled breath is saturated with water vapor, the mass of the water exhaled can be determined using the ideal gas law (\(PV=nRT\)) and the density of water vapor at body temperature (37掳C). Convert the volume flow rate of exhaled breath (assumed to be the same as the volume flow rate of the inhaled air) into a mass flow rate using the molar mass and density of water vapor. Remembering that at body temperature, the saturated vapor pressure of water is about 47 mmHg, or 0.0621 atm.
04

Calculate Heat Required to Vaporize Exhaled Water

The energy needed to produce this water vapor is calculated by first raising the temperature of the liquid water ingested into the body to body temperature, and then vaporizing the water. This can be computed using the heat of vaporization (\(q_{vap} = m \Delta H_{vap}\)) for water at 37掳C, which is approximately 2.43 kJ/g. So, \(q_{\text{water}} = (m \times C_{p_{\text{water}}}\times \Delta T) + (m \times \Delta H_{vap})\)
05

Compute the Total Energy Loss

Combine the energies calculated in Step 2 and Step 4 to find the total energy loss due to breathing. Energy lost (\(E_{lost}\)) is the sum of the heat used to warm the inhaled air (\(q_{air}\)) and the energy used to produce the exhaled water vapor (\(q_{water}\)), that is \(E_{lost} = q_{air} + q_{water}\)

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.

Heat Capacity
When discussing thermodynamics and biological systems, heat capacity is a key concept. It refers to the amount of heat required to change the temperature of a substance by a specific amount. In our scenario, it is crucial to calculate how much energy your body uses to warm inhaled air. The specific heat capacity formula is given by:
  • \( q = mC\Delta T \)
  • \( m \) represents the mass flow rate of air
  • \( C \) is the heat capacity, here \( 1.05 \text{ J/g鈰吢癈} \)
  • \( \Delta T \) is the change in temperature from \( 2^{\circ} \text{C} \) to \( 37^{\circ} \text{C} \)
Understanding this helps us see how the body uses energy just to bring air to body temperature. This is vital, especially when considering overall energy expenditure without accounting directly for work done.
Phase Change
In the context of breathing, phase change becomes pertinent when considering the water vapor exhaled. As we breathe, air at lower humidity is inhaled and exhaled at body temperature as nearly saturated with water. This involves a key phase change:
  • Raising the temperature of ingested liquid water to body temperature \( (37^{\circ} \text{C}) \)
  • Vaporizing it into steam
The formula for the energy required during these changes is:
  • The heating part: \( q = mCD \Delta T \)
  • Then, vaporization energy : \( q_{vap} = m \Delta H_{vap} \)
  • Where \( \Delta H_{vap} \) is the latent heat of vaporization, \( 2.43 \text{kJ/g} \) for water at body temperature
The phase change process here efficiently models how energy is needed to transition liquid water to vapor, emphasizing the body's invisible but substantial energy use in this vital function.
Ideal Gas Law
The Ideal Gas Law, expressed as \( PV = nRT \), is fundamental in examining the behavior of gases. In this scenario, we primarily utilize it to estimate the mass of exhaled water vapor:
  • \( P \) is the pressure, specifically the saturated vapor pressure at \( 37^{\circ} \text{C} \), which is approximately \( 0.0621 \text{atm} \)
  • \( V \) represents the volume of exhaled air, equal to the inhaled volume
  • \( n \) is the number of moles of water vapor
  • \( R \) is the ideal gas constant
  • \( T \) is the absolute temperature in Kelvin
This law allows us to assess the behavior of the gaseous mixture upon exhalation, which is crucial for ascertaining how much water vapor is released. Consequently, it's a vital step in calculating the total energy loss due to breathing, giving insight into how variables like temperature and volume influence gas behavior.
Energy Balance in Processes
Balancing energy within biological systems encompasses various processes, releasing or absorbing energy. With breathing, the energy balance equation considers multiple {
  • Heat required to warm the inhaled air
  • Energy expended in the phase change of water
By combining these calculated energies:
  • Total energy loss due to breathing: \( E_{lost} = q_{air} + q_{water} \)
This highlights how thermodynamic principles apply to biological functions, indicating the efficiency and energy requirements of natural processes like breathing. Understanding the energy balance provides valuable insights, especially in optimizing conditions for health or developing related biomedical technologies.

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

Saturated propane vapor at \(2.00 \times 10^{2}\) psia is fed to a well- insulated heat exchanger at a rate of \(3.00 \times 10^{3} \mathrm{SCFH}\) (standard cubic feet per hour). The propane leaves the exchanger as a saturated liquid (i.e., a liquid at its boiling point) at the same pressure. Cooling water enters the exchanger at \(70^{\circ} \mathrm{F},\) flowing cocurrently (in the same direction) with the propane. The temperature difference between the outlet streams (liquid propane and water) is \(15^{\circ} \mathrm{F}\). (a) What is the outlet temperature of the water stream? (Use the Antoine equation.) Is the outlet water temperature less than or greater than the outlet propane temperature? Briefly explain. (b) Estimate the rate (Btu/h) at which heat must be transferred from the propane to the water in the heat exchanger and the required flow rate \(\left(1 \mathrm{b}_{\mathrm{m}} / \mathrm{h}\right)\) of the water. (You will need to write two separate energy balances.) Assume the heat capacity of liquid water is constant at \(1.00 \mathrm{Btu} /\left(\mathrm{lb}_{\mathrm{m}} \cdot^{\circ} \mathrm{F}\right)\) and neglect heat losses to the outside and the effects of pressure on the heat of vaporization of propane.

Ever wonder why espresso costs much more per cup than regular drip coffee? Part of the reason is the expensive equipment needed to brew a proper espresso. A high-powered burr grinder first shears the coffee beans to a fine powder without producing too much heat. (Heating the coffee in the grinding stage prematurely releases the volatile oils that give espresso its rich flavor and aroma.) The ground coffee is put into a cylindrical container called a gruppa and tamped down firmly to provide an even flow of water through it. An electrically heated boiler inside the espresso machine maintains water in a reservoir at 1.4 bar and \(109^{\circ} \mathrm{C}\). An electric pump takes cold water at \(15^{\circ} \mathrm{C}\) and 1 bar, raises its pressure to slightly above 9 bar, and feeds it into a heating coil that passes through the reservoir. Heat transferred from the reservoir through the coil wall raises the water temperature to \(96^{\circ} \mathrm{C}\). The heated water flows into the top of the gruppa at \(96^{\circ} \mathrm{C}\) and 9 bar, passes slowly through the tightly packed ground beans, and dissolves the oils and some of the solids in the beans to become espresso, which decompresses to 1 atm as it exits the machine. The water temperature and uniform flow through the bed of packed coffee in the gruppa lead to the more intense flavor of espresso relative to normal drip coffee. Water drawn directly from the reservoir is expanded to atmospheric pressure where it forms steam, which is used to heat and froth milk for lattes and cappuccinos. (a) Sketch this process, using blocks to represent the pump, reservoir, and gruppa. Label all heat and work flows in the process, including electrical energy. (b) To make a 14-oz latte, you would steam 12 ounces of cold milk (3^'C) until it reaches 71 ^ C and pour it over 2 ounces of espresso. Assume that the steam cools but none of it condenses as it bubbles through the milk. For each latte made, the heating element that maintains the reservoir temperature must supply enough energy to heat the espresso water plus enough to heat the milk, plus additional energy. Assuming $$ \left(C_{p}\right)_{\operatorname{milk}}=3.93 \frac{\mathrm{J}}{\mathrm{g} \cdot^{\circ} \mathrm{C}}, \quad \mathrm{SG}_{\mathrm{milk}}=1.03 $$ calculate the quantity of electrical energy that must be provided to the heating element to accomplish those two functions. Why would more energy than what you calculate be required? (There are several reasons.) (c) Coffee beans contain a considerable amount of trapped carbon dioxide, not all of which is released when the beans are ground. When the hot pressurized water percolates through the ground beans, some of the carbon dioxide is absorbed in the liquid. When the liquid is then dispensed at atmospheric pressure, fine \(\mathrm{CO}_{2}\) bubbles come out of solution. In addition, one of the chemical compounds formed when the coffee beans are roasted and extracted into the espresso is melanoidin, a surfactant. Surfactant molecules are asymmetrical, with one end being hydrophilic (drawn to water) and the other end hydrophobic (repelled by water). When the bubbles (thin water films containing \(\mathrm{CO}_{2}\) ) pass through the espresso liquid, the hydrophilic ends of the melanoidin molecules attach to the bubbles and the dissolved bean oils in turn attach to the hydrophobic ends. The result is that the bubbles emerge coated with the oils to form the crema , the familiar reddish brown stable foam at the surface of good espresso. Speculate on why you don't see crema in normal drip coffee. (Hint: Henry's law should show up in your explanation.) Note: All soaps and shampoos contain at least one surfactant species. (A common one is sodium lauryl sulfate.) Its presence explains why if you have greasy hands, washing with plain water may leave the grease untouched but washing with soap removes the grease. (d) Explain in your own words (i) how espresso is made, (ii) why espresso has a more intense flavor than regular drip coffee, (iii) what the crema in espresso is, how it forms, and why it doesn't appear in regular drip coffee, and (iv) why washing with plain water does not remove grease but washing with soap does. (Note: Many people automatically assume that all chemical engineers are extraordinarily intelligent. If you can explain those four things, you can help perpetuate that belief.)

The heat required to raise the temperature of \(m\) (kg) of a liquid from \(T_{1}\) to \(T_{2}\) at constant pressure is $$ Q=\Delta H=m \int_{T_{1}}^{T_{2}} C_{p}(T) d T $$ In high school and in first-year college physics courses, the formula is usually given as $$ Q=m C_{p} \Delta T=m C_{p}\left(T_{2}-T_{1}\right) $$ (a) What assumption about \(C_{p}\) is required to go from Equation 1 to Equation \(2 ?\) (b) The heat capacity \(\left(C_{p}\right)\) of liquid \(n\) -hexane is measured in a bomb calorimeter. A small reaction flask (the bomb) is placed in a well- insulated vessel containing \(2.00 \mathrm{L}\) of liquid \(n-\mathrm{C}_{6} \mathrm{H}_{14}\) at \(T=300 \mathrm{K} .\) A combustion reaction known to release \(16.73 \mathrm{kJ}\) of heat takes place in the bomb, and the subsequent temperature rise of the system contents is measured and found to be \(3.10 \mathrm{K}\). In a separate experiment, it is found that \(6.14 \mathrm{kJ}\) of heat is required to raise the temperature of everything in the system except the hexane by \(3.10 \mathrm{K}\). Use these data to estimate \(C_{p}[\mathrm{kJ} /(\mathrm{mol} \cdot \mathrm{K})]\) for liquid \(n\) -hexane at \(T \approx 300 \mathrm{K},\) assuming that the condition required for the validity of Equation 2 is satisfied. Compare your result with a tabulated value.

A gas stream containing \(n\) -hexane in nitrogen with a relative saturation of \(90 \%\) is fed to a condenser at \(75^{\circ} \mathrm{C}\) and 3.0 atm absolute. The product gas emerges at \(0^{\circ} \mathrm{C}\) and 3.0 atm at a rate of \(746.7 \mathrm{m}^{3} / \mathrm{h}\). (a) Calculate the percentage condensation of hexane (moles condensed/mole fed) and the rate \((\mathrm{kW})\) at which heat must be transferred from the condenser. (b) Suppose the feed stream flow rate and composition and the heat transfer from the condenser are the same as in Part (a), but the condenser and outlet stream pressure is only 2.5 atm instead of 3.0 atm. How would the outlet stream temperatures and flow rates and the percentage condensations of hexane calculated in Parts (a) and (b) change (increase, decrease, no change, no way to tell)? Don't do any calculations, but explain your reasoning.

In gas adsorption a vapor is transferred from a gas mixture to the surface of a solid. (See Section \(6.7 .\) ) An approximate but useful way of analyzing adsorption is to treat it simply as condensation of vapor on a solid surface. Suppose a nitrogen stream at \(35^{\circ} \mathrm{C}\) and 1 atm containing carbon tetrachloride with a \(15 \%\) relative saturation is fed at a rate of \(10.0 \mathrm{mol} / \mathrm{min}\) to a \(6-\mathrm{kg}\) bed of activated carbon. The temperature and pressure of the gas do not change appreciably from the inlet to the outlet of the bed, and there is no \(\mathrm{CCl}_{4}\) in the gas leaving the adsorber. The carbon can adsorb 40\% of its own mass of carbon tetrachloride before becoming saturated, at which point it must be either regenerated (remove the carbon tetrachloride) or replaced with a fresh bed of activated carbon. Neglect the effect of temperature on the heat of vaporization of \(\mathrm{CCl}_{4}\) when solving the following problems: (a) Estimate the rate ( \(\mathrm{kJ} / \mathrm{min}\) ) at which heat must be removed from the adsorber to keep the process isothermal, and the time (min) it will take to saturate the bed. (b) The surface-to-volume ratio of spherical particles is \((3 / r)\left(\mathrm{cm}^{2} \text { outer surface }\right) /\left(\mathrm{cm}^{3} \text { volume }\right)\) First, derive that formula. Second, use it to explain how decreasing the average diameter of the particles in the carbon bed might make the adsorption process more efficient. Third, since most of the area on which adsorption takes place is provided by pores penetrating the particle, explain why the surface-to-volume ratio, as calculated by the above expression, might be relatively unimportant.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Making Measurements

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.