/*! 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] | 91Ó°ÊÓ

91Ó°ÊÓ

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°C to 37°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°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°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 91Ó°ÊÓ!

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 \( 1.05 \text{ J/g⋅°C} \)
  • \( \Delta T \) 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

The brakes on an automobile act by forcing brake pads, which have a metal support and a lining, to press against a disk (rotor) attached to the wheel. Friction between the pads and the disk causes the car to slow or stop. Each wheel has an iron brake disk with a mass of \(15 \mathrm{lb}_{\mathrm{m}}\) and two brake pads, each having a mass of \(11 \mathrm{b}_{\mathrm{m}}\). (a) Suppose an automobile is moving at 55 miles per hour when the driver suddenly applies the brakes and brings the car to a rapid halt. Take the heat capacity of the disk and brake pads to be \(0.12 \mathrm{Btu} /\left(\mathrm{lb}_{\mathrm{m}} \cdot^{\circ} \mathrm{F}\right)\) and assume that the car stops so rapidly that heat transfer from the disk and pads has been insignificant. Estimate the final temperature of the disk and pads if the car is (i) a Toyota Camry, which has a mass of about \(3200 \mathrm{Ib}_{\mathrm{m}},\) or (ii) a Cadillac Escalade, which has a mass of about \(5.900 \mathrm{lb}_{\mathrm{m}}.\) (b) Why are the linings on brake pads no longer made of asbestos? Your answer should provide information on specific issues or concerns caused by the use of asbestos.

Propane gas enters a continuous adiabatic heat exchanger \(^{17}\) at \(40^{\circ} \mathrm{C}\) and \(250 \mathrm{kPa}\) and exits at \(240^{\circ} \mathrm{C}\). Superheated steam at \(300^{\circ} \mathrm{C}\) and 5.0 bar enters the exchanger flowing countercurrently to the propane and exits as a saturated liquid at the same pressure. (a) Taking as a basis 100 mol of propane fed to the exchanger, draw and label a process flowchart. Include in your labeling the volume of propane fed \(\left(\mathrm{m}^{3}\right),\) the mass of steam fed \((\mathrm{kg}),\) and the volume of steam fed \(\left(\mathrm{m}^{3}\right)\) (b) Calculate values of the labeled specific enthalpies in the following inlet-outlet enthalpy table for this process. $$\begin{array}{|l|cc|cc|} \hline \text { Species } & n_{\text {in }} & \hat{H}_{\text {in }} & n_{\text {out }} & \hat{H}_{\text {out }} \\ \hline \mathrm{C}_{3} \mathrm{H}_{8} & 100 \mathrm{mol} & \hat{H}_{\mathrm{a}}(\mathrm{kJ} / \mathrm{mol}) & 100 \mathrm{mol} & \hat{H}_{\mathrm{c}}(\mathrm{kJ} / \mathrm{mol}) \\ \mathrm{H}_{2} \mathrm{O} & m_{\mathrm{w}}(\mathrm{kg}) & \hat{H}_{\mathrm{b}}(\mathrm{kJ} / \mathrm{kg}) & m_{\mathrm{w}}(\mathrm{kg}) & \hat{H}_{\mathrm{d}}(\mathrm{kJ} / \mathrm{kg}) \\ \hline \end{array}$$ (c) Use an energy balance to calculate the required mass feed rate of the steam. Then calculate the volumetric feed ratio of the two streams ( \(\mathrm{m}^{3}\) steam fed \(/ \mathrm{m}^{3}\) propane fed). Assume ideal-gas behavior for the propane but not the steam and recall that the exchanger is adiabatic. (d) Calculate the heat transferred from the water to the propane ( \(k J / m^{3}\) propane fed). (Hint: Do an energy balance on either the water or the propane rather than on the entire heat exchanger.) (e) Over a period of time, scale builds up on the heat-transfer surface, resulting in a lower rate of heat transfer between the propane and the steam. What changes in the outlet streams would you expect to see as a result of the decreased heat transfer?

Saturated steam at \(300^{\circ} \mathrm{C}\) is used to heat a countercurrently flowing stream of methanol vapor from \(65^{\circ} \mathrm{C}\) to \(260^{\circ} \mathrm{C}\) in an adiabatic heat exchanger. The flow rate of the methanol is 6500 standard liters per minute, and the steam condenses and leaves the heat exchanger as liquid water at \(90^{\circ} \mathrm{C}.\) (a) Calculate the required flow rate of the entering steam in \(\mathrm{m}^{3} / \mathrm{min}\). (b) Calculate the rate of heat transfer from the water to the methanol ( \(\mathrm{kW}\) ). (c) Suppose the outlet temperature of the methanol is measured and found to be \(240^{\circ} \mathrm{C}\) instead of the specified value of \(260^{\circ} \mathrm{C}\). List five possible realistic explanations for the \(20^{\circ} \mathrm{C}\) difference. 7 An adiabatic heat exchanger is one for which no heat is exchanged with the surroundings. All of the heat lost by the hot stream is transferred to the cold stream.

Molten sodium chloride is to be used as a constant-temperature bath for a high-temperature chemical reactor. Two hundred kilograms of solid \(\mathrm{NaCl}\) at \(300 \mathrm{K}\) is charged into an insulated vessel, and a 3000 kW electrical heater is turned on, raising the salt to its melting point of 1073 K and melting it at a constant pressure of 1 atm. (a) The heat capacity \(\left(C_{p}\right)\) of solid \(\mathrm{NaCl}\) is \(50.41 \mathrm{J} /(\mathrm{mol} \cdot \mathrm{K})\) at \(T=300 \mathrm{K},\) and \(53.94 \mathrm{J} /(\mathrm{mol} \cdot \mathrm{K})\) at \(T=500 \mathrm{K},\) and the heat of fusion of \(\mathrm{NaCl}\) at \(1073 \mathrm{K}\) is \(30.21 \mathrm{kJ} / \mathrm{mol} .\) Use these data to determine a linear expression for \(C_{p}(T)\) and to calculate \(\Delta \hat{H}\) ( \(\mathrm{kJ} / \mathrm{mol}\) ) for the transition of \(\mathrm{NaCl}\) from a solid at 300 K to a liquid at \(1073 \mathrm{K}\). (b) Write and solve the energy balance equation for this closed system isobaric process to determine the required heat input in kilojoules. (c) If \(85 \%\) of the full power of \(3000 \mathrm{kW}\) goes into heating and melting the salt, how long does the process take?

The heat capacity at constant pressure of hydrogen cyanide is given by the expression $$ C_{p}\left[J /\left(\mathrm{mol} \cdot^{\circ} \mathrm{C}\right)\right]=35.3+0.0291 T\left(^{\circ} \mathrm{C}\right) $$ (a) Write an expression for the heat capacity at constant volume for HCN, assuming ideal-gas behavior. (b) Calculate \(\Delta \hat{H}(\mathrm{J} / \mathrm{mol})\) for the constant- pressure process $$ \mathrm{HCN}\left(\mathrm{v}, 25^{\circ} \mathrm{C}, 0.80 \mathrm{atm}\right) \rightarrow \mathrm{HCN}\left(\mathrm{v}, 200^{\circ} \mathrm{C}, 0.80 \mathrm{atm}\right) $$(c) Calculate \(\Delta \hat{U}(\mathrm{J} / \mathrm{mol})\) for the constant- volume process $$\mathrm{HCN}\left(\mathrm{v}, 25^{\circ} \mathrm{C}, 50 \mathrm{m}^{3} / \mathrm{kmol}\right) \rightarrow \mathrm{HCN}\left(\mathrm{v}, 200^{\circ} \mathrm{C}, 50 \mathrm{m}^{3} / \mathrm{kmol}\right)$$ (d) If the process of Part (b) were carried out in such a way that the initial and final pressures were each 0.80 atm but the pressure varied during the heating, the value of \(\Delta \hat{H}\) would still be what you calculated assuming a constant pressure. Why is this so?

See all solutions

Recommended explanations on Chemistry Textbooks

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.