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In very cold weather a significant mechanism for heat loss by the human body is energy expended in warming the air taken into the lungs with each breath. (a) On a cold winter day when the temperature is \(-20^{\circ} \mathrm{C},\) what amount of heat is needed to warm to body temperature \(\left(37^{\circ} \mathrm{C}\right)\) the \(0.50 \mathrm{~L}\) of air exchanged with each breath? Assume that the specific heat of air is \(1020 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and that \(1.0 \mathrm{~L}\) of air has mass \(1.3 \times 10^{-3} \mathrm{~kg} .\) (b) How much heat is lost per hour if the respiration rate is 20 breaths per minute?

Short Answer

Expert verified
The total heat loss by the body due to respiration on a day when the temperature is -20 degrees Celsius is approximately 45288 J/hour.

Step by step solution

01

Convert Temperatures to Kelvin

Before calculations, convert all temperature values into Kelvin, as it is the standard unit used in thermodynamics. Temperature in Kelvin can be obtained by adding 273.15 to the temperature in Celsius. Thus, \(T_{initial} = -20^{\circ}C + 273.15 = 253.15 K \) and \(T_{final}= 37^{\circ}C + 273.15 = 310.15 K\).
02

Calculate the Change in Temperature

Calculate the change in temperature (\(\Delta T\)) which is the difference between the final temperature and the initial temperature. \(\Delta T = T_{final} - T_{initial} = 310.15 K - 253.15 K = 57 K\).
03

Calculate the Mass of Air Per Breath

Given that 1.0 L of air has mass 1.3 x 10^-3 kg, then 0.50 L of air (the volume exchanged with each breath) would have mass \(0.50 L * 1.3 x 10^{-3} kg/L = 0.65 x 10^{-3} kg\).
04

Calculate Heat Energy Per Breath

Using the formula \(q = mc\Delta T\), where \(m\) is the mass of the air, \(c\) is the specific heat of air and \(\Delta T\) is the change in temperature, plug the known values and calculate the heat energy per breath: \(q = 0.65 x 10^{-3} kg * 1020 J/kg.K * 57 K = 37.74 J\).
05

Calculate Total Heat Energy Lost Per Hour

To calculate the heat energy lost per hour, multiply the heat energy per breath, by the number of breaths taken per minute, and then by the number of minutes in one hour. Given that the respiration rate is 20 breaths per minute, then the total heat loss per hour is \(37.74 J/breath * 20 breaths/minute * 60 minutes/hour = 45288 J/hour\).

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Key Concepts

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

Understanding Specific Heat Capacity
Specific heat capacity is a fundamental concept in thermodynamics. It refers to the amount of heat required to raise the temperature of one kilogram of a substance by one Kelvin. In this exercise, the specific heat capacity of air is given as 1020 J/kg·K. This means that for every kilogram of air, you need 1020 Joules of energy to increase its temperature by one Kelvin.

To compute the heat needed to warm the air in the lungs, you multiply the specific heat capacity by the mass of air and the change in temperature. This tells us how much energy is required for each breath, focusing on the warming process. Understanding the concept of specific heat capacity helps in calculating energy transfer in various thermodynamic processes, not just in biological systems like respiration.
Respiration and Heat Loss
Respiration is more than just a process of exchanging gases; it's also a notable source of heat loss in cold environments. When you breathe in cold air, your body must expend energy to heat it up to body temperature. This energy loss can be significant, especially in cold climates, and needs to be replenished to maintain a stable core body temperature.

In the exercise, the problem requires calculating the heat loss associated with warming 0.50 liters of air each breath from -20°C to 37°C. By understanding the rate of breathing and the heat required per breathing cycle, we can determine the total heat loss over time, such as per hour, which reflects the body's efforts to maintain homeostasis. This illustrates a practical application of thermodynamics in explaining biological processes.
Temperature Conversion in Thermodynamics
Temperature conversion is crucial when dealing with thermodynamic problems because calculations often require using the Kelvin scale. The Kelvin scale is the standard in thermodynamics due to its absolute nature, starting from absolute zero, where particles have minimum thermal movement.

Converting from Celsius to Kelvin involves adding 273.15 to the Celsius temperature. For example, -20°C becomes 253.15 K, and 37°C becomes 310.15 K. This conversion allows for proper calculation of temperature change, which is used in further calculations such as determining the heat energy. Understanding how and why to convert temperatures is essential for accurate thermodynamic analysis and ensures consistency across calculations.
Calculating Energy Requirements
The energy calculation is the culmination of understanding specific heat capacity, temperature conversion, and mass of the substance. To find how much heat energy is required to increase the temperature of the air, we use the formula: \[ q = mc\Delta T \]In this formula, \(q\) represents the heat energy, \(m\) is the mass of air, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature.

In the given exercise, substituting in the appropriate numbers allows us to calculate the heat per breath. We then scale this up to find the total heat loss per hour based on breathing rates. This systematic approach highlights the interdependence of multiple thermodynamic principles and illustrates how energy conservation is maintained in physical and biological systems.

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Most popular questions from this chapter

A metal sphere with radius \(3.20 \mathrm{~cm}\) is suspended in a large metal box with interior walls that are maintained at \(30.0^{\circ} \mathrm{C}\). A small electric heater is embedded in the sphere. Heat energy must be supplied to the sphere at the rate of \(0.660 \mathrm{~J} / \mathrm{s}\) to maintain the sphere at a constant temperature of \(41.0^{\circ} \mathrm{C}\). (a) What is the emissivity of the metal sphere? (b) What power input to the sphere is required to maintain it at \(82.0^{\circ} \mathrm{C}\) ? What is the ratio of the power required for \(82.0^{\circ} \mathrm{C}\) to the power required for \(41.0^{\circ} \mathrm{C}\) ? How does this ratio compare with \(2^{4}\) ? Explain.

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?

You have \(1.50 \mathrm{~kg}\) of water at \(28.0^{\circ} \mathrm{C}\) in an insulated container of negligible mass. You add \(0.600 \mathrm{~kg}\) of ice that is initially at \(-22.0^{\circ} \mathrm{C}\). Assume that no heat exchanges with the surroundings. (a) After thermal equilibrium has been reached, has all of the ice melted? (b) If all of the ice has melted, what is the final temperature of the water in the container? If some ice remains, what is the final temperature of the water in the container, and how much ice remains?

An aluminum tea kettle with mass \(1.10 \mathrm{~kg}\) and containing \(1.80 \mathrm{~kg}\) of water is placed on a stove. If no heat is lost to the surroundings, how much heat must be added to raise the temperature from \(20.0^{\circ} \mathrm{C}\) to \(85.0^{\circ} \mathrm{C} ?\)

A \(25,000 \mathrm{~kg}\) subway train initially traveling at \(15.5 \mathrm{~m} / \mathrm{s}\) slows to a stop in a station and then stays there long enough for its brakes to cool. The station's dimensions are \(65.0 \mathrm{~m}\) long by \(20.0 \mathrm{~m}\) wide by \(12.0 \mathrm{~m}\) high. Assuming all the work done by the brakes in stopping the train is transferred as heat uniformly to all the air in the station, by how much does the air temperature in the station rise? Take the density of the air to be \(1.20 \mathrm{~kg} / \mathrm{m}^{3}\) and its specific heat to be \(1020 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

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