/*! 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 34 An average sleeping person metab... [FREE SOLUTION] | 91Ó°ÊÓ

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An average sleeping person metabolizes at a rate of about 80 \(W\) by digesting food or burning fat. Typically, 20% of this energy goes into bodily functions, such as cell repair, pumping blood, and other uses of mechanical energy, while the rest goes to heat. Most people get rid of all this excess heat by transferring it (by conduction and the flow of blood) to the surface of the body, where it is radiated away. The normal internal temperature of the body (where the metabolism takes place) is 37\(^\circ\)C, and the skin is typically 7 C\(^\circ\) cooler. By how much does the person's entropy change per second due to this heat transfer?

Short Answer

Expert verified
The entropy change per second is approximately 0.211 J/K.

Step by step solution

01

Understanding the Problem

The problem requires us to calculate the change in entropy for a sleeping person who transfers heat to the skin at a temperature of 30°C (since the internal temperature is 37°C and the skin is typically 7°C cooler). The metabolism rate is given, and we know that 80% of it translates into heat transfer. We should express temperatures in Kelvin for entropy calculations.
02

Identify Energy Transfer

The metabolism rate of a sleeping person is 80 W, of which 80% becomes excess heat: \(80 \times 0.8 = 64\) W. This indicates the rate at which heat is being transferred from the person's body.
03

Convert Temperature to Kelvin

The skin temperature in Celsius is 30°C (37°C - 7°C). Convert this to Kelvin: \(T_{skin} = 30 + 273.15 = 303.15\) K.
04

Calculate Heat Transfer in Terms of Entropy

The change in entropy \(\Delta S\) per second due to heat transfer can be calculated using: \[\Delta S = \frac{Q}{T} \times \Delta t\]where \(Q\) is the heat transfer rate (64 J/s), \(T\) is the temperature of the skin in Kelvin (303.15 K), and \(\Delta t\) is one second.
05

Perform the Entropy Change Calculation

Substitute known values into the entropy change formula:\[\Delta S = \frac{64}{303.15} \times 1 = 0.211 \text{ J/K}\]This is the rate of entropy change per second for the heat transfer from the body to the skin.

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Key Concepts

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

Heat Transfer
Heat transfer is a fundamental concept that occurs when thermal energy is exchanged between physical systems. In the context of the provided exercise, it involves the movement of heat from a person's body to their skin. This process is essential for regulating body temperature and maintaining homeostasis.
In a sleeping person, 80% of the metabolism energy is converted into heat. This means that if the total metabolic rate is 80 watts, 64 watts of energy is transferred as heat to the skin. This heat is then dissipated into the surrounding environment, primarily by thermal radiation.
Understanding heat transfer in this way helps us appreciate how our bodies naturally manage to stay cool even when generating significant amounts of heat. This heat transfer is crucial to keeping our body at an optimal operating temperature.
Metabolism
Metabolism is the set of life-sustaining chemical reactions that occur within our bodies. It helps convert food into energy that powers bodily functions, from basic cellular processes to physical movement.
In the exercise example, even while sleeping, a person metabolizes at a rate of about 80 watts. This metabolic rate reflects the energy the body utilizes per second to keep vital functions going, such as cell repair and blood circulation. Not all energy from metabolism is used directly; a substantial portion is converted into heat.
This heat is what the body needs to manage beyond basic physiological tasks, and managing this heat through transfer to the skin is part of the body's natural cooling system. Also, these processes illustrate how metabolism not only provides energy but also influences thermal regulation.
Thermal Radiation
Thermal radiation is a mode of heat transfer that involves the emission of electromagnetic waves. It occurs when the body's surface emits infrared radiation, allowing heat to dissipate into the surrounding environment.
The exercise highlights this process as the primary way a person's excess metabolic heat is released. As heat moves to the skin, it radiates into the air, reducing the body's internal heat load.
Understanding thermal radiation is vital as it plays a key role in how living organisms regulate temperature. This natural mechanism does not require direct contact with other objects or conductors, making it an effective and efficient means of heat transfer.
Temperature Conversion
Temperature conversion is an important aspect when dealing with thermodynamic calculations, like the entropy change due to heat transfer. In the exercise, we need to convert temperature from Celsius to Kelvin to use in entropy calculations.
Kelvin is the SI unit for temperature used in scientific calculations, especially when dealing with formulas that involve thermodynamics. The conversion from Celsius to Kelvin is straightforward: simply add 273.15 to the Celsius temperature.
In our example, the skin temperature at 30°C converts to 303.15 Kelvin. By using these converted temperatures, we can accurately apply thermodynamic equations to determine the entropy change due to heat transfer. This ensures standardized results across various scientific disciplines.

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Most popular questions from this chapter

Digesting fat produces 9.3 food calories per gram of fat, and typically 80% of this energy goes to heat when metabolized. (One food calorie is 1000 calories and therefore equals 4186 J.) The body then moves all this heat to the surface by a combination of thermal conductivity and motion of the blood. The internal temperature of the body (where digestion occurs) is normally 37\(^\circ\)C, and the surface is usually about 30\(^\circ\)C. By how much do the digestion and metabolism of a 2.50-g pat of butter change your body's entropy? Does it increase or decrease?

(a) Calculate the theoretical efficiency for an Otto-cycle engine with \(\gamma\) = 1.40 and \(r\) = 9.50. (b) If this engine takes in 10,000 J of heat from burning its fuel, how much heat does it discard to the outside air?

A box is separated by a partition into two parts of equal volume. The left side of the box contains 500 molecules of nitrogen gas; the right side contains 100 molecules of oxygen gas. The two gases are at the same temperature. The partition is punctured, and equilibrium is eventually attained. Assume that the volume of the box is large enough for each gas to undergo a free expansion and not change temperature. (a) On average, how many molecules of each type will there be in either half of the box? (b) What is the change in entropy of the system when the partition is punctured? (c) What is the probability that the molecules will be found in the same distribution as they were before the partition was punctured- that is, 500 nitrogen molecules in the left half and 100 oxygen molecules in the right half?

A 15.0-kg block of ice at 0.0\(^\circ\)C melts to liquid water at 0.0\(^\circ\)C inside a large room at 20.0\(^\circ\)C. Treat the ice and the room as an isolated system, and assume that the room is large enough for its temperature change to be ignored. (a) Is the melting of the ice reversible or irreversible? Explain, using simple physical reasoning without resorting to any equations. (b) Calculate the net entropy change of the system during this process. Explain whether or not this result is consistent with your answer to part (a).

A person with skin of surface area 1.85 m\(^2\) and temperature 30.0\(^\circ\)C is resting in an insulated room where the ambient air temperature is 20.0\(^\circ\)C. In this state, a person gets rid of excess heat by radiation. By how much does the person change the entropy of the air in this room each second? (Recall that the room radiates back into the person and that the emissivity of the skin is 1.00.)

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