/*! 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

A Carnot engine whose high-temperature reservoir is at 620 K takes in 550 J of heat at this temperature in each cycle and gives up 335 J to the low- temperature reservoir. (a) How much mechanical work does the engine perform during each cycle? What is (b) the temperature of the low-temperature reservoir; (c) the thermal efficiency of the cycle?

A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a large tub of ice and water. In 5 minutes of operation, the heat rejected by the engine melts 0.0400 kg of ice. During this time, how much work \(W\) is performed by the engine?

A gasoline engine takes in 1.61 \(\times\) 10\(^4\) J of heat and delivers 3700 J of work per cycle. The heat is obtained by burning gasoline with a heat of combustion of 4.60 \(\times\) 10\(^4\) J/g. (a) What is the thermal efficiency? (b) How much heat is discarded in each cycle? (c) What mass of fuel is burned in each cycle? (d) If the engine goes through 60.0 cycles per second, what is its power output in kilowatts? In horsepower?

(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 typical coal-fired power plant generates 1000 MW of usable power at an overall thermal efficiency of 40%. (a) What is the rate of heat input to the plant? (b) The plant burns anthracite coal, which has a heat of combustion of 2.65 \(\times\) 10\(^7\) J/kg. How much coal does the plant use per day, if it operates continuously? (c) At what rate is heat ejected into the cool reservoir, which is the nearby river? (d) The river is at 18.0\(^\circ\)C before it reaches the power plant and 18.5\(^\circ\)C after it has received the plant's waste heat. Calculate the river's flow rate, in cubic meters per second. (e) By how much does the river's entropy increase each second?

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