/*! 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 52 A person with skin of surface ar... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 20: Problem 52

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.)

Short Answer

Expert verified
The entropy of the air increases by the rate determined by \( \Delta S = \frac{P}{T_2} \) using the radiated power.

Step by step solution

01

Understand the Concept of Radiation

When a person is resting in a room with an air temperature lower than their skin temperature, they lose heat through radiation. According to the Stefan-Boltzmann law, the power radiated by a blackbody is proportional to the fourth power of its absolute temperature.
02

Use the Stefan-Boltzmann Law

The Stefan-Boltzmann law is given by the equation \( P = \sigma \epsilon A (T_1^4 - T_2^4) \), where \( P \) is the power radiated, \( \sigma = 5.67 \times 10^{-8} \text{ W/m}^2 \text{K}^4 \) is the Stefan-Boltzmann constant, \( \epsilon = 1.00 \) is the emissivity, \( A = 1.85 \text{ m}^2 \) is the surface area, \( T_1 = 303 \text{ K} \) (person's skin temperature), and \( T_2 = 293 \text{ K} \) (room temperature).
03

Calculate the Radiated Power

Substitute the values into the Stefan-Boltzmann equation: \( P = 5.67 \times 10^{-8} \times 1 \times 1.85 (303^4 - 293^4) \). Compute the result to get the power in watts.
04

Calculate the Change in Entropy of the Air

The change in entropy \( \Delta S \) of the air can be calculated using the formula \( \Delta S = \frac{P}{T_2} \) where \( P \) is the power radiated and \( T_2 = 293 \text{ K} \). Solve it using the result from Step 3.

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.

Radiation
Radiation is a fundamental form of heat transfer that occurs through electromagnetic waves. This process does not require a medium, meaning it can even happen in a vacuum. When considering humans, we radiate heat energy into our surroundings naturally.
  • Blackbody Radiation: A perfect emitter, known as a blackbody, emits the maximum possible radiation at a given temperature. Human skin is not a perfect blackbody, but for ease of calculations, we often use it as such in physics exercises. In these cases, skin emissivity (\(\epsilon\)) is considered to be 1.00.
  • Stefan-Boltzmann Law: This law helps compute the power radiated by a body, given by the equation \(P = \sigma \epsilon A (T_1^4 - T_2^4)\). Here,
    • \(\sigma = 5.67 \times 10^{-8} \text{ W/m}^2 \text{K}^4\)
    • \(A\) is the surface area of the body
    • \(T_1\) and \(T_2\) are the absolute temperatures of the body and surroundings respectively.
The difference \(T_1^4 - T_2^4\) accounts for the net heat loss due to the body radiating more than it receives from its cooler surroundings.
Entropy Change
Entropy is a measure of disorder or randomness in a system and is a central concept in thermodynamics.

When heat flows from a hot body to a cooler one, the total entropy of the system increases, as heat energy disperses.
  • Calculating Entropy Change: The change in entropy (\(\Delta S\)) is determined using the power (\(P\)) and the temperature of the surroundings (\(T_2\)). The formula is \(\Delta S = \frac{P}{T_2}\). Here,
    • \(P\) is the power calculated using the Stefan-Boltzmann Law.
    • \(T_2\) is the temperature of the surroundings in Kelvin, ensuring the units are consistent with the power value.
Entropy change is key to understanding how systems reach equilibrium, enhancing our grasp of energy efficiency and natural processes.
Heat Transfer
Heat transfer is a process where energy in the form of heat moves from a warmer object to a cooler one.

There are three main mechanisms for heat transfer: conduction, convection, and radiation. However, when it comes to radiation, this transfer happens through electromagnetic waves.
  • Methods of Heat Transfer: While conduction requires direct contact between materials, convection involves the movement of fluid or gas. Radiation, on the other hand, can happen without any medium. It鈥檚 why we feel sun's heat even across space.
  • Real-Life Application: Humans constantly exchange heat via radiation with their surroundings. It explains how we can feel our body warmth even in a room that is cooler than our body temperature.
In the exercise context, understanding these three modes of heat transfer is essential to appreciate how we naturally cool down in an environment where the air temperature is lower than our body temperature.

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

You make tea with 0.250 kg of 85.0\(^\circ\)C water and let it cool to room temperature (20.0\(^\circ\)C). (a) Calculate the entropy change of the water while it cools. (b) The cooling process is essentially isothermal for the air in your kitchen. Calculate the change in entropy of the air while the tea cools, assuming that all of the heat lost by the water goes into the air. What is the total entropy change of the system tea + 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?

An ideal Carnot engine operates between 500\(^\circ\)C and 100\(^\circ\)C with a heat input of 250 J per cycle. (a) How much heat is delivered to the cold reservoir in each cycle? (b) What minimum number of cycles is necessary for the engine to lift a 500-kg rock through a height of 100 m?

A certain brand of freezer is advertised to use 730 kW \(\cdot\) h of energy per year. (a) Assuming the freezer operates for 5 hours each day, how much power does it require while operating? (b) If the freezer keeps its interior at -5.0\(^\circ\)C in a 20.0\(^\circ\)C room, what is its theoretical maximum performance coefficient? (c) What is the theoretical maximum amount of ice this freezer could make in an hour, starting with water at 20.0\(^\circ\)C?

A Carnot engine is operated between two heat reservoirs at temperatures of 520 \(K\) and 300 \(K\). (a) If the engine receives 6.45 \(kJ\) of heat energy from the reservoir at 520 \(K\) in each cycle, how many joules per cycle does it discard to the reservoir at 300 \(K\)? (b) How much mechanical work is performed by the engine during each cycle? (c) What is the thermal efficiency of the engine?
See all solutions

Recommended explanations on Physics Textbooks

Radiation

Read Explanation

Kinematics Physics

Read Explanation

Thermodynamics

Read Explanation

Particle Model of Matter

Read Explanation

Classical Mechanics

Read Explanation

Waves Physics

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.