/*! 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 3 The normal body-temperature of a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 28: Problem 3

The normal body-temperature of a person is \(97^{\circ} \mathrm{F}\). Calculate the rate at which heat is flowing out of his body through the clothes assuming the following values. Room temperature \(=47^{\circ} \mathrm{F}\), surface of the body under clothes \(=1 \cdot 6 \mathrm{~m}^{2}, \quad\) conductivity of the cloth \(=0 \cdot 04 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-10} \mathrm{C}^{-1}\), thickness of the cloth \(=0.5 \mathrm{~cm}\)

Short Answer

Expert verified
The rate of heat flow is approximately -356.57 J/s.

Step by step solution

01

- Identify the variables

First, identify the variables that are provided and those that are needed for calculating the rate of heat flow. We need:- Surface area under clothes, \(A = 1.6 \, \text{m}^2\).- Thermal conductivity, \(k = 0.04 \, \text{J s}^{-1} \text{m}^{-1} \text{C}^{-1}\).- Temperature difference, \(\Delta T = 97^{\circ} \text{F} - 47^{\circ} \text{F}\).- Thickness of cloth, \(d = 0.5 \, \text{cm} \) or \(0.005 \, \text{m}\) (converted to meters).
02

- Convert temperature difference to degrees Celsius

The temperature difference will be calculated in Celsius, because the formula for heat flow rate requires temperatures in this unit. Remember that Fahrenheit to Celsius conversion can be done by \( T_{C} = \frac{5}{9}(T_{F} - 32) \). Calculate for each temperature and find the difference.
03

- Apply the heat flow rate formula

Use the formula for the rate of heat flow: \[ \frac{Q}{t} = -k A \frac{dT}{dx} \] where \( \frac{dT}{dx} = \frac{\Delta T}{d} \). Substitute the given values: \( k = 0.04, A = 1.6, \Delta T = \frac{5}{9}(97 - 47), d = 0.005 \).
04

- Calculate temperature difference in Celsius

Using the conversion \( \Delta T = \frac{5}{9}(97 - 47) = \frac{5}{9} \times 50 = 27.78^{\circ} \text{C}\)
05

- Calculate the heat flow rate

Plug in all values: \[ \frac{Q}{t} = -0.04 \times 1.6 \times \frac{27.78}{0.005} \] Calculate: \[ \frac{Q}{t} \approx -0.04 \times 1.6 \times 5556 = -356.57 \, \text{J/s} \]

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.

Thermal Conductivity
Thermal conductivity is a property of a material that indicates its ability to conduct heat. For example, metal has high thermal conductivity, which means it transfers heat quickly. In contrast, materials like wool or cotton have low thermal conductivity and are good insulators. This property is important in our scenario, where clothing is the barrier between body heat and the cooler room environment.

When you have a material like cloth, its thermal conductivity determines how easily heat can move through it. The formula for calculating heat flow through a material involves its thermal conductivity (\(k\)), surface area (\(A\)), and thickness (\(d\)). These factors are crucial when determining how much heat is transferred from one side of the cloth to the other. In our exercise, the cloth鈥檚 thermal conductivity is given as 0.04 J s\(^{-1}\) m\(^{-1}\) C\(^{-1}\). This value helps us understand how well the cloth can transfer heat.

Understanding thermal conductivity can assist in designing clothes for different climates or ensuring comfortable indoor temperatures. It is paramount in physics, particularly when assessing heat exchange efficiency between objects.
Temperature Conversion
Temperature conversion is crucial when solving problems involving heat transfer, especially when working with formulas that require temperatures in a specific unit like Celsius. In the United States, Fahrenheit is often used, whereas, in most other countries Celsius is preferred. Thus, converting between these two scales is frequently necessary.

To convert from Fahrenheit to Celsius, the equation used is:\[ T_{C} = \frac{5}{9}(T_{F} - 32) \]where \(T_{F}\) is the temperature in Fahrenheit and \(T_{C}\) is the temperature in Celsius. This formula helps standardize measurements so they can be effectively used in scientific calculations.
  • For instance, converting 97掳F (body temperature) results in about 36.11掳C.
  • Similarly, room temperature of 47掳F converts to 8.33掳C.
Calculating the difference between these values in Celsius (\(\Delta T\)) is essential for applying the heat flow equation correctly, as seen in this exercise. Mastering temperature conversion is a fundamental skill in physics that ensures precise measurements and accurate results.
Heat Flow Rate Formula
The heat flow rate formula quantifies how much heat is transferred through a material over time. This is particularly relevant for scenarios where a temperature difference exists across a medium, such as clothing, walls, or building insulation. The basic formula is:\[ \frac{Q}{t} = -k A \frac{dT}{dx} \]where:
  • \( \frac{Q}{t} \) is the heat flow rate (J/s), representing how much heat is transferred per second.
  • \(k\) is the material鈥檚 thermal conductivity (J s\(^{-1}\) m\(^{-1}\) C\(^{-1}\)).
  • \(A\) is the surface area through which heat is transferring (m\(^2\)).
  • \(\frac{dT}{dx} = \frac{\Delta T}{d}\) represents the temperature gradient, where \(\Delta T\) is the temperature difference (C) and \(d\) is the thickness of the material (m).
In the exercise, knowing these variables allows us to calculate that the rate at which heat escapes through clothing is approximately -356.57 J/s. The negative sign indicates that heat is leaving the body. This formula is a cornerstone in thermodynamics and helps us understand how heat moves in various contexts. Recognizing the components of this equation allows better manipulation and understanding of heat transfer scenarios.

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

Three rods of lengths \(20 \mathrm{~cm}\) each and area of cross section \(1 \mathrm{~cm}^{2}\) are joined to form a triangle \(A B C\). The conductivities of the rods are \(K_{A B}=50 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-10} \mathrm{C}^{-1}\), \(K_{B C}=200 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1_{0}} \mathrm{C}^{-1}\) and \(K_{A C}=400 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1_{0}} \mathrm{C}^{-1}\). The junctions \(A, B\) and \(C\) are maintained at \(40^{\circ} \mathrm{C}, 80^{\circ} \mathrm{C}\) and \(80^{\circ} \mathrm{C}\) respectively. Find the rate of heat flowing through the rods \(A B, A C\) and \(B C\)

The left end of a copper rod (length \(=20 \mathrm{~cm}\), area of cross section \(\left.=0-20 \mathrm{~cm}^{2}\right)\) is maintained at \(20^{\circ} \mathrm{C}\) and the right end is maintained at \(80^{\circ} \mathrm{C}\). Neglecting any loss of heat through radiation, find (a) the temperature at a point \(11 \mathrm{~cm}\) from the left end and (b) the heat current through the rod. Thermal conductivity of copper \(=385 \mathrm{~W} \mathrm{~m}^{-1}{ }^{\circ} \mathrm{C}^{-1}\).

An amount \(n\) (in moles) of a monatomic gas at an initial temperature \(T_{0}\) is enclosed in a cylindrical vessel fitted with a light piston. The surrounding air has a temperature \(T_{s}\left(>T_{0}\right)\) and the atmospheric pressure is \(p_{a} .\) Heat may be conducted between the surrounding and the gas through the bottom of the cylinder. The bottom has a surface area \(A\), thickness \(x\) and thermal conductivity \(K\). Assuming all changes to be slow, find the distance moved by the piston in time \(t\).

A hollow tube has a length \(l\), inner radius \(R_{1}\) and outer radius \(R_{2}\). The material has a thermal conductivity \(K\). Find the heat flowing through the walls of the tube if (a) the flat ends are maintained at temperatures \(T_{1}\) and \(T_{2}\left(T_{2}>T_{1}\right)\) (b) the inside of the tube is maintained at temperature \(T_{1}\) and the outside is maintained at \(T_{2}\)

A cubical block of mass \(1 \cdot 0 \mathrm{~kg}\) and edge \(5 \cdot 0 \mathrm{~cm}\) is heated to \(227^{\circ} \mathrm{C}\). It is kept in an evacuated chamber maintained at \(27^{\circ} \mathrm{C}\). Assuming that the block emits radiation like a blackbody, find the rate at which the temperature of the block will decrease. Specific heat capacity of the material of the block is \(400 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\).
See all solutions

Recommended explanations on Physics Textbooks

Astrophysics

Read Explanation

Scientific Method Physics

Read Explanation

Mechanics and Materials

Read Explanation

Classical Mechanics

Read Explanation

Translational Dynamics

Read Explanation

Oscillations

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.