convection conduction radiationnordlund/courses/ph201/links/...convection = process by which heat is...

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surrounding cooler and denser air pushes heated and less

dense air upwards; resulting updraft called a “thermal”

Example of forced convection

cool air duct

warm air duct

CONDUCTION OF HEAT THROUGH A MATERIAL

heat Q conducted during a time t through a bar of length

L and cross-sectional area A is

L

tTkAQ

Δ

SI Units of thermal conductivity: J/(s·m·oC)

thermal

conductivity

excellent thermal

insulator

excellent thermal

conductors

Materials such as styrofoam,

goose, and wool contain small,

dead-air spaces.

Net effect is very poor thermal

conductor (excellent thermal

insulator).

Dead-air spaces inhibit heat

transfer by convection currents,

and air itself has very low

thermal conductivity.

QUESTION: Find the amount of heat per second

conducted through a glass picture window (1.5 m x 2.0 m)

that is 4.0-mm thick when the outside temperature is 5oC

(41 F) and the room is at 22.2oC (72 F)?

SOLUTION:

2.0 m

1.5 m

4.0 mm

QUESTION: One wall of a house consists of plywood

backed by insulation. The thermal conductivities of the

insulation and plywood are, respectively, 0.030 and 0.080

J/(s·m·oC), and the area of the wall is 35 m2.

Find the amount of heat

conducted through the wall

in one hour.

SOLUTION

radiation = process in which

energy is transferred by means of

electromagnetic waves

A material that is a good absorber

is also a good emitter.

A material that absorbs

100% of the received

radiation is called a

perfect blackbody.

THE STEFAN-BOLTZMANN LAW OF RADIATION

radiant energy Q, emitted in a time t by an object that has a

Kelvin temperature T, a surface area A, and an emissivity e, is

given byAtTeQ 4

emissivity e = dimensionless number between zero and one

= ratio of what an object radiates to what the

object would radiate if it were a perfect emitter

Stefan-Boltzmann constant

481067.5 K

2msJ

e = 1.0 for blackbody radiator

QUESTION:

The supergiant star Betelgeuse has a surface

temperature of about 2900 K and emits a power of

approximately 4x1030 W (1 W = 1 J/s).

Assuming that Betelgeuse is a perfect emitter and

spherical, find its radius.

SOLUTION:

ratings of thermal insulation = R-values

R

TA

kL

TA

L

TkAtQ

D

D

D

//

R value = L/k

ADDITIONAL APPLICATIONS OF HEAT TRANSFER

larger R values reduce heat per unit time Q/t flowing

through material and, therefore, mean better insulation

increase R by increase thickness (length) of insulation or

using a material (combination of materials in series) with

smaller thermal conductivities k

Thermos bottle minimizes heat transfer

via conduction, convection, and radiation.

Vacuum space

reflects radiation

no air convection

poor conductor

Chapter 14.4 Diffusion

diffusion = process in which molecules move from a region

of higher concentration to one of lower concentration

Why Diffusion is Relatively Slow

A gas molecule has a translational rms speed of hundreds of

meters per second at room temperature. At such speed, a

molecule could travel across an ordinary room in just a fraction

of a second. Yet, it often takes several seconds, and sometimes

minutes, for the fragrance of a perfume to reach the other side

of the room. Why does it take so long?

zigzag path result of

millions of collisions =

Brownian motion

A Transdermal Patch

Diffusion

Conduction

Similarity between Diffusion and Conduction

L

tCDAm

D

FICK’S LAW OF DIFFUSION

mass m of solute that diffuses in a time t through a solvent

contained in a channel of length L and cross sectional area A

concentration gradient

between ends, DCdiffusion

constant, D

SI Units for the Diffusion Constant: m2/s

Water Given Off by Plant Leaves

Large amounts of water can be given

off by plants. Inside the leaf, water

passes from the liquid phase to the

vapor phase at the walls of the

mesophyll cells.

The diffusion constant for water is

2.4x10-5m2/s. A stomatal pore has a

cross sectional area of about

8.0x10-11m2 and a length of about

2.5x10-5m. The concentration on the

interior side of the pore is roughly

0.022 kg/m3, while that on the outside

is approximately 0.011 kg/m3.

Determine the mass of water that

passes through the stomatal pore in

one hour.

SOLUTION:

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