temperature & heat...diffusivity value: recognises heat is absorbed in raising temp of material,...
TRANSCRIPT
When there is a temperature difference
between two objects, heat transfer can occur.
Heat transfer methods
Three methods of heat transfer :
• Conduction
• Convection
• Radiation
In general all three processes occur
simultaneously.
One method normally dominates
Direct physical contact
results in collisions between molecules
Hot Cold Conduction
Same ideas applies to heat conduction within a
single object containing regions at different
temperatures.
Conduction is the transfer of heat by direct
physical contact.
Conduction
Molecules in object at higher temperature have
higher average kinetic energy and vice versa;
Result
energy transfer
hot cold
•Temperature difference DT (T1-T2)
•Area of contact A
•Thickness of the object d
•coefficient of thermal conductivity k
of the substance
Rate of heat conduction
The rate of heat transfer (Q/t) by
conduction depends on:
Consider an object of area A, thickness d,
opposite faces at temperatures T1 and T2
T1 T2
Q
A
d
Large k value: good thermal conductor
Small k value: poor thermal conductor
→ insulator
(Q/t) =rate of heat flow (Watts, W)
k = coefficient of thermal conductivity (Wm-1K-1)
A = area of object m2
DT = temperature difference (K or oC)
d= thickness of object (m)
T1-T2 = DT
Q kA T
t d
D
Rate of heat conduction
T1 T2
Q
A
d
Steady state:
constant heat flow
Equilibrium condition
Rate of heat conduction
Thermal conduction occurs at different
rates in different materials
Wooden stick burning at one end remains
relatively cold at the other end.
Metal spoon transmits heat rapidly from
one end to the other.
In metals there are electrons that can move
freely→carry thermal energy. Metals are good
thermal conductors.
Thermal conductivity of an object determines
how hot or cold it feels to the touch.
Example:
Tiles and carpet at
the same temperature
Material k (Wm-1K-1)
Copper 400
Glass 0.8
Brick 0.6
Floor tile 0.7
Wood 0.08
Muscle 0.042
Air 0.02
Carpet 0.04
Fat 0.021
Thermal Conductivities
Thermal Conductivities
Thermal conductivity of Dental Materials
Enamel 0.93 Wm-1K-1
Dentine 0.58 Wm-1K-1
Zinc phosphate cement 1.17 Wm-1K-1
Amalgam ≈ 55 Wm-1K-1
Composite 2 Wm-1K-1
Gold 290 Wm-1K-1
Low Thermal Conductivity layer
(e.g. zinc phosphate cement) needed underneath
amalgam filling to protect pulp from a temperature
rise.
Enamel and dentine are effective
thermal insulators
•reduces thermal shock to pulp (hot or cold foods)
Amalgam : High Thermal Conductivity.
If layer of dentine between bottom of cavity floor
and pulp is thin » »» thermal shock
Transfer of thermal energy depends on
Thermal conductivity
Thermal diffusivity
Thermal Diffusivity
Time
Tem
pera
ture
Thermal Diffusivity
kh
c
k (thermal conductivity)
(density)
c (specific heat) units m2s-1
Thermal diffusivity (h):
measures how rapidly a material can change
its temperature to reach a steady state
Measures transient thermal response of material
More important thermal characteristic in dentistry
than thermal conductivity
Since temperature changes rapidly in oral cavity;
thermal stimuli are transient
Thermal Diffusivity (10-6 m2s-1)
Enamel 0.41
Dentine 0.24
Zinc phosphate
cement
0.88
Amalgam 9.6
Gold 116
Thermal Diffusivity
Thermal diffusivity value
usually more appropriate in deciding dental
materials suitability
Examples
Ability of restorative base material
to protect pulp from thermal damage
Denture base material: should have high
thermal diffusivity so that wearers are
“immediately” aware of hot/cold food
Diffusivity value: recognises heat is absorbed in
raising temp of material, thus reducing the heat
available to pass through material kh
c
How much thermal energy is lost in a period
of 24hours by conduction through a window
of area 1.0m² and thickness 0.4cm if the
temperatures at the outer and inner surfaces
are 5.0°C and 15°C respectively?
kglass = 0.84 W.m-1.°C-1
Q/t = [(0.84*1.0*10)/0.004]W = 2100W
In 24hours:
Q=2100 x 24 x 3600 J = 181.44MJ
Rate of heat conduction
Example.
Q kA T
t d
D
(2100W = 2100 Joules per second)
In general, hot fluids
have a lower density
than cold fluids
(thermal expansion),
Convection is the transfer of heat by mass
movement of fluid (liquid or gas).
Convection
Natural convection.
heated fluid naturally
rises and the cold fluid
moves downward,
complex patterns (such
as convections rolls in
a pot of water or hot air
rising above a fire).
Water in pot
Air in room
Convection occurs when fluid is unevenly heated.
Convection can also be forced
Convection
Examples:
•Blood circulation.
•Engine cooled by pumped air or water.
Gases and liquids are not good thermal
conductors
however they can transfer heat rapidly by
convection.
Natural convection in atmosphere
plays major role in determining the
daily weather conditions
Natural convection in oceans
Important global heat transfer mechanism
Convection
Body temp ≈ 37ºC
Blood circulation:
Blood flow regulated according to need.
Overheated person, blood vessels to
surface dilate and so carry more blood to
the surface for cooling.
Layers of fat beneath the skin
help to maintain body temperature
Fat
Poor thermal conductor:
few blood vessels to carry blood to surface
where energy losses by convection can occur
Radiation
The sun is our major source of heat.
It warms the earth: How?
Heat transfer by conduction and convection
not possible.
Very little material (relatively few molecules)
between us and the sun
Heat transfer from sun is by radiation
Energy in the form of electromagnetic waves
All objects emit electromagnetic radiation.
At ordinary temperatures this radiation
is mainly at infrared wavelengths.
Travel and carry energy through empty space
10-13m 10-10 10-7 10-6 10-5 10-2 10m
Electromagnetic Spectrum
Radiation
Infrared radiation is strongly absorbed by
water molecules including those in our body cells.
Infrared radiation is converted to heat
as it is absorbed by our bodies.
energy
Wavelength
Radiation is a property of a
single object: it does not
depend on a temperature
difference.
Radiation
Hot Cold
Given 2 objects, the higher temperature one
radiates more than the lower temperature
one so there is a net transfer from the former to
the latter.
Radiation
Fundamentally different from conduction and
convection which involve molecular collisions;
------ non-contact!
Does not require a material substance for its
transmission. Can transmit in vacuum.
The rate of heat radiation (Q/t) from a single
object is proportional to:
•Temperature (T) K
•Surface area of the object (A)
•emissivity of the substance at the
surface (e)
= Stefan-Boltzmann constant
= 5.67 x 10-8 W m-2 K-4
e =emissivity (0 ≤ e ≤ 1) (depends on surface)
Q/t = rate of heat radiation (W)
Q/t increases very strongly with temperature!!
Rate of heat radiation
Any object with a temperature greater than
absolute zero (0K) emits radiant heat.
Hence a fire will radiate heat into a room but
the room will also transfer heat to the fire by
radiation
4QeAT
t
Rate of heat radiation
emissivity (0 ≤ e ≤ 1) (depends on surface)
Black, rough surface e→1
Bright Shiny surface e→0
Bright shiny surfaces reflect most of the
radiation falling of them and so are poor
absorbers and poor emitters
Black surfaces
are good emitters and absorbers.
The net rate of radiation by a object at T1 in
surroundings T2 is:
Net rate of heat transfer depends only on the
properties of the object and the temperature
of the surroundings
If temperature of the object and surrounding
are equal there is no net heat transfer.
4 4
1 2( )Q
eA T Tt
What is the rate of heat loss by radiation from
a black roof of area 250m², if its temperature
is 25°C and that of the surroundings is 0°C?
Assume the emissivity of the roof is 0.95.
Q/t =
5.67x10-8 Wm-2 K-4* 0.95 * 250m2 * [(298 K) 4-(273 K)4]
Q/t =1346.6x10-8*23.3x108 W = 31,376 W
T1=(273+25)K=298K
T2=(273+0)K=273K
4 4
1 2( )Q
eA T Tt
Rate of heat radiation:
Example.
Compare the rate of heat loss by conduction
and by radiation through a window of area
2m² and thickness 5mm, if the inside
temperature is 20°C and the outside 5°C.
The emissivity of the room is 0.5;
kglass = 0.84 W.m-1.°C-1 .
Example
Rate of heat loss by conduction:
Q/t = 0.84*2.0*15/0.005 W = 5,040W
Rate of heat loss by radiation:
Q/t = 5.67.10-8 * 0.5 * 2 * (2934-2784)
Q/t = 79W
T1=273+20K=293K
T2=273+5K=278K
Heat lost by conduction is much greater
than that lost by radiation.
Rate of heat radiation:
4 4
1 2( )Q
eA T Tt
Q kA T
t d
D
= 5.67 x 10-8 W m-2 K-4
Regulation of body temperature
Body temperature
Maintain at nearly constant value ≈ 37oC
Basic metabolic action
Most of consumed energy → heat inside body
Efficiency of muscles (external work) ≈ 20%
Physical activity
80% of consumed energy → heat inside body
Heat must be removed to maintain temp ≈ 37oC
Example: cycling for one hour
energy consumption 360kcal 80% =288kcal
Q = c.m. TD QT =
c.mD
288 4186T = 4.6
3500 75
oC
D
Q=288x4186 Joules
Mass=75kg
Specific heat c ≈3500 J.kg-1.°C-1
Temperature rise
Regulation of Body Temperature
Removal of heat from body (conduction)
Heat flow from inside body to skin surface
Temperature difference required
Skin temperature lower
Warm environment 35 oC
Cold environment 28 oC
Body tissue:
poor conductor without blood flow
Thermal Conductivity
k ≈ 0.2 Wm-1 oC-1
Q kA T
t d
D
1 1 21.2( ) 1.5 2
200.03
oQ Wm C m CJs
t m
17 /Q
kcal hrt
without blood flow Insufficient conduction rate
to remove excess heat from body
Regulation of Body Temperature
Heat transported
inside body to skin
by blood movement (circulatory system)
capillaries leading to the skin dilate
Skin to outer skin surface by conduction
Thermal Conductivity ( k) of air ≈ 0.02 Wm-1C-1
(confined by clothing) low
Heat must be quickly removed from skin
Heat removed from skin by
Convection (air colder than skin)≈(10kcal/hour)
Radiation ( environment colder than skin)
evaporation
4 4
1 2( )Q
eA T Tt
The net rate of radiation from skin at T1 = 32oC
to surroundings at T2 = 25oC is:
8 2 4 2 4 45.67 10 1 1.5 305 298Q
Wm K mt
165 56 /Q
Js kcal hrt
Perspiration
Principally cooling off when surrounding
temperature is greater than skin temperature.
Perspiration leads to cooling of the body by
evaporation of water:
Preferential evaporation of high energy
molecules reduces the average temperature of
the remaining molecules.
So 2.26x106 J (540 kcal ) of heat is lost for every
Kg (litre) of sweat that evaporates from skin
Latent heat of vaporisation of water
(Lv)=2.26x106 J.kg-1
(Lv)=540 kcal.kg-1
Large value of latent heat of vaporisation
QT =
c.mD
6
o -1
2.26 10 JT = 9.2
3500 J(kg. C) 70 kg
oC
D
Evaporation of 1kg(1litre) water from70kg person
(body mainly composed of water) lowers body
temperature by
Example.
The temperature of the sun is approximately
6x103K at its surface. Assuming it is spherical
with a radius of 6.95x108 m and an emissivity
of 0.92, calculate the total power radiated from
its surface. = 5.67 x 10-8 W m-2 K-4
4QeAT
t
Stefan’s law
Power is the rate at which energy is used
P= Q/t
Surface area A of sphere (sun) = 4pr2
A = 4p*(6.95x108 m)2 ≈ 6 x1018 m2
Total power radiated from sun
= 5.67x10-8 Wm-2K-4 *(0.92)* 6 x1018m2 *6x103K)4
=4.06x1026 W
Perspiration
Efficient cooling by evaporation may employ a
liquid other than water, e.g. alcohol rub in
hospitals.
When relative humidity is high, this process
is inefficient and body overheats; water
gathers on skin surface because it is not
removed by evaporation.
Relative humidity is concerned with the saturation
of air with water vapour and hence does not
affect the evaporation of alcohol
Evaporation rate also depends on
Environmental temperature
Wind speed
On hot day fan circulating air at ambient
temperature feels cool because air from it is drier
compared to sweaty body and therefore enables
increased evaporation
Heat loss in real situations
Thermos flask
Hot or cold
liquid
Rubber support
Vacuum
Container
Spring
centering
device
Glass walls
with silvered
surfaces
Vacuum minimises heat lost (or gained) by
conduction & Convection
Silvered surfaces minimise heat lost
(or gained) by radiation
Heat loss in real situations
Room
Conduction
Convection around
windows & doors
(cold air)
Co
nve
ctio
n (
ho
t a
ir)