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PHYS1001 Physics 1 REGULAR Module 2 Thermal Physics HEAT CAPACITY LATENT HEAT. What is cooking all about?. PHYSICS: FUN EXCITING SIMPLE. ptC_heat .ppt. Overview of Thermal Physics Module:

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slide1

PHYS1001

Physics 1 REGULAR

Module 2 Thermal Physics

HEAT CAPACITY

LATENT HEAT

What is cooking all about?

PHYSICS: FUN EXCITING SIMPLE

ptC_heat .ppt

slide2

Overview of Thermal Physics Module:

  • Thermodynamic Systems:Work, Heat, Internal Energy0th, 1st and 2nd Law of Thermodynamics
  • Thermal Expansion
  • Heat Capacity, Latent Heat
  • Methods of Heat Transfer:Conduction, Convection, Radiation
  • Ideal Gases, Kinetic Theory Model
  • Second Law of ThermodynamicsEntropy and Disorder
  • Heat Engines, Refrigerators
slide3

HEAT CAPACITY

LATENT HEAT

§17.5 p582

Heat

Heat capacity (specific heat capacity)

Phase changes

Conservation of energy

calorimetry

References: University Physics 12th ed Young & Freedman

slide4

What happens when we heat a substance?

How does the temperature change?

When does the state of matter change (phase changes)?

slide5

Phases of matter

Gas - very weak

intermolecular forces,

rapid random motion

high temp

low pressure

Liquid- intermolecular

forces bind closest neighbours

low temp

high pressure

Solid- strong

intermolecular forces

slide7

Simple model for heating a substance at a constant rate

T(t)

boiling

point

g

l /g

melting

point

l

s / l

s

Q(t)

mcsDT

mclDT

mcgDT

mLf

mLv

slide8

Q= ± m LS

Phase changes

at constant temperature

deposition

LS latent heat of sublimation

or heat of sublimation

sublimation

freezing

condensation

gas

liquid

solid

evaporation

melting

Q= ± m Lf

Q= ± m LV

At melting point: Lf latent heat of fusion or heat of fusion

At boiling point: LV latent heat of vaporization or heat of vaporization

Q > 0 energy absorbed by substance during phase change

Q < 0 energy released by substance during phase change

slide9

Specific heat or specific heat capacity, c

Tf

DT= Tf - Ti

Ti

Mass of object m

Specific heat (capacity) c

heat

Q

NO phase change during temperature change

slide10

Specific heat Latent heats

Latent heat – phase change

(formation or breakage of chemical bonds requires or releases energy)

Water - large values of latent heats at atmospheric pressure

Lf= 3.34x105 J.kg-1 (273 K)

Lv= 2.26x106 J.kg-1(373 K)

Substance c (J.kg-1.K-1)

Aluminum 910

Copper 390

Ice 2100

Water 4190

Steam 2010

Air 1000

Soils / sand ~500

You can be badly scolded by steam – more dangerous than an equivalent amount of boiling water WHY?

slide11

Latent heat and phase changes

As a liquid evaporates it extracts energy from its surroundings and hence the surroundings are cooled.

When a gas condenses energy is released into the surroundings.

Steam heating systems are used in buildings. A boiler produces steam and energy is given out as the steam condenses in radiators located in rooms of the building.

slide12

Evaporation and cooling

Evaporation rates increase with temperature, volatility of substance, area and lower humidity. You feel uncomfortable on hot humid days because perspiration on the skin surface does not evaporate and the body can't cool itself effectively.

The circulation of air from a fan pushes water molecules away from the skin more rapidly helping evaporation and hence cooling.

Evaporative cooling is used to cool buildings.

Why do dogs pant?

When ether is placed on the skin it evaporates so quickly that the skin feels frozen. Ethyl chloride when sprayed on the skin evaporates so rapidly the skin is "frozen" and local surgery can be performed.

slide13

Problem C.1

The energy released when water condenses during a thunderstorm can be very large.

Calculate the energy released into the atmosphere for a typical small storm.

Where did the water come from?

slide14

Solution

Identify / Setup

Assume 10 mm of rain falls over a circular area of radius 1 km

h = 10 mm = 10-2 m r = 1 km = 103 m

volume of water V = r2h =  (106)(10-2) = 3104 m3

mass of water m = ? kg density of water  = 103 kg.m-3

m =  V = (103)(3104) = 3107 kg

Latent heat – change of phase

Q = m L Lv = 2.26106J .kg-1

Energy released in atmosphere due to condensation of water vapour

r

h

slide15

Execute

Q = m LV = (3107)(2.26106) J = 71013 J

Evaluate

The energy released into the atmosphere by condensation for a small thunder storm is more than 10 times greater then the energy released by one of the atomic bombs dropped on Japan in WW2. This calculation gives an indication of the enormous energy transformations that occur in atmospheric processes.

slide16

Problem C.2

For a 70 kg person (specific heat 3500 J.kg-1.K-1), how much extra released energy would be required to raise the temperature from 37 °C to 40 °C?

Solution

Identify / Setup

m = 70 kg c = 3500 J.kg-1.K-1 T = (40 – 37) °C = 3 °C

Specific heat capacityQ = m c T

Execute

Q = m c T

= (70)(3500)(3) J = 7.4105 J

= 0.74 MJ Evaluate

slide17

Waterhas a very large specific heat capacity compared to other substances cwater = 4190 J.kg-1.K-1

The large heat capacity of water makes it a good temperature regulator, since a great amount of energy is transferred for a given change in temperature.

Why is there a bigger difference between the max and min daily temperatures at Campbelltown compared to Bondi?

Why is water a good substance to use in a hot water bottle?

Why is the high water content of our bodies (c ~ 3500 J.kg-1.K-1) important in relation to the maintenance of a constant core body temperature?

slide18

CALORIMETRY PROBLEMS

This is a very common type of problem based upon the conservation of energy. It involved changes in temperature and phase changes due to heat exchanges.

Setup

All quantities are taken as positive in this method.

Identify the heat exchanges (gained or lost), phase changes and temperature changes.

Conservation of energy

energy gained = energy lost

slide19

Problem C.3

How much ice at –10.0 °C must be added to 4.00 kg of water at 20.0 °C to cause the resulting mixture to reach thermal equilibrium at 5.0 °C.

Sketch two graphs showing the change in temperature of the ice and the temperature of the water as functions of time.

Assume no energy transfer to the surrounding environment, so that energy transfer occurs only between the water and ice.

slide20

Solution

Identify / Setup

ICE gains energy from the water

Ice/water 0 oC

Water 5 oC

Ice -10 oC

WATER losses energy to the ice

Water 20 oC

Water 5 oC

Heat gained by ice Qice = heat lost by water Qwater

Q = m cTQ = m L

slide21

mice = ? kgmwater = 4.00 kg

temperature rise for ice to melt

Tice1 = 0 – (–10) °C = 10 °C

temperature rise of melted ice

Tice2 = (5 – 0) °C = 5 °C

temperature fall for water

Twater = (20 – 5) °C = 15 °C

cice = 2100 J.kg-1.K-1cwater = 4190 J.kg-1.K-1

Lf = 3.34105 J.kg-1

NB all temperature changes are positive

slide22

Execute

T

ice

0 oC

t

equilibrium temperature reached

T

water

0 oC

t

slide23

heat gained by ice

Qice =mice ciceTice1 + miceLf + mice cwaterTice2

Qwater = mwater cwaterTwater

Qice = Qwater

miceciceTice1 + miceLf + mice cwaterTice2 = mwatercwaterTwater

heat lost by water

conservation of energy

alternatively can calculate each term

Evaluate

slide24

T

A

B

t

Problem C.4 June 2007 Exam Question (5 mark)

A sample of liquid water A and a sample of ice B of identical masses, are placed in a thermally isolated container and allowed to come to thermal equilibrium. The diagram below is a sketch of the temperature T of the samples verses time t. Answer each of the following questions and justify your answer in each case.

1 Is the equilibrium temperature above, below or at the freezing point of

water?

2 Has the liquid water partly frozen,

fully frozen, or not at all?

3 Does the ice partly melt,

or does it undergo no melting?

slide25

Solution

Identify / Setup

Phase change – temperature remains constant

Q =  m L

Ice melts at 0 oC and liquid water freezes at 0 oC

Temperature change – no change in phase

Q = m c T

Ice warms and liquid water cools

Energy lost by liquid water (drop in temperature)

= Energy gained by ice (rise in temperature + phase change)

slide26

Execute

(1)

Ice increase in temperature initially and then remains constant when there is a change in phase. Therefore, the equilibrium temperature reached is the freezing point.

(2)

The ice reaches the freezing point first and the then the temperature remains constant. As the water cools, the ice melts. The temperature never rises above the freezing point, therefore, only part of the ice melts.

(3)

The temperature of the water falls to its freezing point and never falls below this and hence it is most likely that no liquid freezes.

slide27

Y&F Example 17.10 What’s Cooking

Cu pot

m = 2.0 kg T = 150 oC

Water added

m = 0.10 kg T = 25 oC

Final temperature of water and pot Tf?

energy lost by pot = energy gained by water: 3 possible outcomes

1 none of water boils 25 oC < Tf < 100 oC

2 some of the water boils Tf = 100 oC

3 all water boils to steam 100 < Tf < 150 oC

have to becareful with such problems

slide28

How do we measure a person’s metabolic rate?

Santorio Santorio weighed himself before and after a meal, conducting the first

controlled test of metabolism, AD 1614.

slide29

A fever represents a large amount of extra energy released. The metabolic rate depends to a large extent on the temperature of the body.

The rate of chemical reactions are very sensitive to temperature and even a small increase in the body's core temperature can increase the metabolic rate quite significantly. If there is an increase of about 1 °C then the metabolic rate can increase by as much as 10%. Therefore, an increase in core temperature of 3% can produce a 30% increase in metabolic rate. If the body's temperature drops by 3 °C the metabolic rate and oxygen consumption decrease by about 30%.

This is why animals hibernating have a low body temperature.

During heart operations, the person's temperature maybe lowered.