Environmental Physics Chapter 4: Heat Copyright © 2008 by DBS.

Environmental Physics

Chapter 4:Heat

Copyright © 2008 by DBS

Introduction

Figure 4.1: U.S. household energy consumption by end use. 1 Quad = 1015 Btu.

Residential sector uses 50% for space heating

~20% of all the energy in US is used for heating and cooling buildings

Energy (conservation) efficiency should be first step in dealing with environmental impacts…

Introduction

• Thermodynamics – study of heat and work

Heat and Work and the First Law

• Total Energy, E:

E = KE + PE + TE + chemical energy + electrical energy

• Heat (Q) and work (W) are the only ways to add energy to an object to change its total E,

1st law of thermodynamics:

Won+ Qto = Δ(KE + PE + TE + chemical energy + electrical energy) = ΔE

• Law of Conservation of Energy: The work done on a system plus the heat added to it is equal to the change in the total energy of that system

or Energyin = Energyout


• Important discovery of the 18th century:heat is the transfer of energy between two bodies due to temperature differences

• Previously, heat was mistakenly thought to be a material fluid, called “caloric” that would flow from a hot body into a cold one, causing an increase in temperature and mass

• British physicist James Joule, in a series of highly accurate experiments, provided conclusive evidence that heat is a form of energy in transit and that it can cause the same changes in a body as work


• Equivalence between mechanical work and heat

• Joule measured the increase in temperature of a water bath when a paddle wheel was turned

• Observed the same effect (rise in water temperature) either with the performance of work or by addition of heat

• Units are Joules or ft.lbs (heat is a form of Energy)

Figure 4.2: Relationship between work and heat. A temperature change in the water can be caused either by letting the weight drop (causing the blades to rotate) or by adding heat from the gas burner.

Temperature and Heat

Temperature

• Property of an object, much like color and shape

• measurement of average KE of molecules

• Cannot tell us the amount of energy in a substance, since it is independent of mass


• Because the temperature scales have different zero points, formulas must be used to carry out the conversions

K = ºC + 273.15

ºC = K - 273.15

5

9

9

5

(ºF - 32) or (ºF - 32)

1.8

ºF= (ºC) + 32 or 1.8(ºC) + 32

ºC =

Melting point ice

Farenheit, Celcius and Kelvin Scales

Melting point ice

Example Temperature Conversions

1. Convert 350 oF to oC and K

oC = (350 - 32 )(5/9) = (318)(5/9) = 177 oC

K = 177 + 273 = 450 K

2. Convert -40 oC to oF

oF = (9/5)(-40) + 32 = 9(-8) + 32 =

-72 + 32= - 40o F

3. Convert 298 K to oC

oC = 298 - 273 = 25 oC


Heat

• measure of total energy content of vibrating molecules (KE and PE)

• can tell us the amount of energy in a substance, since it is dependent on mass

• governed by law conservation of energy – Joule’s exp.

Heat is energy that flows from one object to another when there is a difference in temperature between the objects

Temperature determines the direction of heat flowhotter → cooler object

Heat and Temperature

• objects can have the same temperature but different amounts of heat

Steam burn – temp. is equal

but heat content is greater

Figure 4.4: Thermal energy. (a) If both brick assemblies are heated in a kiln for several hours, they will have the same temperature, but the larger array will store nine times as much thermal energy as the smaller one.


Specific heat capacity (c)• When heat is added to a substance, we usually find an increase in temperature• heat energy (joules) required to raise 1 g of a substance up by 1 °C (or 1 K)• different substances require different amounts of heat• Large c of water makes it an excellent coolant• substances with small specific heats absorb little energy when warming and give off little

energy when cooled

Why is c for H2O so much more than Cu?

Heat and Temperature Heat Capacity

When the samples are both heated by 1 °C, the addition to the KE (motion of molecules) is the same

For water, more energy must be added to the PE (energy from intermolecular forces) part of internal energy


• What factors are important?– Mass (m)– Temperature change, T = Tf –Ti

– I.D. of substance, steel, water, (specific heat capacity, c)

• Heat gained or lost, Q = mcT

Iron’s ability to store heat is less than waters

Question

What is more effective in cooling your cup of coffee, 100 g aluminum or 100 g milk?

Aluminum has lower c than milk (which is mostly water). The aluminum absorbs less heat from the coffee for each degree of temperature that it changes than the milk does.

Question

Calculate the heat energy required to raise the temperature of 24.5 g of mercury from 5.0 oC to 35.2 oC

T = 35.2 - 5.0 = 30.2m = 24.5 g

c = 0.14 kJ kg K

T = 35.2 - 5.0 = 30.2

Q = mcT Q = (0.14 kJ) (0.0245 kg) (30.2 K ) = 0.10 kJ kg K

Question

A kettle full of water is rated at 2 kW and starting at room temperature (25 ºC) takes 4 minutes to boil.

(i) How much energy is used?

(ii) How much water was boiled?

2 kW = 2 kJ / sE = Pt = 2 kJ / s x 240 s = 480 kJ

Q = mcT Q = 480 kJ = (m) (4.2 kJ ) (75 K )

kg K

m = 1.5 kg

Question

Which object experiences the greatest temperature change?

Assume equal masses and heat losses.

Substance Specific Heat Capacity

Marble 0.88

Aluminum 0.91

Copper 0.380

ΔT = Q mc


• Adding heat may not increase the temperature!

• May change state of matter

At the boiling / melting temperature, adding heat energy changes state WITHOUT RAISING THE TEMPERATURE

Temperature and HeatLatent Heat

• Latent heat = energy needed to change state (solid, liquid, gas) without affecting temperature

e.g. Energy needed to evaporate water is released when water condensese.g. Energy needed to melt ice is released when water freezes

• Sensible heat = heat that results in temperature change

Latent heats are high compared with specific heat capacity – intermolecular bonds must be broken

Substance Latent heat fusion

(kJ / kg)

Latent heat evaporation

(kJ / kg)

Water 335 2260

Lead 23 858

Aluminum 393 10,500 Q = m Lf

Question

How much thermal energy will be released when50 g of water freezes?

How much thermal energy in joules must be absorbed by 50 g of ice at 0 ºC to melt it?

Q = m Lf

= 0.050 kg x 334 kJ/kg = 16700 J

16700 J


At the boiling / melting temperature, adding heat energy changes state WITHOUT RAISING THE TEMPERATURE

Heat absorbed

Heat liberated

Question

If the specific heat capacity of ice is 2.1 kJ kg-1 K-1, how much heat would have to be added to 200 g of ice, initially at -10 °C to raise the ice to the melting point and completely melt the ice?

Total energy = Qraise + Qmelt

Total energy = mc T + m Lf

= (0.200 kg x 2.1 kJ kg-1 K-1 x 10 K) + (0.200 x 334 kJ kg-1)

= 4.2 kJ + 66.8 kJ = 71 kJ

End

• Review

Heat Transfer Principles

Figure 4.7: Heat flows when there is a temperature difference ΔT.

In this case,

ΔT = 70° − 50° = 20°F

Heat Transfer• One of two ways in which energy can be transferred to a body• Occurs when there is a temperature difference• Occurs through conduction, convection and/or radiation

Heat Transfer PrinciplesConduction

• Conduction – movement of heat through a solid substance, exchange of thermal energy between atoms

• Most important in solids

• Block demonstration

Figure 4.8: Heat is transferred by conduction through the metal spoon from the hot coffee to the colder fingers.

Demo

Metal conducts heat more readily than wood, so more heat flows from your hand into the metal than the wood. Since contact with the metal cools your hand more rapidly the metal block feels colder.

Why does B melt the ice quicker than the warm block?


• Depends on temperature gradient, size of conductor and conductivity

• Rate of heat transfer by conduction (Qc/t)

Where Q = heat (J) transferred in time t (s), k = thermal conductivity (W m-1 K-1), A = surface area, δ = thickness, T1 and T2 are temperatures on each side



Q = kA(T2 – T1)

t δ

• Good insulators e.g. polystyrene, wool jumpers rely on incorporating air into structure

Substance Thermal Conductivity

W m-1 K-1

Diamond 1000

Copper 401

Aluminum 210

Iron 76

Glass 1.1

Brick 0.13

Water 0.62

Air 0.024



Q = kA(T2 – T1)

t δ

• To reduce heat loss:

– Reduce T2

– Reduce A

– Increase δ

Figure 4.10: Percentage of energy saved by lowering the thermostat from 72°F to the values shown on the curved lines, for the time periods shown.

Heat Transfer PrinciplesConvection

• Gases and fluids molecules are too far apart for heat to conduct

Figure 4.11: Convection currents in water.

Heat Transfer Principles Convection

Demo

• Galilean thermometer

– Liquids change density when heated

Transmission of HeatConvection

• Warm fluid expands, density decreases and it tends to rise

• Ocean currents and winds redistribute heat from the tropics to the poles


• May be natural (density differences) or assisted by wind

Figure 4.12: Heat transfer through a double-pane window.


• Convection currents are important in some types of solar heating systems

Figure 4.13: Solar air heater for use in a window.

Transmission of HeatRadiation

• EM radiation is transferred not through matter, but through electrical and magnetic fields

http://www.edumedia.fr/a185_l2-transverse-electromagnetic-wave.html

-Self propagating as it moves through space

- Electrical charges are accelerated

- Carries energy and momentum which may be imparted on interaction with matter

- does not require a medium in which to travel

http://www.edumedia.fr/a185_l2-transverse-electromagnetic-wave.html


• Classified according to frequency

v = f λ

Where:

1 Hz

v = speed of light = 3 x 108 (m/s), f = frequency (Hz or cycles s-1), λ = wavelength (m)

3 Hz

Different types of EM radiation all have the same velocity in a vacuum

3.0 x 108 m/s = 1.1 billion km/h = 671 million mph

Question

What is the wavelength of a cell phone using the microwave frequency (GHz)?

f x λ = 3.0 x 108

λ = 3.0 x 108 /109 = 0.3 m

Fig. 4-15, p. 112

Figure 4.15: The electromagnetic spectrum, shown as a function of wavelength.


• All objects above absolute zero (0 K) emit radiation

• Amount of energy emitted from an object is proportional to its temperature

• Humans, animals, the Earth etc. and basically anything < 1000 °C emit IR

• Sun’s surface ~ 6000 °C emits primarily visible radiation + some IR and UV

cf.

Stefan-Boltzman and Wien’s laws

Earth emits majority long-wave (LW) radiation

= Infra-red

Sun emits majority short-wave (SW) radiation = visible


• For a body to maintain a certain temperature, Energy in=Energyout

Figure 4.17: The equilibrium temperature of an object is maintained if the energy input is equal to the energy output.

At night a body continues to radiate heat – radiative cooling

Fig. 4-18, p. 115

Figure 4.18: A hot-water radiator as an illustration of heat transfer via conduction, convection, and radiation.

• Convection – movement of heat through a fluid (liquids and gases) brought about by changes in temperature affecting density

• Conduction – movement of heat through a solid substance, exchange of thermal energy between atoms

• Radiation – transfer of heat energy via electro-magnetic waves through a vacuum

End

• Review

Heat Engines

• “heat engines” – devices in which heat is converted into useful work

e.g. automobile, electrical generating plant

• Requires a source of heat, e.g. burning a fuel, nuclear, solar etc.

• The flow of heat proceeds through the “working fluid” (gas or liquid)

Heat Engines

• Energy flow diagram for a heat engine:

Since energy is conserved heat leaving the source (QH) is equal to the heat entering the sink (QC) plus work done (W)

QH = QC + W

and W = QH - QC

• Higher TH and lower TC the higher the work output

• Energy available for work comes from a decrease in the total energy of the working fluid

Figure 4.19: A heat engine transforms heat into work.

Heat Engines

• Open cycle – working fluid is exhausted into environment

e.g. 4-stroke gasoline– Intake– Compression– Power (volume expansion of gas)– Exhaust

Heat Engines

• Closed cycle – working fluid is sent back to the heat source to start the cycle over

e.g. steam turbine – working fluid = water

QH = W + QC

Heat into plant = net work out + net heat out (fuel combustion) (electricity) (of condenser)

W comes from dec. in ΔE of the steam, condenser provides low-temp. sink

Heat Engines

Types of heat engines:

Working fluid changes state

Working fluid remains a gas (air)

Heat Engines

• Ocean Thermal Conversion (OTEC)

Figure 4.20: Ocean Thermal Energy Conversion (OTEC). The temperature difference between waters on the top of the water and down deep allows one to construct a heat engine.

The Second Law of Thermodynamics

• Why doesn’t book lying on a table take thermal energy from the table and convert it into kinetic energy (work)?

• 1st law of thermodynamics doesn’t prevent this form happening!

Figure 4.21: Impossibilities according to the second law of thermodynamics. (a) Heat withdrawn from the table is converted into mechanical energy—the kinetic energy of the block, (b) Heat from sea water is converted into electrical energy (the resulting ice cubes are discarded).


• Second law of thermodynamics: for any spontaneous process, the entropy (disorder) of an isolated system can only increase or stay the same, but never decrease

• Important statements that follow from the second law:

– Heat can flow spontaneously only from a hot source to a cold source

– No heat engine can be constructed in which heat from a hot source is converted entirely to work. Some heat has to be discharge to a sink at a lower temperature (cf. previous examples)


• The efficiency (η) of an energy conversion process is defined as:

η = Eout/Ein x 100 %

• Principle of conservation of energy says that the work output (energy out) equals the heat input minus the heat transferred out (W = QH – QC):

Efficiency = W = QH – QC x 100%

QH QH

= 1 – (QC / QH) x 100%

If some heat is transferred out to a cold sink, then we can never have a 100% efficient process

Therefore we will never have perpetual motion machines…

The Second law of ThermodynamicsMaximum Efficiency

• If some heat has to be discarded, what is the best we can do?

The Second law of ThermodynamicsMaximum Efficiency

• Upper temperature limit is 500 ºC (restricted by engineering, pollution and corrosion), exhaust temperature 100 º C

• Max possible efficiency for a heat engine (Carnot efficiency):

η = TH – TC x100 % = 1 – TC/TH x100 %

TH

η = 61% (must be computed in K)

• Theoretical upper limit, other losses (e.g. friction, heat loss from boilers, transmission losses)

• At best we can convert 35 % of the thermal energy in burning fossil fuels to mechanical/electrical

• More than ½ lost as waste heat

Question

A heat engine takes in 1200 J of heat from the high temperature heat source in each cycle and does 400 J of work in each cycle. What is the efficiency of the engine? How much heat is released to the environment in each cycle?

η = W/QH = 400 J / 1200 J = 33%

From 1st law: W = QH - QC = 1200 J - QC = 400 J

QC = 800 J

Question

Calculate the efficiency of a power station located in a colder climate (Tc = 0 ºC). Why wouldn’t it be beneficial to generate power in colder climates for use in warmer areas?

η = 69 %

The Second law of ThermodynamicsAvailable Energy

• It is impossible to convert a given quantity of heat energy completely into work. In an energy conversion process, energy is always degraded in quality, so that its ability to do work is reduced

Summary

• 1st law – energy is conserved, QH = W + QC (heat in = work and heat out)

• Heat engines make use of a flow of heat from hot to cold in order to do work

• 2nd law limits the amount of work obtainable from a heat engine

• Heat energy that flows from the hot source cannot be entirely converted into work; some heat has to be discharged into the environment

• Maximum efficiency may be calculated using the Carnot efficiency equation

• Total entropy of a system increases in a physical process

Environmental Physics Chapter 4: Heat Copyright © 2008 by DBS.

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