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Physics Subject Area Test Thermodynamics

Physics Subject Area Test

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Physics Subject Area Test. Thermodynamics. There are three commonly used temperature scales, Fahrenheit, Celsius and Kelvin. . Temperature Conversions . Converting Between the Kelvin and Celsius Scales . Converting Between the Fahrenheit and Celsius Scales . Thermal expansion - PowerPoint PPT Presentation

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Page 1: Physics Subject Area Test

Physics Subject Area Test

Thermodynamics

Page 2: Physics Subject Area Test

* Temperature Conversions

There are three commonly used temperature scales, Fahrenheit, Celsius and Kelvin.

Page 3: Physics Subject Area Test

Converting Between the Kelvin and Celsius Scales

Converting Between the Fahrenheit and Celsius Scales

Page 4: Physics Subject Area Test

Thermal expansion

• Expanding solids maintain original shape• Expanding liquids conform to the container

Linear expansion

L = Δ αL TΔ L = length α = coefficient of liner expansionΔT = temperature change

Page 5: Physics Subject Area Test

Example:The highest tower in the world is the steel radio mast of Warsaw Radio in Poland, which has a height if 646m. How much does its height increase between a cold winter day when the temperature is -35 C and a hot summer day when the ⁰temperature is +35 C ?⁰

L = Δ αL TΔ = 12x10-6/ ⁰C x 646 x 70 C ⁰ = 0.54m

Page 6: Physics Subject Area Test

• Volume expansion

V= Δ βV TΔ L = length β = coefficient of liner expansionΔT = temperature changecold hot

Β = 3 α

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Thermal Conduction

= heat flowk = proportionality constant, thermal conductivityA = cross sectional areaΔT = change in temperatureΔx = distance crossed, thickness of material

Heat flow: the heat current; the amount of heat that passes by some given place on the rod per unit time

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•  

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*convectionHeat is stored in a moving fluid and is carried from one place to another by the motion of this fluid

*radiation The heat is carried from one place to another by electromagnetic waves

* conduction the process of handing on energy from one thing to the next

Page 10: Physics Subject Area Test

Stefan-Boltzmann Law

* Amount of power radiated (I) by a body at temperature T and having a surface area A is given by the Stefan-Boltzmann law

I = power radiated

e = emissivity (between 0 and 1)

σ = Stefan’s constant = 5.6703 x 10-8 W/m2·K

A = surface area

T = temperature

Shiny objects are not good absorbers or radiators & have emissivity close to 0

Black objects have emissivity close to 1

Page 11: Physics Subject Area Test

Changes of StateLatent heat ( heat of transformation) – the heat absorbed during the change of state ΔQ = quantity of heat transferredm = mass of the material

Heat of fusion - heat absorbed when changing from a solid to a liquid

Heat of vaporization - heat absorbed when changing from a liquid to a gas

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GAS LAWS

Boyle’s LawP1V1=P2V2

P= pressureV = volume

Page 14: Physics Subject Area Test

Charles LawV1/V2=T1/T2

V = volumeT = temperature

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Gay-Lussac’s LawP1/T1=P2/T2

P = pressureT = temperature

Page 16: Physics Subject Area Test

Combined Gas Law

Ideal Gas Law

PV = n R T P= pressureV = volumeT = temperaturen = molesR = Gas constant = 0.08206 L-atm/mol K

=

Page 17: Physics Subject Area Test

*Gas Density

PV = nRT n/V = P/RT Molarity = n/VDensity D = m/VMolecular Wt M = m/n D=Mn/V = PM/RT M= DRT/P

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*First Law of Thermodynamics

Energy can be neither created nor destroyed but only transformed

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THE GENERAL ENERGY EQUATION

Energy In = Energy Out or U2 - U1 = Q -W

whereU1: internal energy of the system at the

beginningU2: internal energy of the system at the endQ : net heat flow into the systemW : net work done by the system

Q = ΔU + ΔW

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A closed tank has a volume of 40.0 m2 and is filled with air at 25 C and 100 kPa. We want ⁰to maintain the temperature in the tank at 25 C as ⁰water is pumped into it. How much heat will have to be removed from the air in the tank to fill it half full?

Q gas=W gas=PgasV 1 ln( V 2

V T )=PgasV T ln ( 12 V T

V T )=PgasV T ln 12

= (100kPa) (40.0 m2)(-0.69314) = -2772.58kJ

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*STATE CHANGES• Isobaric

– the pressure of and on the working fluid is constant

– represented by horizontal lines on a graph• Isothermal

– temperature is constant – Temperature doesn’t change, internal energy remains

constant, & the heat absorbed by the gas = the work done by the gas

– The PV curve is a hyperbola• Adiabatic

– there is no transfer of heat to or from the system during the process

– Work done = decrease in internal energy & the temperature falls as the gas expands

– -the PV curve is steeper than that of and isothermal expansion

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* Quasi-Static Processes

Quasi-static (quasi-equilibrium) processes – sufficiently slow processes, any intermediate state can be considered as an equilibrium state (the macroparamers are well-defined for all intermediate states).

Examples of quasi-equilibrium processes:

isochoric: V = const isobaric: P = const isothermal: T = const adiabatic: Q = 0

For quasi-equilibrium processes, P, V, T are well-defined – the “path” between two states is a continuous lines in the P, V, T space.

P

V T

1

2

Advantage: the state of a system that participates in a quasi-equilibrium process can be described with the same (small) number of macro parameters as for a system in equilibrium (e.g., for an ideal gas in quasi-equilibrium processes, this could be T and P). By contrast, for non-equilibrium processes (e.g. turbulent flow of gas), we need a huge number of macro parameters.

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* Work

The sign: if the volume is decreased, W is positive (by compressing gas, we increase its internal energy); if the volume is increased, W is negative (the gas decreases its internal energy by doing some work on the environment).

2

1

),(21

V

VdVVTPW

The work done by an external force on a gas enclosed within a cylinder fitted with a piston:

W = (PA) dx = P (Adx) = - PdV

x

P

W = - PdV - applies to any shape of system boundary

The work is not necessarily associated with the volume changes – e.g., in the Joule’s experiments on determining the “mechanical equivalent of heat”, the system (water) was heated by stirring.

dU = Q – PdV

A – the piston area

force

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*Specific Heat the heat absorbed during the change of state

Q = nCv ΔT Q = amount of heat required

n = number of moles

Cv = specific heat at a constant volume

ΔT = Change in temperature

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How to calculate changes in thermal energy

Specific heat is the amount of heat required to raise the temperature of 1 kg of a material by one degree (C or K).C water = 4184 J / kg C

Q = m x T x Cp

Q = change in thermal energym = mass of substance

T = change in temperature (Tf – Ti)

Cp = specific heat of substance

Page 27: Physics Subject Area Test

Second Law of Thermodynamics

Entropy = the transformation of energy to a more disordered state- can be thought of as a measure of the randomness of a system

- related to the various modes of motion in moleculesThe second law of thermodynamics: entropy of

an isolated system not in equilibrium tends to increase over time

• No machine is 100% efficient• Heat cannot spontaneously pass from a colder

to a hotter object

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Kinetic molecular theory

The relationship between kinetic energy and intermolecular forces determines whether a collection of molecules will be a solid, liquid or a gas*Pressure results from collisions*The # of collisions and the KE contribute to pressure*Temperature increase KE