Upload
rhian
View
31
Download
0
Embed Size (px)
DESCRIPTION
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
Citation preview
Physics Subject Area Test
Thermodynamics
* Temperature Conversions
There are three commonly used temperature scales, Fahrenheit, Celsius and Kelvin.
Converting Between the Kelvin and Celsius Scales
Converting Between the Fahrenheit and Celsius Scales
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
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
• Volume expansion
V= Δ βV TΔ L = length β = coefficient of liner expansionΔT = temperature changecold hot
Β = 3 α
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
•
*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
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
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
GAS LAWS
Boyle’s LawP1V1=P2V2
P= pressureV = volume
Charles LawV1/V2=T1/T2
V = volumeT = temperature
Gay-Lussac’s LawP1/T1=P2/T2
P = pressureT = temperature
Combined Gas Law
Ideal Gas Law
PV = n R T P= pressureV = volumeT = temperaturen = molesR = Gas constant = 0.08206 L-atm/mol K
=
*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
*First Law of Thermodynamics
Energy can be neither created nor destroyed but only transformed
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
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
*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
* 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.
* 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
*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
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
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
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