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Physics Subject Area Test

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

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Physics Subject Area Test

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  1. Physics Subject Area Test Thermodynamics

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

  3. Converting Between the Kelvin and Celsius Scales • Converting Between the Fahrenheit and Celsius Scales

  4. Thermal expansion • Expandingsolidsmaintain original shape • Expandingliquidsconform to the container • Linear expansion ΔL = αLΔT L = length α = coefficient of liner expansion ΔT = temperature change

  5. 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 -35⁰C and a hot summer day when the temperature is +35 ⁰C ? ΔL = αLΔT = 12x10-6/ ⁰C x 646 x 70⁰C = 0.54m

  6. Volume expansion ΔV= βVΔT L = length β = coefficient of liner expansion ΔT = temperature change cold hot Β = 3 α

  7. Heat flow: the heat current; the amount of heat that passes by some given place on the rod per unit time = heat flow k = proportionality constant, thermal conductivity A = cross sectional area ΔT = change in temperature Δx = distance crossed, thickness of material Thermal Conduction

  8. convection Heatisstored in a movingfluid and iscarriedfrom one place to another by the motion of thisfluid • radiation The heatiscarriedfrom one place to another by electromagneticwaves • conduction the process of handing on energy from one thing to the next

  9. 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 Stefan-Boltzmann Law

  10. Latent heat ( heat of transformation) – the heat absorbed during the change of state ΔQ = quantity of heat transferred m = 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 Changes of State

  11. P1V1=P2V2 P= pressure V = volume GAS LAWSBoyle’s Law

  12. V1/V2=T1/T2 Charles Law • V = volume • T = temperature

  13. P1/T1=P2/T2 Gay-Lussac’s Law • P = pressure • T = temperature

  14. = Ideal Gas Law • P= pressure • V = volume • T = temperature • n = moles • R = Gas constant • = 0.08206 L-atm/mol K • PV = n R T Combined Gas Law

  15. PV = nRT n/V = P/RT Molarity = n/V Density D = m/V Molecular WtM = m/n D=Mn/V = PM/RT M= DRT/P Gas Density

  16. Energy can be neither created nor destroyed but only transformed First Law of Thermodynamics

  17. Energy In = Energy OutorU2- U1= Q -W where U1: internal energy of the system at the beginning U2: internal energy of the system at the end Q : net heat flow into the system W : net work done by the system Q = ΔU + ΔW THE GENERAL ENERGY EQUATION

  18. A closed tank has a volume of 40.0 m2 and is filled with air at 25⁰C and 100 kPa. We want to maintain the temperature in the tank at 25⁰C as water is pumped into it. How much heat will have to be removed from the air in the tank to fill it half full? = (100kPa) (40.0 m2)(-0.69314) = -2772.58kJ

  19. 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 STATE CHANGES

  20. 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). 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. For quasi-equilibrium processes, P, V, T are well-defined – the “path” between two states is a continuouslines in the P, V, T space. Examples ofquasi- equilibrium processes: Quasi-Static Processes • isochoric:V = const • isobaric:P = const • isothermal:T = const • adiabatic:Q = 0 P 2 V 1 T

  21. The workdone by an external forceon a gas enclosed within a cylinder fitted with a piston: A – the piston area W = (PA) dx = P (Adx) = - PdV force 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). x P Work W = - PdV - applies to any shape of system boundary dU = Q – PdV 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.

  22. SpecificHeat the heatabsorbedduring the change of state Q = nCvΔT Q = amount of heat required n = number of moles Cv = specific heat at a constant volume ΔT = Change in temperature

  23. 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 energy m = mass of substance T = change in temperature (Tf – Ti) Cp = specific heat of substance

  24. 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 molecules The 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

  25. 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 Kinetic molecular theory

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