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Physics Subject Area Test Thermodynamics. There are three commonly used temperature scales, Fahrenheit, Celsius and Kelvin.

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Presentation on theme: "Physics Subject Area Test Thermodynamics. There are three commonly used temperature scales, Fahrenheit, Celsius and Kelvin."— Presentation transcript:

1 Physics Subject Area Test Thermodynamics

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

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

4 Thermal expansion Expanding solids maintain original shape Expanding liquids conform to the container Linear expansion ΔL = αLΔTL = 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 70 ⁰C = 0.54m

6 Volume expansion ΔV= βVΔTL = length β = coefficient of liner expansion ΔT = temperature change coldhot 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


9 * convection Heat 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




13 P 1 V 1 =P 2 V 2 P= pressure V = volume

14 V 1 /V 2 =T 1 /T 2 V = volume T = temperature

15 P 1 /T 1 =P 2 /T 2 P = pressure T = temperature

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

17 PV = nRT n/V = P/RT Molarity = n/V Density D = m/V Molecular Wt M = m/n D=Mn/V = PM/RT M= DRT/P

18 Energy can be neither created nor destroyed but only transformed

19 THE GENERAL ENERGY EQUATION Energy In = Energy Out or U 2 - U 1 = Q -W where U 1 : internal energy of the system at the beginning U 2 : 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

20 A closed tank has a volume of 40.0 m 2 and is filled with air at 25 ⁰C and 100 kPa. We want to maintain the temperature in the tank at 25 ⁰C 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 m 2 )(-0.69314) = -2772.58kJ


22 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

23 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.

24 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). 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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

25 * Specific Heat the heat absorbed during the change of state Q = nC v ΔTQ = amount of heat required n = number of moles C v = specific heat at a constant volume ΔT = Change in temperature

26 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 C p Q = change in thermal energy m = mass of substance  T = change in temperature (T f – T i ) C p = specific heat of substance

27 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

28 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

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