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  1. 1. Chapter 1 Basic Concepts of Thermodynamics 1
  2. 2. What is thermodynamics? • The study of thermodynamics is concerned with ways energy is stored within a body and how energy transformations, which involve heat and work, may take place. • Approaches to studying thermodynamics – Macroscopic (Classical thermodynamics) • study large number of particles (molecules) that make up the substance in question • does not require knowledge of the behavior of individual molecules – Microscopic (Statistical thermodynamics) • concerned within behavior of individual particles (molecules) • study average behavior of large groups of individual particles 2
  3. 3. Applications of Thermodynamics Power plants The human body Air-conditioning systems Airplanes Car radiators Refrigeration systems 3
  4. 4. Thermodynamic Systems •Thermodynamic System – quantity of matter or a region of space chosen for study •Boundary – real or imaginary layer that separates the system from its surroundings •Surroundings – physical space outside the system boundary • Types of Systems – Closed – Open 4
  5. 5. A thermodynamic system is that portion of the Universe that we have selected for investigation •The surroundings are everything outside the system. 5
  6. 6. 6
  7. 7. The boundary separates the system from the surroundings 7
  8. 8. Properties • Any characteristic of a system in equilibrium is called a property. • Types of properties – Extensive properties - vary directly with the size of the system Examples: volume, mass, total energy – Intensive properties - are independent of the size of the system Examples: temperature, pressure, color • Extensive properties per unit mass are intensive properties. specific volume v = Volume/Mass = V/m density ρ = Mass/Volume = m/V 8
  9. 9. 9 State & Equilibrium • State of a system – system that is not undergoing any change – all properties of system are known & are not changing – if one property changes then the state of the system changes • Thermodynamic equilibrium – “equilibrium” - state of balance – A system is in equilibrium if it maintains thermal (uniform temperature), mechanical (uniform pressure), phase (mass of two phases), and chemical equilibrium
  10. 10. 10 Processes & Paths • Process – when a system changes from one equilibrium state to another one – some special processes: • isobaric process - constant pressure process • isothermal process - constant temperature process • isochoric process - constant volume process • isentropic process (adiabatic process) - constant entropy process • Path – series of states which a system passes through during a process
  11. 11. 11 Thermodynamic equilibrium (Zeorth Law) • If two systems are in thermal equilibrium with a third system, they are also in thermal equilibrium with each other. • If A and C are each in thermal equilibrium with B, A is also in equilibrium with C. • Temperature as a quality of heat, by Galileo and Newton • The temperatures are equal for all systems in thermal equilibrium. • Temperature scale • Thermometers
  12. 12. 12 First Law of thermodynamics • The first law of thermodynamics is an extension of the law of conservation of energy • The change in internal energy of a system is equal to the heat added to the system minus the work done by the system ΔU = Q - W
  13. 13. 13 First Law of thermodynamics
  14. 14. 14 • An adiabatic process transfers no heat – therefore Q = 0 • ΔU = Q – W • When a system expands adiabatically, W is positive (the system does work) so ΔU is negative. • When a system compresses adiabatically, W is negative (work is done on the system) so ΔU is positive. Adiabatic process
  15. 15. 15 • An isothermal process is a constant temperature process. Any heat flow into or out of the system must be slow enough to maintain thermal equilibrium • For ideal gases, if ΔT is zero, ΔU = 0 • Therefore, Q = W – Any energy entering the system (Q) must leave as work (W) Isothermal process
  16. 16. 16 Isobaric process • An isobaric process is a constant pressure process. ΔU, W, and Q are generally non-zero, but calculating the work done by an ideal gas is straightforward W = P·ΔV • Water boiling in a saucepan is an example of an isobar process
  17. 17. 17 Isochoric process • An isochoric process is a constant volume process. When the volume of a system doesn’t change, it will do no work on its surroundings. W = 0 ΔU = Q • Heating gas in a closed container is an isochoric process
  18. 18. 18 • The amount of heat required to raise a certain mass of a material by a certain temperature is called heat capacity Q = mcxΔT • The constant cx is called the specific heat of substance x, (SI units of J/kg·K) Heat Capacity
  19. 19. 19 ( ) ( ) V T U U dU dT dV T V       ( ) V V U C T    The heat capacity at constant volume dU=dQ=Qv, at constant volume, no additional work QV=CvdT Heat Capacity
  20. 20. 20 • CV = heat capacity at constant volume CV = 3/2 R • CP = heat capacity at constant pressure CP = 5/2 R • For constant volume Q = nCVΔT = ΔU • The universal gas constant R = 8.314 J/mol·K Heat Capacity for of ideal gas
  21. 21. 21 Latent Heat The word “latent” comes from a Latin word that means “to lie hidden.” When a substance changes phases (liquid  solid or gas  liquid) energy is transferred without a change in temperature. This “hidden energy” is called latent heat. For example, to turn water ice into liquid water, energy must be added to bring the water to its melting point, 0 ºC. This is not enough, however, since water can exist at 0 ºC in either the liquid or solid state. Additional energy is required to change 0 ºC ice into 0 ºC water. The energy increases the internal energy of the water but does not raise its temp. When frozen, water molecules are in a crystalline structure, and energy is needed to break this structure. The energy needed is called the latent heat of fusion. Additional energy is also needed to change water at 100 ºC to steam at 100 ºC, and this is called the latent heat of vaporization.
  22. 22. 22 Latent Heat Formula L is the energy per unit mass needed to change the state of a substance from solid to liquid or from liquid to gas. Ex: Lf (the latent heat of fusion) for gold is 6440 J/kg. Gold melts at 1063 ºC. 5 grams of solid gold at this temp will not become liquid until additional heat is added. The amount of heat needed is: (6440 J/kg)(0.005 kg) = 32 J. The liquid gold will still be at 1063 ºC. Q = mLf or Q = mLv Q = thermal energy m = mass L = heat of fusion or vaporization
  23. 23. 23 Enthalpy – heat of transformation • △U= Qv (Closed system, only expansion work exchanging, constant volume) • Qv , heat transactions at constant volume • Qp, heat transactions at constant pressure (heat of transformation = (heat absorbed / (m or n) ΔU= Q-W=Qp- PΔV ΔU= Qp – P2 V2 + P1V1= U2-U1 QP=(U2+ P2 V2)-(U1+ P1V1) H≡U+PV,∴Qp=H2-H1=ΔH •State function, extensive • the heat of transformation in any change of phase is equal to the difference between the enthalpies of the system in the two phases. Enthalpy
  24. 24. 24 Enthalpy – heat of transformation Let l12 = lS-L : fusion, l23 = lL-V : vaporization, l13 = lS-V : sublimation In a cyclic process, Δh = 0. S –L—V—S Therefore: Δh1 + Δh2 + Δh3 = 0. l13 = l12 + l23 e.g. latent heat of vaporization of water:
  25. 25. 25 Comparing H and U • H=U+PV • State function • Same units
  26. 26. 26 Heat capacity • Heat capacity at constant pressure • Heat capacity at constant volume • Closed system, in equilibrium, PV work only ( ) d p p p Q H C T T      d p p H Q C T D    ( ) d V V V Q U C T T      d V V U Q C T D   
  27. 27. 27 Second law of thermodynamics • The second law of thermodynamics introduces the notion of entropy (S), a measure of system disorder • The 2nd Law can also be stated that heat flows spontaneously from a hot object to a cold object (spontaneously means without the assistance of external work) • The 2nd Law helps determine the preferred direction of a process • A reversible process is one which can change state and then return to the original state • This is an idealized condition – all real processes are irreversible
  28. 28. 28 Heat engines • A device which transforms heat into work is called a heat engine • This happens in a cyclic process • Heat engines require a hot reservoir to supply energy (QH) and a cold reservoir to take in the excess energy (QC) - QH is defined as positive, QC is negative Cold Reservoir, TC Engine Hot Reservoir, TH QH QC W Real engine. QH = QC + W
  29. 29. 29 Thermal Efficiency of a Heat Engine eng 1 h c c h h h W Q Q Q e Q Q Q      Second Law: Kelvin-Planck Form • It is impossible to construct a heat engine that, operating in a cycle, produces no other effect than the absorption of energy from a reservoir and the performance of an equal amount of work – Means that Qc cannot equal 0 Some Qc must be expelled to the environment – Means that e cannot equal 100% Cold Reservoir, TC Engine Hot Reservoir, TH QH QC = 0 W Impossible engine. QH = W
  30. 30. Refrigerators Cold Reservoir, TC Engine Hot Reservoir, TH QH QC W Real fridge. QC + W = QH A refrigerator forces heat from a cold region into a warmer one. It takes work to do this, otherwise the 2nd Law would be violated. Can a fridge be left open in the summer to provide a make shift air conditioner? Nope, since all heat pumped out of the fridge is pumped back into the kitchen. Since QH > QC because of the work done, leaving the refrigerator open would actually make your house hotter!
  31. 31. 31 Impossible fridge. QC = QH Cold Reservoir, TC Engine Hot Reservoir, TH QH QC W = 0 Second Law – Clausius Form • It is impossible to construct a cyclical machine whose sole effect is to transfer energy continuously by heat from one object to another object at a higher temperature without the input of energy by work • Or – energy does not transfer spontaneously by heat from a cold object to a hot object.
  32. 32. 32 COP c Q W  • In cooling mode, • A good refrigerator should have a high COP – Typical values are 5 or 6 • The effectiveness of a heat pump is described by a number called the coefficient of performance (COP). • In heating mode: energy transferred at high temp COP = work done by heat pump h Q W 
  33. 33. 33 Carnot Cycle • AB and CD are isothermal processes • BC and DA are adiabatic processes • The work done by the engine is shown by the area enclosed by the curve, Weng • The net work is equal to |Qh| – |Qc| • DEint = 0 for the entire cycle h c c h c h c T T e and T T Q Q    1 • Temperatures must be in Kelvins • All Carnot engines operating between the same two temperatures will have the same efficiency
  34. 34. 34 Entropy and second Law of thermodynamics • The processes in which the entropy of an isolated system would decrease do not occur. In every process taking place in an isolated system, the entropy of the system either increase or remains constant. • Entropy of Carnot cycle (ΔS =0) • for all cycles ΔS =0. therefore • dS is exact differential . S: extensive properties. Calculation of entropy changes in Reversible Process. 1. reversible adiabatic process: dQ= 0 and dS=0 (S=cte) : isentropic process. 2. reversible isothermal process: from a to b
  35. 35. 35 E.g. : Change in phase at cte pressure and T: 3. reversible isochoric process: no phase change 4. reversible isobaric process: no phase change Since System+ surrounding =universe Suniverse =cte
  36. 36. 36 T-S diagram for Carnot cycle: Qr : Area of the cycle = net heat flow into the system.
  37. 37. 37 Calculation of entropy changes in irreversible Process. for reversible process only • Consider a body at temperature T1 in contact with heat reservoir at T2 (T2 > T1) heat flow into the boby , therefore ΔS> 0 How does Sreservoir changes? Q =cp (T2 –T1) into the body
  38. 38. 38 In every irreversible process, the entropy of the universe increases.
  39. 39. 39 Problems: Chapter 3 (P 90-96) : 3 -6-16-26-28 Chapter 5 (141-146): 4-7-10-14-13-16-20-21.
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