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Thermodynamics

Thermodynamics. Chapter 15. Calculating Work. Work = area under Pressure vs. Volume graph W = Fd F = PA W=PAd W = P D V Calculus link W = - p dV. V 2. V 1. Isochoric (isovolumetric). No change in volume W = 0. Isobaric Process . No change in pressure

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Thermodynamics

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  1. Thermodynamics Chapter 15

  2. Calculating Work Work = area under Pressure vs. Volume graph W = Fd F = PA W=PAd W = PDV Calculus link W = - p dV V2 V1

  3. Isochoric (isovolumetric) • No change in volume • W = 0

  4. Isobaric Process • No change in pressure • Volume expands and does work • WORK = - AREA

  5. Work from Graph: Example 1 A gas expands at a constant pressure of 1 X 105 N/m2 (~100 kPa, ~ 1 atm) from a volume of 10 mL to a volume of 35 mL. Calculate the work done by the gas.

  6. W = PDV W = (1 X 105 N/m2)(0.035 L – 0.010 L) W = 2500 J

  7. Work from Graph: Example 2 A gas is cooled to produce a decrease in pressure. The volume is held constant at 10.0 m3, and the pressure decreases from 1 X 105 N/m2 to 0.5 X 105 N/m2. Calculate the work done on the gas.

  8. W = PDV W = P X 0 W = 0 J (No work is done)

  9. Work from Graph: Example 3 A gas expands and increases in pressure as shown in the next graph. Calculate the work done on the gas. (Remember to include the entire area under the graph).

  10. ANS:

  11. Example 4 Calculate the work done from 500 to 1000 cm3 on the following graph. Remember to go all the way down to the x-axis.

  12. Isothermal Systems • Area under the curve W = - p dV = - nRT dV = - nRT dV V V W = -nRT ln Vf = -piVi ln Vf = pfVf ln Vf Vi Vi Vi

  13. Isothermal Example A cylinder contains 7.0 grams of nitrogen (N2). • Calculate the moles of N2 • Calculate the work that must be done to compress the gas at a constant temperature of 80oC until the volume is halved.

  14. Top Line = at 500K Bottom Line = at 300K

  15. Work and Cyclic Processes • Cyclic process – gas returns to its original state • Important for studying • Steam engine • Car engine DU = 0 Q = W • Work = Area enclosed

  16. Work and Cyclic Processes Calculate the work shown in the previous graph.

  17. Heat (Q) • Heat – Energy transferred from one body to another because of a difference in temperature • Cooking a turkey hot oven  cooler turkey • Extensive Property – depends on amount of material (iceberg vs. water) • Unit • Joules. • Calorie

  18. calories • calorie – amount of heat energy needed to raise the temperature of 1 gram of water by 1 degree Celsius (or Kelvin) • Not a nutritional Calorie 1 nutritional Calorie = 1000 calories (or 1 kilocalorie)

  19. Converting between heat units 1 cal = 4.18 Joules 252 cal = 1 BTU 1054 Joules = 1 BTU

  20. James Prescott Joule • Weight moves paddles • Friction from paddles warms water • Workfalling = Heatpaddles • Mechanical Equivalent of Heat

  21. Heat: Example 1 How high would you have to climb to work off a 500 Calorie ( 500,000 cal) ice cream? Assume you mass 60 kg. (500,000 cal)(4.186J/cal) = 2.09 X 106 J Heat = Work Heat = mgh h = Heat/mg h = 2.09 X 106 J/(60 kg)(9.8 m/s2) = 3600 m h ~ 11,000 ft

  22. Heat: Example 2 A 3.0 gram bullet travels at 400 m/s through a tree. After passing through the tree, the bullet is now only going 200 m/s. How much heat was transferred to the tree?

  23. KE = ½ mv2 KE = ½ mv2 KE = ½ (0.003 kg)(400 m/s)2 KE = ½ (0.003 kg)(200 m/s)2 KE = 240 J KE = 60 J Heat loss = 180 J to the tree

  24. The Laws of Thermodynamics 1st Law • Energy is conserved DE = W + Q 2nd Law • Natural processes tend to move toward a state of greater disorder • Heat goes hot to cold (Clausius) • No device converts all heat to work (Kelvin-Planck) DS >0

  25. The Laws of Thermodynamics 3rd Law • The entropy of a pure crystal at absolute zero is zero DG = DH - TDS

  26. Important Definitions Thermodynamics – Study of the transfer of energy as heat and work System – Objects we are studying

  27. First Law Sign Conventions Heat is added to the system + Heat is lost - Work on the system + Work done by the system -

  28. First Law Example 1 2500 J of heat is added to a system. This heat does 1800 J of work on the system. Calculate the change in internal energy. DE = W + Q DE = 1800 J +2500 J DE = 4300 J

  29. First Law Example 2 2500 J of heat is added to a system. This increase in temperature allows the system to do 1800 J of work. Calculate the change in internal energy. DE = W + Q DE = -1800 J +2500 J DE = 700 J

  30. Adiabatic Systems – No heat exchange (Q=0) • Fast processes • Heat has no time to enter/leave system • Car piston Q = 0 For an ideal gas DE = W + Q DE = 3/2 nRT DE = W DE = 3nRDT 2

  31. Work: Example 1 In an engine, 0.25 moles of gas in the cylinder expand adiabatically against the piston. The temperature drops from 1150 K to 400K. How much work does the gas do? (the pressure is not constant).

  32. Q = 0 DE = Q-W DE = W DE = 3nRDT 2 DE = 3(0.25 mole)(8.315 J/K-mol)(400 K - 1150 K) 2 DU = 2300 J W = 2300 J

  33. Temperature and Internal Energy Temperature – measure of the average kinetic energy of individual molecules Internal (Thermal) Energy – Total energy of all the molecules in an object U = 3 nRT 2

  34. Temperature Suppose we heat mugs of water from 25oC to 90oC. One Mug Two Mugs 25oC to 90oC 25oC to 90oC Same Temp. Change Same Temp. Change Requires less heat Requires more heat

  35. Specific Heat Specific Heat – Amount of heat needed to raise the temperature of one gram of a substance by one degree Celsius or Kelvin • Unit – J/kgoC • Symbol = C • The higher the specific heat, the more energy needed to raise the temperature • Wooden spoon versus a metal spoon

  36. Calculating Heat Examples: • Al has a specific heat of 0.22 kcal/kg oC • Gold has a specific heat of 0.03 kcal/kg oC • Which gets hotter sitting in the sun? • Water’s specific heat: 1.00 kcal/kg oC or 4186 J/kgoC

  37. Calculating Heat Suppose you immerse a hot pan into a dishpan of water to cool it. Which will experience a greater change in temperature (gain or loss)? Heat lost = -Heat gained Q1 + Q2 + Q3 +…. = 0

  38. Calculating Heat Q = mCpDT Q = heat (J) m = mass (kg) Cp = specific heat (J/kgoC) DT = Tfinal – Tinitial

  39. Calculating Heat: Example 1 How much heat must be supplied to a 500.0 gram iron pan (C = 450 J/kgoC) to raise its temperature from 20.0oC to 100oC? Q = mCDT Q = (0.500 kg)(450/kg oC)(100oC -20oC) Q = (0.500 kg)(450/kg oC)( 80oC) Q = 18,000 J or 18 kJ

  40. Calculating Heat: Example 2 Suppose the pan is filled with 400 g of water. What would be the total heat? Q = mCDT Q = (0.400 kg)(4186/kg oC)(100oC -20oC) Qwater =134,000 J Qtotal = 152,000 J or 152 kJ

  41. Calculating Heat: Example 3 200-g of tea at 95oC is poured into a 150-g glass (C = 840 J/kgoC) at 25oC. What will be the final temperature of the cup/glass? Heat lost = -Heat gained Qtea = - Qcup mteaCteaDT = -mcupCcupDT

  42. mteaCteaDT = -mcupCcupDT (0.200 kg)(4186J/kgoC)(T-95oC) = -(0.150 kg)(840J/kgoC)(T-25oC) (837)(T-95oC) = -(126)(T-25oC) 837T – 79,500 = 3150 -126T 963 T = 82,700 T = 86oC

  43. Calculating Heat: Example 4 A 100-g piece of aluminum (C = 900 J/kgoC) is heated to 100 oC and immersed in 250-g of water at 20 oC. What is the final temperature of the system? Heat lost = -Heat gained Qtea = - Qcup mAlCAlDT = -mwaterCwaterDT

  44. mAlCAlDT = -mwaterCwaterDT (0.100 kg)(900J/kgoC)(T-100oC) = -(0.250 kg)(4186J/kgoC)(T-20oC) (90)(T-100) = -(1047)(T-25) 90T – 9,000 = 20,930 - 1047T 1137T = 29,930 T = 26oC

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