Thermodynamics

1. Thermodynamics Basics The study of heat and its transformations to mechanical energy Must be a starting point of kinetic energy--Absolute Zero

2. �th Law of Thermodynamics Two objects in thermal equilibrium with a third object are in thermal equilibrium with each other (same as in math: if a=b and b=c then a=c) System-- collection of objects considered together apart from the rest of the universe Internal energy-- total kinetic and potential energy of the atoms/molecules in a system Heat--transfer of thermal energy between objects in a system

3. First Law of Thermodynamics Same as law of conservation of energy As applied to thermodynamics, becomes as follows: Whenever heat is added to a system, it transforms to an equal amount of some other form of energy. As usually applied, Heat added = increase in internal energy + external work done by system Or Q = DU + W which is rearranged to DU = Q - W

4. First Law of Thermodynamics Combustion engines work this way

5. Four Processes In thermodynamics, the first law allows work and heat energy to combine to make internal energy. This allows four possibilities: Isothermal--where temperature of the system does not change DU= � so Q=W Isochoric(isovolumetric)--where the volume does not change DV= � so DU=Q Isobaric--where the pressure does not change DU = Q - W = Q - P DV Adiabatic-- not identical to the isothermal, since temperature changes, but heat does not Q = � so DU = -W = -P DV

6. Examples 13.2 & 13.3 Gas confined by a piston expands against a constant pressure of 100 kPa, and changes volume from 0.15 to 0.25 m3 when 2 � 104 J of heat are absorbed. What work is done and what is the change in internal energy? Isobaric process-- DU = Q - W = Q - P DV Work = P DV= 100,000 ( .25-.15)= 10,000 J DU = Q - W = 2 � 104 J - 10,000 J = 10,000 J A heat engine has its internal energy decreased by 400 J while doing 250 J of work. What heat is given out or taken in? Q = DU + W = -400 J + 250 J = -150 J heat is given out

7. Thermodynamic Processes Reversible processes-- system is nearly in equilibrium at all times In each cycle, DU ends up approximately =� so Q=W in an ideal case However, there is NEVER more work to come out than heat that goes in

8. Carnot Cycle Developed by Sadi Carnot, who showed that all movements were ultimately due to heat His ideal engine sets the upper limit on efficiency, and he recognized that work could only happen as high temperature moved toward lower His cycle consists of four reversible processes, two isothermal and two adiabatic

9. Example 13.4 The gas in a heat engine expands from its initial volume to twice its former value as pressure changes. Show that the work done is given by W=1/2(P1+P2)(V1) Use the graph to see that Work is the area under a PV graph, which consists of two parts, a triangle and a rectangle. Area of the triangle= � (P1-P2)(V2-V1) Area of rectangle = P2(V2-V1) Add these together to find Work W= � P1 (V2-V1) - � P2(V2-V1)+ P2(V2-V1) = � P1 (V2-V1) + � P2(V2-V1) = 1/2(P1+P2)(V2-V1) Since V2=2V1, this becomes 1/2(P1+P2)(V1)

10. Carnot Efficiency and Heat Engines Thermal efficiency = W/QH W= QH-QC so The= QH-QC QH Since QC/QH = TC/TH, Ideal efficiency = (TH -TC ) = 1- TC TH TH Where TH is the temperature of the engine and TC is the temperature of the exhaust to the surroundings Highest efficiency can only be when absolute zero is exhausted to the surroundings Relatively high efficiency can be created if the ? T is large

11. Examples What is the maximum possible thermal efficiency of a steam engine that takes in steam at 100�C and exhausts it at 27 �C ? CHANGE TO KELVIN!! 373-300 = 0.20 =>20% 373

12. Homework #1 p396ff 13, 17, 19, 22, 27, 31

13. Refrigerators and Heat Pumps Refrigerators and heat pumps take advantage of the evaporation/condensation gain and loss of heat to do the jobs we require of them Heat pumps do it in reverse of below

14. Coefficient of Performance Efficiency of a refrigerator is slightly different from a heat engine and is symbolized c.p. The equation for this involves the cold temperature because the refrigerator is removing heat by using work c.p. = QC = QC = TC W QH-QC TH-TC For a heat pump, exchange QC and TC for QH and TH c.p. = QH = QH = TH W QH-QC TH-TC

15. Example 13.8 A household refrigerator has a coefficient of performance of 6.0. If the room temperature outside is 30�C, what is the coldest temperature the refrigerator can make? c.p. = TC so rearrange to get TC = c.p.(TH) TH-TC 1 + c.p. TC = 6.0(303) = 260 K = -13 �C 7.0

16. Second Law of Thermodynamics Heat will never of itself flow from a cold object to a hot object Energy always flows downhill except if an outside agency acts Heat engines cannot change all of the heat to work, since some heat is wasted

17. Order Tends to Disorder Natural systems tend to proceed toward a state of greater disorder This can be reversed within a system by work from the outside

18. Order Tends to Disorder The measure of disorder in a system is entropy In the universe, entropy always increases; in a system it can be reversed

19. Entropy in Calculations Defining change of entropy for a reversible process, a system at average Kelvin temperature T and absorbing an amount of heat Q DS = Q T

20. Examples A student takes a 2.5-kg block of ice at 0�C, places it on a large rock outcropping, and watches the ice melt. (a) What is the entropy change of the ice (water)? (b) If the source of heat (the rock) is very massive and remains at a constant 21�C, what is the entropy change of the rock? (c) What is the total entropy change? The block of ice melts at 0�C = 273 K. The energy required to melt the ice is Q = mLf = (2.5kg)(3.34 x 105 J/kg) = 8.35 x 105 J. Use the defining equation for entropy directly to get the entropy change of the ice: DSice = Q = 8.35 x 105 J = 3060 J/K T 273 K The entropy change of the rock is also found using the same equation, but in this case the heat is negative and the temperature is 21�C = 294 K. DSrock = Q = 8.35 x 105 J = - 2840 J/K T 294 K The total entropy change is DS = DSice + DSrock = 3060 J/K - 2840 J/K = 220 J/K

21. Examples A student mixes 100 g of water at 60�C (sample 1) with 200 g of water at 40�C (sample 2). Determine the change in entropy of the system. Start with the equation heat lost + heat gained = 0 where the heat gained or lost equals mc DT. The heat lost for sample 1 is Q1 = m1 c DT = (100g)(1 cal/g�C)(T - 60�C) The heat gained in sample 2 is Q2 = m2 c DT = (200g)(1 cal/g�C)(T - 40�C) Solving, final temperature = 46.7�C. The heat gained or lost is found to be Q1 = - 1.33 kcal, Q2 = + 1.33 kcal, The average temperature of the first sample is 53.3�C, or 326 K. The average temperature of the second sample is 43.3 �C, or 316 K. The change in entropy of the initially warm water (sample 1) is then approximately. DS1 = Q1 = - 1.33 kcal = - 4.08 x 10-3 kcal/K T1 326 K The change in the entropy of the initially cool water (sample 2) is DS2 = Q2 = + 1.33 kcal = + 4.21 x 10-3 kcal/K T2 316 K The change in the entropy of the system is the sum of the changes in entropy of the component parts, DS = DS1 + DS2 = 0.13 x 10-3 kcal/K

22. Homework #2 Jones/Childers book go to Assignments on drriley.tk 29, 33, 37, 43, 45