1. That “entopy-nerial” spirit! What is thermodynamics? The movement of heat! (or more properly, energy!) There are 3 Laws of Thermodynamics: These laws may seem remote to most people, but they can be applied to every aspect of science, from astrophysics to meteorology, to zoology, to pizza! The First Law says energy is conserved; The Second Law says everything moves toward equilibrium because of something called entropy; The Third Law says that there is a lowest temperature, called absolute zero, where this entropy stuff is zero. Now, for the details…

2. First Law of Thermodynamics The First Law is usually referred to as the Law of Conservation of Energy. 1. If the state of a system is changed by applying work or heat (or both) then the change in the energy of the system must equal the amount of energy applied. 2. If the system is doing work, then the system cannot do this work without losing energy in the amount equal to the work. YOU EXPERIENCE THIS DAILY! IF YOU ARE DOING WORK, YOU LOSE ENERGY! ALSO, HOW MANY TIMES HAVE YOU SAID “YOU ONLY GET OUT OF IT WHAT YOU PUT INTO IT!”, OR “YOU GET WHAT YOU PAY FOR!” See! You already knew the First Law! Essentially, the First Law states that energy cannot be created or destroyed. In other words, energy is transferred, or, “there’s no free lunch!”!

3. Second Law of Thermodynamics Let’s start this one with an example: If you take two bottles. Fill one with a gas and make the other a vacuum, and then connect the bottles. The gas will quickly reach a state of equilibrium where it fills both bottles with equal pressure and temperature, both values lower than those in the first bottle. Since the First Law states that the energy of the system is the same after the expansion as it was before the expansion, then there must be a new value, we call it entropy, which increased with the loss of heat. Why was there a heat loss? The gas expanded! When a gas expands, there are less molecular collisions. This is why aerosols are cool when you spray them. This is why the nozzle of a fire extinguisher gets cold when in use. And… This is why air in the atmosphere that rises, cools. Less air molecules in the upper parts of the atmosphere means less pressure. Less pressure means expansion. Expansion means cooling.

4. Second Law of Thermodynamics Entropy is defined as the "capacity for change" of a system. If the state of a system is changed but the entropy is not changed, then the process is called reversible (able to be changed back to the original state without any more added energy). Wouldn’t you like to take back something you did or said as a youth? Or maybe as an adult? Unfortunately, taking back something usually requires A LOT of extra energy on our part! Meaning, most processes in nature are irreversible! It is said that universal entropy is always increasing - since entropy is the driving force behind equilibrium, this means that the universe is constantly moving toward a less dynamic state. Simply put, our universe, our atmosphere, etc. are always trying to reach equilibrium! Let’s think about this for a second… Will it ever happen? What would have to happen for the atmosphere to reach equilibrium?

5. Second Law of Thermodynamics You would have to stop transferring energy from a certain external object (sun). That would be a cold proposition! The formal statement of the Second Law is: It is impossible to move heat, by a cyclical process, from something at lower temperature to something at higher temperature unless work is added to the system. Since any two objects at different temperatures brought together will come to equilibrium at the same temperature, with increased entropy for the spontaneous change, to force the heat to move in the opposite direction requires some external source of energy (work) to make up for the change in entropy. A good example is the refrigerator: if you leave the door to your refrigerator open (don't try this with perishables on the shelves), what happens to the temperature in the kitchen?

6. Third Law of Thermodynamics At lower temperatures, the change in entropy for spontaneous reaction decreases. This observation lead to the postulate, that as the temperature of a pure substance approached some lower limit (called absolute zero) the entropy for would approach zero. After plotting the temperatures and energy changes for several spontaneous reactions, it is possible to work backwards to find the value of absolute zero, which is -273C (about -460F). The Kelvin scale places 0K at absolute zero and then uses Celsius for increment size, so that water freezes at 273K and boils at 373K. With further math, the nature of absolute zero was defined as the Third Law: If the entropy of each element at absolute zero can be taken as zero, then all elements above absolute zero must have a finite, positive entropy. However, because entropy cannot be reduced to zero (as per the Second Law), no system can reach absolute zero.

7. Summary of Thermodynamics Energy is transferred; Our systems always tend towards an equilibrium; And we likely won’t ever get there.

8. Let’s now look at pizza! To a first approximation, the pizza is composed of three cylindrical disks or layers. The lowest layer is made of yeast-flour dough, the middle layer is made largely of tomato paste, and the top layer is cheese. Trace materials such as oregano, basil and pepper are also present in small quantities. For now, we will ignore any toppings.

9. Are the ingredients by themselves sufficient to be called a pizza? In order to have a “pizza”, you must first apply some energy! But how much? You have three very different types of ingredients (or constituents) in this layered configuration. Each will likely behave differently when energy is added. What are some outcomes in nature when energy is added to a system?

10. Outcomes when energy is added… The First Law says the system will either heat up or do work. The Second Law says the system will tend toward an equilibrium, but that some irreversible changes are likely to occur. Let’s look at the 1st Law: How will the system heat up when energy is added? Energy is transferred via: CONDUCTION CONVECTION RADIATION

11. Heat Transfer Conduction Occurs in materials when heat flows from the hot part of the substance to the colder part. There is no net movement of matter involved. Radiation This is electromagnetic radiation emitted by a body due only to its temperature. This radiation spans a range of wavelengths and the intensity of the energy for a particular wavelength depends on the temperature of the emitting body. Convection Movement of molecules of a fluid due to a change in density when heated.

12. Effects of Heat Transfer Temperature Change Motion Phase Change Now, let’s apply these changes to our three layered system, the pizza! Three things occur as the stacked disk structure equilibrates with the high temperature oven. (533K, 160K higher than water’s boiling point.)

13. The Crust The dough bakes into bread, a low- water content material with a large number of non-connecting air spaces. The crust becomes an excellent insulator because of the unconnected air spaces throughout. Ever burn a piece of bread on the inside? Try it in a toaster sometime!

14. The Sauce The tomato paste dehydrates. Tomato paste has a high heat capacity and a low conductance. It serves as a buffer between the insulating crust and the cheese.

15. The Cheese The cheese undergoes a complex series of transitions involving protein de-naturation and lipid rearrangement from liquid crystals to more disordered states. (In other words, it melts!) The transitions result in a high heat capacity for the melted cheese.

16. The Cheese Because of the insulating dough and the high heat capacity paste, the cheese layer is well insulated from below. Meaning, most of the energy (heat) loss will be through the top of the cheese. When you eat pizza hot out of the oven, do you burn your tongue? Or the roof of your mouth?

17. The Ensemble Conduction, convection and radiation will be at work transferring energy through the top of the pizza. To eliminate the energy loss through the top, place an insulating layer over it. Air is a great insulator! Just like in our atmosphere! But, to keep the air in place (limit energy loss due to convection), use a “pizza box”. These boxes also tend to have non- connecting air spaces, just like the crust (corrugated).

18. Latent Heat When a solid melts or a liquid boils, energy must be added but the temperature remains constant! (This can be explained by considering that it takes energy to break the bonds holding the material together.) The amount of energy it takes to melt or boil a certain amount of material is called a latent heat.

19. Latent Heat For water, the latent heat of fusion (heat needed to melt ice to water) is 79.7 cal/gm. For water, the latent heat of vaporization (heat needed to boil water) is 540 cal/gm. For alcohol, the latent heat of vaporization is less at 204 cal/gm.

20. Latent Heat - Example Example: how much energy does it take to vaporize 1 liter of water if the water is initially at a temperature of 98 o F ?

21. Latent Heat - Example cont. First we need to find the energy to raise the temperature of the water up to boiling: this involves the heat capacity (which for water is 1 cal/gm* o C) (density of water is 1 gm/cc, 1 liter = 1000 cc): C = Q/(m*  T), with  T = (212-98)*5/9=63 o C Q = (1 cal/gm* o C)*(1 gm/1cc)*1000 cc*63 o C = 63,333 cal * (4.186 J/cal) = 265,000 J.

22. Latent Heat - Example cont. Now we add in the latent heat: (for water, this is 540 cal/gm) Q = L*m = (540 cal/gm)*(1 gm/cc)*(1000 cc) = 540,000 cal * (4.186 J/cal) = 2,260,000 J Total energy required is: 265,000 J + 2,260,000 J = 2,525,000 J.

23. Latent Heat - Example #2 Question: how much water would be needed to keep cool for 4 hours by evaporation if the outside temperature is 100 o F (essentially same as body’s) and a power output of 200 Watts (doing some work) is desired?

24. Latent Heat - Example cont. Since the body generates 200 Watts, or 200 Joules a second, the body must evaporate water to carry this energy away. Q = (200 J/sec)*(4 hs)*(3600 sec/hr) = 2,880,000 J. From the previous considerations, evaporating 1 liter of water carries away 2,525,000 J. Thus we need 2.88MJ / (2.525MJ/liter) = 1.14 liters of water.

25. Latent Heat - Example cont. Would more or less alcohol be needed to keep cool for the same energy output? (The heat capacity of alcohol is 2.4 J/gm* o C; the density of [ethanol].791 gm/cc; the boiling point is 78 o C; latent heat of vaporization is 854 J/gm). From this you should be able to decide whether water or alcohol is better for heat regulation.

26. Heat Transfer There are four ways of moving heat: Evaporation (using latent heat - we’ve already looked at this) Convection (moving heat with a material) Conduction (moving heat through a material) Radiation We’ll develop equations for conduction and radiation and talk about convection.

27. Heat Transfer: Convection Heat Transfer by Convection is when you heat some material and then move that material containing the heat. The amount of heat energy moved depends on the heat in the material (heat capacity times amount of material times the temperature difference) and how much material you move per time. The blood and hot air furnaces use this method.

28. Heat Transfer: Conduction Heat will flow through a solid material from the hot end to the cold end. What is flowing? No matter is flowing! We can think of energy as flowing in this case! We measure the flow of energy as power: 1 Watt = 1 Joule/sec.

29. Heat Transfer: Conduction Power = Q/t = k*A*  T/L where k is a constant that depends on the material, called the thermal conductivity; where A is the cross sectional area; where L is the distance from the hot end to the cold end; and  T is the temperature difference between the hot and cold ends. A L k T hi T low L hotcold

30. Conduction - R values Units of thermal conductivity: from Power = Q/t = k*A*  T/L k has units of W*m/(m 2 K) or J/(sec*m*K), and k depends only on the material. Often a material is given an R value, where R includes both the material and the thickness of the material: R = L/k, and R has the units of m 2 *sec*K/J (or ft 2 * o F*hr/BTU, where 1 ft 2 * o F*hr/BTU = 0.176 m 2 *sec*K/J )

31. Conduction - R values P = Q/t = k*A*  T/L = A*  T / R where we define R = L / k. where if we have several different materials and thicknesses, we can simply add the individual R’s to get the total R: R total =  R i.

32. Conduction - Conductivity approximate k values for some materials: –metals: k = 1 cal/(sec*cm* o C) –glass: k = 2 x 10 -3 cal/(sec*cm* o C) –wood, brick, fiberglass: k = 1 x 10 -4 cal/(sec*cm* o C) –air: k = 5 x 10 -5 cal/(sec*cm* o C)

33. Conduction - Example Let’s calculate the R value of brick 4 inches in thickness: L = 4 in * (.0254 m/in) =.10 m k = 1.5 x 10 -4 cal/(sec*cm* o C) * (4.186 J/cal)*(100 cm/m) =.063 J/(sec*m* o C) R = L/k =.10 m /.063 J/(sec*m* o C) = 1.6 m 2 *K/Watt * (1 ft 2 * o F*hr/BTU) /(.176 m 2 *K/Watt) = 9 (1 ft 2 * o F*hr/BTU)

34. Conduction - Example What is the heat loss through the brick walls (assume no other insulation) of a house that is 50 ft x 30 ft (floor area: 1500 ft 2 ) x 8 ft when the temperature inside is 72 o F and the temperature outside is 20 o F ?

35. Conduction - Example P = Q/t = k*A*  T/L = A*  T / R A = 50ft * 8ft + 30ft * 8ft + 50ft * 8ft + 30ft*8ft = 1280 ft 2 * (1 m 2 /10.76 ft 2 ) = 120 m 2.  T = (72-20) o F * (5K/9 o F) = 29 K R = 1.6 m 2 *K/Watt P = 120 m 2 * 29 K / 1.6 m 2 *K/Watt = 2175 Watts = 2.175 kW.

36. Blackbody Radiation: What is a blackbody? A BLACK object absorbs all the light incident on it. A WHITE object reflects all the light incident on it, usually in a diffuse way rather than in a specular (mirror-like) way.

37. Blackbody Radiation: The light from a blackbody then is light that comes solely from the object itself rather than being reflected from some other source. A good way of making a blackbody is to force reflected light to make lots of reflections: inside a bottle with a small opening.

38. Blackbody Radiation: If very hot objects glow (such as the filaments of light bulbs and electric burners), do all warm objects glow? Do we glow? (Are we warm? Are you HOT?)

39. Blackbody Radiation: What are the parameters associated with the making of light from warm objects? –Temperature of the object. – Surface area of the object. – Color of the object ? (If black objects absorb better than white objects, will black objects emit better than white objects?)

40. Blackbody Radiation: Color Experiment Consider the following way of making your stove hot and your freezer cold:

41. Color Experiment Put a white object in an insulated and evacuated box with a black object. The black object will absorb the radiation from the white object and become hot, while the white object will reflect the radiation from the black object and become cool. Put the white object in the freezer, and the black object in the stove.

42. Color Experiment Does this violate Conservation of Energy?

43. Color Experiment Does this violate Conservation of Energy? NO Does this violate the Second Law of Thermodynamics (entropy tends to increase) ?

44. Color Experiment Does this violate Conservation of Energy? NO Does this violate the Second Law of Thermodynamics (entropy tends to increase) ? YES This means that a good absorber is also a good emitter, and a poor absorber is a poor emitter. Use the symbol  to indicate the blackness (  =1) or the whiteness (  =0) of an object.

45. Blackbody Radiation: What are the parameters associated with the making of light from warm objects? – Temperature of the object, T. – Surface area of the object, A. – Color of the object, 

46. Blackbody Radiation: Is the  for us close to 0 or 1? (i.e., are we white or black?) We emit light in the IR, not the visible. So what is our  for the IR?

47. Blackbody Radiation: So what is our  for the IR? Have you ever been near a fire on a cold night? Have you noticed that your front can get hot at the same time your back can get cold? Can your hand block this heat from the fire? Is this due to convection or radiation?

48. Blackbody radiation: For humans in the IR, we are all fairly good absorbers (black). An estimated value for  for us then is about.97.

49. Blackbody Radiation: Experimental Results At 310 Kelvin, only get IR Intensity wavelength UV IRblue yellow red

50. Blackbody Radiation: Experimental Results At much higher temperatures, get visible look at blue/red ratio to get temperature Intensity wavelength UV IRblue yellow red

51. Blackbody Radiation: Experimental Results P total =  AT 4 where  = 5.67 x 10 -8 W/m 2 *K 4  peak = b/T where b = 2.9 x 10 -3 m*K Intensity wavelength UV IRblue yellow red

52. Blackbody Radiation: Example Given that you eat 2000 Calories/day, your power output is around 100 Watts. Given that your body temperature is about 90 o F, and Given that your surface area is about 1.5 m 2,

53. Blackbody Radiation: Example Given P total = 100 Watts Given that T body = 90 o F Given that A = 1.5 m 2 WHAT IS THE POWER EMITTED VIA RADIATION?

54. Blackbody Radiation: Example P emitted =  AT 4 –  =.97 –  = 5.67 x 10 -8 W/m 2 *K 4 – T = 273 + (90-32)*5/9 (in K) = 305 K – A = 1.5 m 2 P emitted = 714 Watts (compared to 100 Watts generated!)

55. Blackbody Radiation: Example need to consider power absorbed at room T P absorbed =  AT 4 –  =.97 –  = 5.67 x 10 -8 W/m 2 *K 4 – T = 273 + (72-32)*5/9 (in K) = 295 K – A = 1.5 m 2 P absorbed = 625 Watts (compared to 714 Watts emitted!)

56. Blackbody Radiation: Example Total power lost by radiation = 714 W - 625 W = 89 Watts (Power generated is 100 Watts.) Power also lost by convection (with air) and by evaporation.

57. Blackbody Radiation: Example At colder temperatures, our emitted power stays about the same while our absorbed power gets much lower. This means that we will get cold unless –we generate more power, or –our skin gets colder, or –we reflect the IR back into our bodies. Use metal foil for insulation!