1. This Week  Heat and Temperature  Water and Ice Our world would be different if water didn’t expand  Engines We can’t use all the energy! Why is a diesel engine more efficient?  Geysers: You have to be faithful 6/12/2016Physics 214 Fall 20101

2. 6/12/2016Physics 214 Fall 20102 Heat Heat is a form of energy and any object has internal energy in the form of kinetic energy of the atoms or molecules and in potential energy connected with the molecular structure and the electromagnetic forces among the constituents. If we add energy to an object the internal energy increases. Energy can be added in many ways. For example we can do work such as  the frictional forces on a moving object.  electromagnetic radiation from the sun.  a source of heat like a hair dryer.

3. 6/12/2016Physics 214 Fall 20103 First Law of Thermodynamics The increase in the internal energy of a system is equal to the amount of heat added to the system minus the amount of work done by the system. U = internal energy Q = heat that is added or removed W system = is the work done by the system Looking at the pictures you can see that the force pushing on the piston does positive work and therefore is putting energy into the gas. The work done by the gas is negative because the force the piston exerts is to the right and the movement of the piston is to the left. ENERGY CONSERVATION ΔU = Q – W system or ΔU = Q + W external

4. 6/12/2016Physics 214 Fall 20104 Temperature Heat flows from a hot body to a cold body until they reach thermal equilibrium. Temperature is the quantity that measures whether one body is hotter than another and at thermal equilibrium both bodies have the same temperature. The simplest picture is a gas of free molecules where the energy is kinetic. A higher temperaturemeans higher velocities and more stored energy. Increase U and T increases http://jersey.uoregon.edu/vlab/Thermodynamics/index.html http://www2.biglobe.ne.jp/~norimari/science/JavaApp/Mole/e-gas.html

5. 6/12/2016Physics 214 Fall 20105 Measuring temperature Physical properties of an object change with temperature. For example mercury expands as the temperature increases and we can use that expansion to measure T. There are three temperature scales, Celsius, Fahrenheit and Kelvin Mixture of ice and water 0 0 C 32 0 F 273 0 K Boiling point of water 100 0 C 212 0 F 373 0 K T c = 5/9(T f – 32) T f = 9/5T c + 32 T K = T c + 273 Note that the temperature at which water boils depends on the value of the atmospheric pressure so that at higher altitudes water boils at a lower temperature

6. 6/12/2016Physics 214 Fall 20106 Absolute zero An ideal gas obeys the law PV = constant x T if we keep V constant the pressure is proportional to T. If we measure P versus T at constant volume we find a temperature where the pressure is zero. This is absolute zero where the stored internal energy has it’s lowest possible value. This is Zero degrees Kelvin

7. 6/12/2016Physics 214 Fall 20107 Heat capacity For a given object at a specific temperature the total internal energy depends on how big the object is. If we add a fixed amount of energy the temperature will rise but the rise will be smaller the larger the object is. So we define a quantity called specific heat c = is the quantity of heat required to raise unit mass of a substance by one degree. Q = mcΔT New unit 1 calorie = heat required to raise 1 gram of water 1 0 C 1calorie = 4.186 joules Note to raise 1kg of water 1 0 C requires 1000cal Note that internal energy can change without a temperature change. This is due to internal rearrangement of the molecules Ice Water

8. 6/12/2016Physics 214 Fall 20108 Water and Ice Water has an unusual behavior as the temperature is changed. As we lower the temperature from room temperature it shrinks but it 4 0 C it starts to expand and that is why ice floats. Remember density ρ = mass/volume so below 4 0 C the volume increases and the density decreases. Above 4 0 C the molecules are tightly packed. Below 4 0 C the molecules form a crystalline structure that has holes.

9. 6/12/2016Physics 214 Fall 20109 Change of state We are used to solids, liquids and gases and that most substances can be in any of the three states depending on temperature and pressure. Solids have more orderly structure and the atoms/molecules are tightly joined. In a liquid the atoms/molecules are loosely joined. In a gas the atoms/molecules are “free”. When we tear apart something that is joined energy is required and this is true as we go from solid to liquid and liquid to gas. Taking water as an example 80 calories/gram are required to melt ice at 0 o C to water at 0 o C and 540 calories/gram to convert water to gas at 100 o C. When water freezes 80cal/gm has to be removed and when water condenses 540 cals/gm is released. 80 cals/gm is the latent heat of fusion 540 cals/gm is the latent heat of vaporization http://jersey.uoregon.edu/vlab/Balloon/

10. 6/12/2016Physics 214 Fall 201010 Gases A quantity of gas can be described by three variables. Pressure, volume and temperature U = internal energy (kinetic energy of the molecules) U determines the temperature Q = heat energy that is put in or taken out W gas = work done by the gas when it expands or is compressed W gas is + if it expands – if it is compressed Work done by gas = -Fd = -PAd ΔV = Ad W = -PΔV External work done = W = P ΔV

11. 6/12/2016Physics 214 Fall 201011 Physical Laws for a gas Energy conservation ΔU = Q – W gas Ideal gas law PV = NkT (T in degrees Kelvin) Adiabatic no heat in or out Q = 0 Compression work is + (ΔU, T increase) Expansion work is – (ΔU, T decrease) Isothermal T does not change ΔU = 0 Q = W = PΔV put in heat gas expands take out heat gas must be compressed Isobaric pressure is kept constant Put in heat T increases gas expands (hot air balloon

12. 6/12/2016Physics 214 Fall 201012 Solids and liquids Temperature depends on the internal energy Put in heat T rises except at a change of state We define the specific heat c as the amount of heat required to raise unit mass of a substance by one degree. Q = mcΔT The usual unit is calories/gram/ 0 C 1 cal = 4.186 joules The specific heat of water is 1cal/gm/ 0 C Change of state ice water steam Requires 80 cal/gm to melt ice at 0 C to water at 0 C Requires 540cal/gm to make steam at 100 0 C To condense steam or freeze water requires the removal of 540 cal/gm or 80cal/gm. Q

13. 6/12/2016Physics 214 Fall 201013 Heat engines The use of energy and the conversion of energy is essential not only in our practical everyday life but is a requirement for life to exist. The sun puts energy into the earth and we use such energy sources as gasoline, coal and nuclear power. The conversion of one form of energy into another form of energy is governed by the physical laws. The generic term used is a heat engine for any system that takes an energy source and uses it to produce work. A car engine is a practical example. Of course there is a practical aspect that an energy source can become depleted, like oil but we will find that there is a fundamental law that prevents the utilization of all the available energy.

14. 6/12/2016Physics 214 Fall 201014 2 nd Law of Thermodynamics Heat engine uses energy Q H and produces work W and releases Q c Efficiency = ε = W/Q H W = Q H – Q c (c = environment) 2 nd law No engine working in a continuous cycle can take heat at a single temperature and convert that heat completely to work.

15. 6/12/2016Physics 214 Fall 201015 Carnot Cycle We can design a “perfect engine” without friction. The Carnot cycle is is an ideal engine and an analysis reveals that ε = (T H –T c )/T H (T in o K) For example in a gasoline engine T H is the temperature of combustion and Tc is outside temperature. If we take T H as 1500 o and T c as 300 o then ε = 80%. In practice the efficiency is maybe 35%.

16. 6/12/2016Physics 214 Fall 201016 Car engines In an engine with spark plugs a gasoline/air mixture is compressed and the temperature rises and then the spark ignites the mixture before it reaches the combustion temperature. For increased efficiency one would like to reach as high a temperature as possible so fuel injection solves this problem by compressing air and injecting the fuel after the combustion temperature is reached. Turbo charged engines force more air into the chamber when the piston reaches it’s lowest point using power from the engine. Super chargers do the same thing but with an extra electrical motor. This causes the engine to run hotter. Diesel engines use a lower grade of petroleum and also runs at a much higher temperature

17. 6/12/2016Physics 214 Fall 201017 Review Chapters 10 and 11 W = PΔV for the piston shown W external = + W system = - ΔU = Q – W system or ΔU = Q + W external Temperature scales Celsius, Fahrenheit Kelvin Mixture of ice and water 0 0 C 32 0 F 273 0 K Boiling point of water 100 0 C 212 0 F 373 0 K T c = 5/9(T f – 32) T f = 9/5T c + 32 T K = T c + 273 c = is the quantity of heat required to raise unit mass of a substance by one degree. Q = mcΔT 1 calorie = heat required to raise 1 gram of water 1 0 C = 4.186 joules Change of state (internal energy changes temperature is constant) 80 cals/gm is the latent heat of fusion 540 cals/gm is the latent heat of vaporization Work done on gas = Fd = PAd ΔV = Ad W = PΔV Work done by gas is -W

18. 6/12/2016Physics 214 Fall 201018 Gases Energy conservation ΔU = Q – W gas Ideal gas law PV = NkT (T in degrees Kelvin) Adiabatic no heat in or out Q = 0 Compression work is + (ΔU, T increase) Expansion work is – (ΔU, T decrease) Isothermal T does not change ΔU = 0 Q = W = PΔV put in heat gas expands take out heat gas must be compressed Isobaric pressure is kept constant Put in heat T increases gas expands (hot air balloon) Work done by gas = -Fd = -PAd ΔV = Ad W = -PΔV External work done = W =P ΔV

19. 6/12/2016Physics 214 Fall 201019 Heat engines Efficiency = ε = W/Q H Change in internal energy in one cycle is zero W = Q H – Q c (c = environment) 2 nd law No engine working in a continuous cycle can take heat at a single temperature and convert that heat completely to work.

20. 6/12/2016Physics 214 Fall 201020 Geysers (Old Faithful) Geysers produce a jet of water very often at equal time intervals. Old Faithful in Yellowstone erupts every 90 minutes or so. What is required is a source of heat, a source of water and a constricted vertical channel that the water flows into. As the column of water increases in height the pressure increases at the bottom of the water column where the heat is. The increase in pressure raises the boiling point. This continues until the water at the bottom starts to boil. This causes an expansion of the column which reduces the pressure and immediately takes the temperature of all the water below the boiling point and there is an explosive boiling resulting in the eruption. Heat P = ρgh

21. 6/12/2016Physics 214 Fall 201021 3E-03 Fire Syringe RAPID COMPRESSION IS ADIABATIC GIVING RAPID RISE OF AIR TEMPERATURE IN THE CHAMBER WHICH EXCEEDS THE IGNITION TEMPERATURE OF THE FLAMMABLE MATERIAL. Compression and rise in air temperature This system is analogous to the combustion cycle within a diesel engine or any fuel injected engine. Can you guess the every-day application of this phenomenon ? What will happen to the combustible material when the plunger is rapidly pushed down ?

22. 6/12/2016Physics 214 Fall 201022 Questions Chapter 10 Q1 Is an object that has a temperature of 0°C hotter than, colder than, or at the same temperature as one that has a temperature of 0°F? Q2 Which spans a greater range in temperature, a change in temperature of 10 Fahrenheit degrees or a change of 10 Celsius degrees? Water freezes at 0 o C and 32 o F so 0 o F is colder There is 100 o C between water freezing and boiling and 180 o F so 1 o C = 1.8 o F so a change of 10 o C is larger

23. 6/12/2016Physics 214 Fall 201023 Q4 We sometimes attempt to determine whether another person has a fever by placing a hand on their forehead. Is this a reliable procedure? What assumptions do we make in this process? Q5 Is it possible for a temperature to be lower than 0°C? Body temperatures internally are very similar for all people so in principle we can tell if a temperature is elevated just like any other hot object. The assumption is that your hand is at body temperature. Yes. Absolute zero is – 273 0 which one can think of as the place where the molecules of an ideal gas have zero energy Q6 Is it possible for a temperature to be lower than 0 K on the Kelvin temperature scale? Yes. Absolute zero is – 273 0 K

24. 6/12/2016Physics 214 Fall 201024 Q8 Two objects at different temperatures are placed in contact with one another but are insulated from the surroundings. Will the temperature of either object change? They will exchange heat until they both reach the same temperature Q10 Two objects of the same mass, but made of different materials, are initially at the same temperature. Equal amounts of heat are added to each object. Will the final temperature of the two objects necessarily be the same? No. The specific heat, which is the heat energy required to raise the temperature one degree, is different for each material Q13 What happens if we add heat to water that is at the temperature of 100°C? Does the temperature change? Explain. The water turns into steam at 100 o C

25. 6/12/2016Physics 214 Fall 201025 Q21 An ideal gas is compressed without allowing any heat to flow into or out of the gas. Will the temperature of the gas increase, decrease, or remain the same in this process? Q23 Heat is added to an ideal gas, and the gas expands in the process. Is it possible for the temperature to remain constant in this situation? The temperature will increase ΔU = W Yes. This is an isothermal expansion and W = Q

26. 6/12/2016Physics 214 Fall 201026 Q25 Heat is added to a hot-air balloon causing the air to expand. Will this increased volume of air cause the balloon to fall? Q27 A block of wood and a block of metal have been sitting on a table for a long time. The block of metal feels colder to the touch than the block of wood. Does this mean that the metal is actually at a lower temperature than the wood? Archimedes principle states that the buoyant force is equal to the weight of liquid displaced. So if the balloon stays the same size and as the air expands it leaves the balloon it will rise faster because the weight of the air inside will be less. If no air escapes but the balloon increases in size it will also rise because the buoyant force is larger No. What you feel is heat flow and the thermal conductivity of metal is much bigger than that of wood. This is also why you feel much colder when the air is damp and why trapped dry air in fiberglass is used for insulation. It is also because room temperature is lower than body temperature

27. 6/12/2016Physics 214 Fall 201027 Ch 10 E 2 Temperature is 14° F. What is the temperature in Celsius? T c = 5/9 (T F – 32) = 5/9 (14 - 32)=5/9 (-18) = -10° C

28. 6/12/2016Physics 214 Fall 201028 Ch 10 E6 How much heat does it take to raise the temperature of 70 g of H 2 O from 20°C to 80°C? C= 1 cal/gram/ o C. Q = mc  T=(70)(1)(80-20) = 4200 cal

29. 6/12/2016Physics 214 Fall 201029 Ch 10 E 12 Add 600 J to 50 g of H 2 O initially at 20°C a)How many calories? b)What is the final temperature of the H 2 O a)600 /4.186 = 143.3 cal = Q b) Q=mc  T  T=Q/mc=143.3/(50)(1)=2.87°C T F =22.87°C

30. 6/12/2016Physics 214 Fall 201030 Ch 10 E 16 Add 500 cal of heat to gas. Gas does 500 J of work on surroundings. What is the change in internal energy of gas?  U = Q-W Q = 500/ 4.186 = 2093 J W=500 J  U = 2093 – 500 = 1593 J

31. 6/12/2016Physics 214 Fall 201031 Ch 10 CP 2 A student’s temperature scale = 0°s = ice point of H 2 O; 50° s = boiling point of H 2 O. The student then measures a temperature of 15°s. a)What is this temperature in degrees Celsius? b)What is this temperature in degrees Farenheit? c)What is this temperature in Kelvins? d)Is the temperature range spanned by 1°s larger or smaller than spanned by 1°C? a) Ice point H 2 O = 0°s=0°C Boiling point H 2 O = 50°s=100°C Obvious relationship: T s =1/2 T c, T c =2T s T c =2T s =2(15) = 30°C b) T F = 9/5 T c + 32 = 9/5(30) + 32 = 86°F c) T k = T c + 273.2 = 30 + 273.2 = 303.2K d)1°s spans 2°C, so that the range spanned by 1°S is LARGER than that spanned by 1°C. TsTs TcTc 50°s100°c 0°s0°c

32. 6/12/2016Physics 214 Fall 201032 Ch 10 CP 4 150 g of metal at 120°C is dropped in a beaker containing 100 g H 2 O at 20°C. (Ignore the beaker). The final temperature of metal and water is 35°C. a) How much heat is transferred to the H2O? b) What is the specific heat capacity of metal? c) Use the same experimental setup – how much metal at 120°C to have final temperature of metal and water = 70°C? a) Q = mc  T = (100)(1)(35-20) = 1500 cal b)Q = mc  T → c = Q/m  T = c = -1500/(150)(35-120) = 0.12 cal/gram/°C c) Q H2O = mc  T = (100)(1)(70-20) = 5000 cal Q metal = - Q H2O = - 5000 Q metal = mc  T; m = Q metal/c  T = -5000/(0.12)(70-120) m = 833 g

33. 6/12/2016Physics 214 Fall 201033 Ch 11 E 6 A Carnot engine takes in heat at 650 K and releases heat to a reservoir at 350K. What is the efficiency? e c = (T H – T c) / T H = 650-350 / 650 = 0.46

34. 6/12/2016Physics 214 Fall 201034 Ch 11 E 8 Carnot engine operates b/w 600 K and 400 K and does 200 J of work in each cycle. a) What is the efficiency? b)How much heat does it take in from the high-temp reservoir during each cycle? a) ε c = (T H - T c) /T H = (600-400)/600 = 0.33 b) ε = W/Q H, Q H = W/ε = 200/0.33 = 600J

35. 6/12/2016Physics 214 Fall 201035 Ch 11 CP 2 Carnot engine operates b/w 500° C and 150° C and does 30 J of work in each cycle. a) What is the efficiency? b) How much heat is taking in from the high-temp reservoir each cycle? c) How much heat is released to low-temp reservoir each cycle? d) What is the change, if any, in the internal energy of gas each cycle? a) ε c = (T H -T c) / T H = (500+273.3)-(150+273.3)/(500+273.3) =.045 b) ε = W/Q H, Q H = W/ε = 30/0.45 = 66.3 J c) W = Q H -Q c, Q c = Q H -W = 66.3J - 30J = 36.3J d)  U = Q-W, W = 30J Q = Q H - Q c = 66.3 – 36.3 = 30 J  U = 30-30 = 0