The First Law of Thermodynamics

Transcription

1 The First Law of Thermodynamics Modern Physics August 31,

2 Energy Conservation In this section, we will discuss the concepts of heat, internal energy, and work. In PHY 140, we had talked about conservation of energy (Chapters 7 and 8 of Serway) and had said that for a system ΔKE + ΔPE + ΔE int = W + Q + T MW + T MT + T ET + T ER Change in kinetic energy + potential E. + internal E. = work + heat + mechanical waves, e.g., sound + matter + electrical + radiation We ll explore the relationship between these terms, in particular these two: Internal Energy (E int ): All the energy of a system associated with its microscopic components, e.g., atoms/molecules Heat (Q): This is the process of transferring energy across the boundary of a system due to a temperature difference between the system and its surroundings 2

3 Work Units of work: Force * distance = N-m = Joule Heat: Originally, was defined by means of 1 calorie = energy required to raise the temperature of 1 g of water from 14.5 to 15.5 C Now, we use the Joule [1 cal = J] Note that when we eat food or do exercise, we talk about consuming/burning off Calories. Actually, this usage of Calorie (food Calorie) = 1,000 calories (c) We did not understand this fully initially as same 3

4 Example If you lift 50 kg (~110 pounds) through 2.0 m (~6.6 feet), then you re performing work (a) How much work? (In J, cal, and kcal/cal) (b) To burn off 2,000 (food) Calories, how many times would you have to lift this weight? (c) How long would part (b) take? (Estimate) 4

5 Smaller the value it means it heats up faster, and can transfer heat away from a hot object efficiently (quickly!) c Water (liq) 4186 Iron 448 Copper 387 Silver 234 Gold 129 Mercury (liq) 140 Specific Heat When we add / subtract heat to a system, usually only its temperature changes Except when you have a phase transition: solid->liquid, liquid->gas, etc. Also, assuming KE and PE remain the same The amount of energy needed to raise the temperature of a system by, say, 1 C, depends on what it is made out of. For 1 kg of water, we need 4,186 J(oules) For 1 kg of copper, we need 387 J We define Q = C 0 * ΔT, where C 0 is the heat capacity Change in E int of system, if no work et al. done OR, Q = m * c * ΔT, where m is mass and c is called the SPECIFIC HEAT (not light speed..) That is, c = Q / ( m * ΔT ) -> Units of J / ( kg * C) Think: if very high, what are the implications? And if is low? 5

6 Calorimetry How do we measure the SPECIFIC HEAT of something? Consider this: Q cold Q hot m water T water, c water Boundary, which isolates the system This object has m X, T X, c X (unknown). We immerse an object of mass m X and temperature T X and specific heat c X in a bath of water with m W, T W, and c W If T X > T W, then heat will be transferred from the object to the water, and after a while they will come to equilibrium, i.e., the same temperature T f So, -Q hot = Q cold -> the negative sign just means that Q hot is *leaving from* m X 6

7 continued Energy transfer for water = m W * c W * ( T f T W ) [ > 0 ] For hot object = m X * c X * ( T f T X ) [ < 0 ] Equating the two m W * c W * ( T f T W ) = -m X * c X * ( T f T X ) = m X * c X * ( T X T f ) Re-arranging to get c X, we have c X = m W * c W * ( T f T W ) / [ m X * ( T X T f ) ] (Here, we have ignored the container. If you include the mass/temperature/specific heat of the container, then the equations will become slightly modified) You will be getting a homework problem like this 7

8 Example We put a metal block of mass = 0.05 kg and at a temperature of 200 C into a bath of water with mass = kg and initial temperature of 20 C. The equilibrium temperature is 22.4 C. (a) What is c of the metal? (b) From our table, it appears it is made of? (c) Why didn t we get the correct answer? 8

9 Latent Heat In previous discussions, temperature increased when we transferred heat into a system But, when we have a phase transition, i.e., the state of a system changes from solid -> liquid, or liquid->gas, the heat goes into changing the state without changing the temperature. Imagine that you have 1 kg of ice at -5 C. Now, transfer heat into the ice. The temperature of the ice goes from -5 C to 0 C Ice -> 0 C (PHASE TRANSITION, which is governed by a quantity called LATENT HEAT) Temperature of water goes from 0 C -> final value Q = L * m i or L = Q / m i (mass of 0 C) Similarly, we have latent heat when 100 C turns into steam/water 100 C 9

10 Examples Note that 1 kg of steam at 100 C will burn you more severely than 1 kg of water at 100 C. Why? We have 25.0 liters of LN 2 [density = 0.81 its boiling point. We put a 25 W heating element in this container and leave it there for 4 hours. How much of the LN 2 boils off? (Note 1 Watt = 1 J/sec) MelBng Pt. Latent heat Boiling Pt. Latent heat of of fusion vaporizabon [J per kg] [J per kg] Water (Liq) 0 C 3.33E C 2.26E+06 Copper 1,083 C 1.34E+05 1,187 C 5.06E+06 Silver C 8.82E+04 2,193 C 2.33E+06 Liquid Nitrogen (LN2) E+05 10

11 Work and Heat and the First Law Some definitions State variables define the state of a system, e.g., Pressure, Temperature, Volume, E int (also KE/PE). A state of a system can be specified only when it is in thermal equilibrium internally, i.e., all parts of the system are at the same temperature/pressure Transfer variables describe the transfer of energy in / out of the system => these variables cause a change in the state of the system So far, we have discussed heat as a transfer variable some How gas about work? What would happen if we were to push down on the piston, compress the gas inside it? piston 11

12 Work as it Relates to a Gas dy area = A gas P, V We push the piston quasi-statically, that is, the system is assumed to be in internal and thermal equilibrium at all times How much work is done on the gas? At any instant F = P * A (force due to gas on piston, and vice versa) -> -> dw = F * dr (vector dot product symbol * here) ^ ^ dw = -F j * dy j = -P * A * dy = - P * dv If dv is negative then the gas is being compressed => work done on gas, i.e., dw > 0 If the gas were to expand, dv > 0 and dw < 0 If the volume remains the same, then dw = 0 V f W = dw = - P dv. In general, P will depend on V. V i 12

13 Examples on Board Area under the curve is the total work performed/done. Work done depends on initial / final points, but it also depends on the path At constant volume, no work is done When we increase the pressure at constant volume, we can do so by heating up the gas This is the connection between work and heat -- both may be used for changing the state of a system 13

14 Part II: 1 st Law of Thermo ΔE int = Q + W (this is a special case of the energy conservation law, where only Q and W are permitted in the equation) In an isolated system, no work is done on it and no heat is transferred => ΔE int = 0 or E int_initial = E int_final Consider a non-isolated system that starts and ends up at same state (cyclic process) ΔE int = 0. Remember that E int is a state variable Q = -W for a cyclical process. Law! 14

15 Example on Board (a) Find the net energy transferred to the system by heat as we go from A->B->C->A, i.e., a cyclic process Units: 1 atm = 1.013e5 N / m ^ 2 = 1.013e5 Pa = kpa (Remember: e or E is short-hand or calculator/computer notation for x 10 ^ ) As we go around the closed curve, ΔE int = 0 (we start from point A and end up at point A) Let s analyze each leg separately now (b) Exercise for you: what happens if we go the other way, i.e., A->C->B->A Would the answer change? Why or why not?? 15

16 Applications and Terminology ADIABATIC process: One in which no heat enters or leaves the system, that is Q = 0 => ΔE int =W (from first law) ISOTHERMAL process: One which occurs at constant temperature For an ideal gas, ΔT = 0 => ΔE int = 0 or Q = W Heat can enter a system and leave as work, keeping the temperature constant ISOBARIC process: At constant pressure. Both Q and W are, in general NOT equal to zeroes See A<-> C from the previous example ISO-VOLUMETRIC process: At constant volume => W = 0 and ΔE int = Q. 16

17 Energy Transfer Mechanisms 3 main -- Conduction -> Two systems in contact with each other and one is hotter than the other. Heats flows from hot->cold. Usually due to atomic vibrations Convection -> Energy transfer by movement of hot materials Take a pot of water and put it on an electric stove Stove transfers energy to bottom of pot by conduction. Bottom of pot heats up a thin layer of water right next to it (again mainly by conduction) This warm water has lower density (than cool water), so it rises and the denser, colder water at the top of the water sinks and gets heated This process is convection Radiation Can mix 17

18 Radiation Why do you get warmed by the sun? After all, there is vacuum between the earth and sun! ALL objects give off electromagnetic radiation The frequency/wavelength of this radiation depends on the temperature of the object The temperature of the outer surface of the sun is ~6,000 K, and the peak of its EM radiation is ~yellow The rate at which a body radiates energy Power=Energy/Time=σ * A * ε * T 4 -> Stefan s law. Sigma: constant = 5.67e-8 W / ( m^2 * K^4 ) A: Area Epsilon: emissivity (=absorptivity) of object (0 to 1) T: temperature in Kelvin When ε = 1 -> Known as BLACKBODY We ll return to this later. A BIG DEAL! 18

19 Final Example Assume that a 100-Watt light bulb has a tungsten filament, and radiates off 2 W of power (other 98 W is carried away by conduction and convection). If the area of a filament is mm^2, and the emissivity is 0.950, what is its temperature? (This is a simple plug & chug problem.) Note that the melting point of Tungsten is 3,683 K (fun fact not needed to solve problem) Supposedly, it took Thomas Edison a while to perfect the light bulb, because he could not find an element that would be stable for a long period of time! (Did not invent) 19

20 Homework 5 problems due next Wednesday Copies available here 20