11/13/2003 PHY Lecture 19 1
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- Madeleine Clarke
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1 Announcements 1. Schedule Chapter 19 macroscopic view of heat (today) Chapter 20 microscopic view of heat (Tuesday 11/18) Review Chapters (Thursday 11/20) Exam 3 (Tuesday 11/25) 2. Physics colloquium today Professor Gary Pielak from UNC will speak about Protein Biophysics in Cells. 3. Today s lecture Chapter 19 Temperature Heat The first law of thermodynamics 11/13/2003 PHY Lecture 19 1
2 Dictionary definition: temperature a measure of the the warmth or coldness of an object or substance with reference to some standard value. The temperature of two systems is the same when the systems are in thermal equilibrium. Equilibrium: Not equilibrium: T 1 T 2 T 3 Zeroth law of thermodynamics: If objects A and B are separately in thermal equilibrium with a third object C, then objects A and B are in thermal equilibrium with each other. 11/13/2003 PHY Lecture 19 2
3 Temperature scales T F =9/5 T C T F Kelvin scale: T = T C o T 0 T C 11/13/2003 PHY Lecture 19 3
4 There is a lowest temperature: T 0 = o C = 0 K Kelvin ( absolute temperature ) scale T C = T K Example Room temperature = 68 o F = 20 o C = K 11/13/2003 PHY Lecture 19 4
5 11/13/2003 PHY Lecture 19 5
6 Effects of temperature on matter Solids and liquids Model of a solid composed of atoms and bonds Thermal exansion: L = α L i T L L i (equilibrium bond length at T i ) 11/13/2003 PHY Lecture 19 6
7 Typical expansion coefficients at T C = 20 o C: Linear expansion: L = α L i T Steel: α = 11 x 10-6 / o C Concrete: α = 12 x 10-6 / o C Volume expansion: V=L 3 V = 3α V i T = β V i T Alcohol: β = 1.12 x 10-4 / o C Air: β = 3.41 x 10-3 / o C 11/13/2003 PHY Lecture 19 7
8 Brass Steel 11/13/2003 PHY Lecture 19 8
9 Heat Q -- the energy that is transferred between a system and its environment because of a temperature difference that exists between them. Sign convention: Q<0 if T 1 < T 2 Q>0 if T 1 > T 2 T 1 T 2 Q 11/13/2003 PHY Lecture 19 9
10 Heat withdrawn from system Thermal equilibrium no heat transfer Heat added to system 11/13/2003 PHY Lecture 19 10
11 Units of heat: Joule Other units: calorie = J Kilocalorie = 4186 J = Calorie Heat capacity: C = amount of heat which must be added to the system to raise its temperature by 1K (or 1 o C). Q = C T More generally: Q = C( T ) dt i f Heat capacity per mass: C=mc T T Heat capacity per mole (for ideal gas): C=nC v f i C=nC p 11/13/2003 PHY Lecture 19 11
12 Some typical specific heats Material Water (15 o C) Ice (-10 o C) Steam (100 o C) Wood Aluminum Iron Gold J/(kg oc) cal/(g oc) /13/2003 PHY Lecture 19 12
13 Heat and changes in phase of materials Example: A plot of temperature versus Q added to 1g = kg of ice (initially at T=-30 o C) Q=2260J Q=333J 11/13/2003 PHY Lecture 19 13
14 Some typical latent heats Material Ice Water (0 o C) Water Steam (100 o C) Solid N Liquid N (63 K) Liquid N Gaseous N 2 (77 K) Solid Al Liquid Al (660 o C) Liquid Al Gaseous Al (2450 o C) J/kg /13/2003 PHY Lecture 19 14
15 Peer instruction question Suppose you have a well-insulated cup of hot coffee (m=0.25kg, T=100 o C). In order to get to class on time you add 0.25 kg of ice (at 0 o C). When your cup comes to equilibrium, what will be the temperature of the coffee. (A) 0 o C (B) 10 o C (C) 50 o C (D) 100 o C 11/13/2003 PHY Lecture 19 15
16 Review question Suppose you have a well-insulated cup of hot coffee (m=0.25kg, T=100 o C). In order to get to class on time you add 0.25 kg of ice (at 0 o C). When your cup comes to equilibrium, what will be the temperature of the coffee. Q = m c w (T f 100 o C) + m L ice + m c w (T f 0 o C) = 0 2 c w T f = c w 100 L ice T f = c w 100/(2 c w ) L ice /(2 c w ) T f = /(2 4186) = 10 o C 11/13/2003 PHY Lecture 19 16
17 Energy in the form of work: Work done by the system due to volume change W = x f f f F Fdx = Adx = PdV A x i x x i V V i Sign convention: W>0 if V f > V i (expansion) W<0 if V f < V i (contraction) 11/13/2003 PHY Lecture 19 17
18 Work done by a gas: Isobaric (constant pressure process) P (1.013 x 10 5 ) Pa P o W= P o (V f -V i ) V i V f 11/13/2003 PHY Lecture 19 18
19 Work done by ideal gas: Isovolumetric (constant volume process) P f P (1.013 x 10 5 ) Pa P i W=0 V i 11/13/2003 PHY Lecture 19 19
20 Work done by a gas: Isothermal (constant temperature process) P (1.013 x 10 5 ) Pa P i For ideal gas: PV = nrt W i f = V f V f nrt V PdV = dv = nrt ln V V V i i V f = PV i i ln Vi V f i V i V f 11/13/2003 PHY Lecture 19 20
21 Work done by a gas: Adiabatic (no heat flow in the process process) For ideal gas: Q i f = 0 PV γ = P i V i γ P (1.013 x 10 5 ) Pa P i W i f = γ 1 f V f γ PV i i PV V i i f PdV = dv = 1 γ V V V γ 1 V i i i V V i V f 11/13/2003 PHY Lecture 19 21
22 Thermodynamic statement of conservation of energy First Law of Thermodynamics E int = Q - W Heat added to system Work done by system Internal energy of system 11/13/2003 PHY Lecture 19 22
23 Internal energy E int =E int (T,V,P) (for an ideal gas, E int =E int (T) can be changed by interaction with the system T sys W system Q environment Tenv E int = Q - W Note: Thermal equilibrium implies a uniform temperature. In this example, the system and the environment are presumed to be in thermal equilibrium within themselves but not in thermal equilibrium with eachother. 11/13/2003 PHY Lecture 19 23
24 Examples process: P (1.013 x 10 5 ) Pa P f P i B A C D W net =W (ABCDA) =W BC +W DA =(P f P i ) (V f V i ) E int(abcda) = E int(ab) + E int(bc) + E int(cd) + E int(da) =0 Q net =Q (ABCDA) =W net V i V f 11/13/2003 PHY Lecture 19 24
25 1 m 3 6m 3 30 Pa 10 Pa 11/13/2003 PHY Lecture 19 25
26 11/13/2003 PHY Lecture 19 26
27 11/13/2003 PHY Lecture 19 27
28 Mechanisms of heat transfer: 1. Thermal convection Heat transmitted by heated fluid flow. (Ex. -- home heating units, stove top cooking) 2. Radiation Heat transmitted by electromagnetic radiation. (Ex. sun, microwave cooking) 3. Thermal conduction Heat transfer from one side of a material to another due to a temperature gradient. 11/13/2003 PHY Lecture 19 28
29 Quantitative statement of thermal conduction: Q T = ka Units: 1 J/s = 1 Watt (W) t x cross-sectional area thermal conductivity coefficient Material Copper Glass Water Air k (W/(m oc)) /13/2003 PHY Lecture 19 29
30 11/13/2003 PHY Lecture Examples of thermal conduction T L T T A k dt dq = / / k L k L L L k L L T T A k dt dq eff eff + + = + =
31 Example: T 1 =20 o C, T2= 0 o C Single pane glass window: A=1m 2, L 2 = 0.002m, k 2 =0.8 W/m oc dq dt = k T A T L = W Double pane glass window: L 1 =0.01m, k 1 = W/m oc (air) dq dt k eff = k = L eff A L 2 T + 2 T1 L + L 1 L2 + L1 + L0 k + L / k + L / k 2 / = 46W 11/13/2003 PHY Lecture 19 31
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