Chapter 2 Heat, Temperature and the First Law of Thermodynamics
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1 Chapter 2 Heat, Temperature and the First Law of Thermodynamics 2.1. Temperature and the Zeroth Law of Thermodynamics 2.2. Thermal Expansion 2.3. Heat and the Absorption of Heat by Solids and Liquids 2.4. Wor and Heat in Thermodynamic Processes 2.5. The First Law of Thermodynamics and Some Special Cases 2.6. Heat Transfer Mechanisms
2 2.2. Thermal Expansion 1. Linear expansion: (solids) ΔL LαT 2. Area expansion: (solids) ΔA Aα AT 3. Volume expansion: (solids and liquids) ΔV VβT 2.3. Heats of transformation Heat capacity (C) Q C ΔT Q cm ΔT C (T Specific Heat (c) cm(t f f Molar Specific Heat T ) i T ) 1 mol = 6.02 x e units [atom or molecule] i Phase change Q Lm heat of vaporization L V is the amount of energy per unit mass that must be added to vaporize a liquid or that must be removed to condense a gas. heat of fusion L F is the amount of energy per unit mass that must be added to melt a solid or that must be removed to freeze a liquid.
3 2.4. Wor and Heat in Thermodynamic Processes A gas confined to a cylinder with a movable piston,from an initial state p i, V i, T i to a final state p f, V f, T f : The procedure for changing the system from its initial state to its final one is called a thermodynamic process. During a thermodynamic process, Energy as heat may be transferred into the system from a thermal reservoir. Wor can also be done by the system by raising (positive) or lowering the piston (negative wor). The wor dw done by the system for a differential displacement dw Fds (pa)(ds) p(ads) pdv F : the force exerted by the gas on the piston; A: the cross-sectional area of the piston; ds
4 The total wor done by the gas is given by the integral : W dw V f pdv V i If the gas expands : dv > 0 the wor done by the gas : dw > 0 If the gas were compressed : dv < 0 the wor done by the gas (which can be interpreted as wor done on the gas) : dw < 0 When the volume remains constant No wor is done on the gas
5 From state i to f, there are many ways to change the gas The wor done depends on the particular path W dw V f V i pdv (f): The net wor done by the gas for a complete cycle W net >0.
6 Checpoint 4: The p-v diagram here shows 6 curved paths (connected by vertical paths) that can be followed by a gas. Which two of the curved paths should be part of a closed cycle (those curved paths plus connecting vertical paths) if the net wor done by the gas during the cycle is to be at its maximum positive value? c and e gives a maximum area enclosed by a clocwise cycle
7 2.5. The First Law of Thermodynamics and Some Special Cases The first law of thermodynamics: - When a system changes from state i to state f: + The wor W done by the system depends on the path taen. + The heat Q transferred by the system depends on the path taen. However, the difference Q-W does NOT depend on the path taen. It depends only on the initial and final states. The quantity Q-W therefore represents a change in some intrinsic property of the system and this property is called the internal energy E int. ΔE int E For a differential change: de int int,f dq E int, i dw Q W The internal energy E int of a system tends to increase if energy is added as heat Q and tends to decrease if energy is lost as wor done by the system.
8 Let W on be the wor done on the system: W on = - W ΔE Q int W on Checpoint 5: In the figure below, ran the paths according to (a) E int, (b) W done by the gas, (c) Q; greatest first. ΔE int Q W (a) all tie (only depending on i and f) (b) (c) Q = E int + W so the raning is
9 Some special cases: ΔE int Q W 1. Adiabatic processes: Q=0 (no transfer of energy as heat) - A well-insulated system. - Or a process occurs very rapidly. ΔE int W 2. Constant-volume (isochoric) processes: W=0 (no wor done by the system) ΔE int Q 3. Cyclical processes: E int =0 In these processes, after some interchanges of heat and wor, the system is restored to its initial state. Q W 4. Free expansion: Q=W=0 ΔE int 0
10 Checpoint 6: One complete cycle is shown (see figure). Are (a) E int for the gas and (b) the net energy transferred as heat Q positive, negative, or zero? (a) zero (b) the cycle direction is counterclocwise, so W < 0, thus Q < 0
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13 2.6. Heat Transfer Mechanisms There are three types of transfer of energy as heat between a system and its environment: conduction, convection, and radiation Conduction: Example: Leaving the end of a metal poer in a fire its handle gets hot because energy is transferred from the fire to the handle by conduction. Physical mechanism: Due to the high temperature of the poer s environment, the vibration amplitudes of the atoms and electrons of the metal are relatively large, and thus the associated energy are passed along the poer, from atom to atom during collisions between adjacent atoms.
14 We consider a slab of face area A, thicness L, in thermal contact with a hot reservoir T H and a cold reservoir T C : Let Q be the energy transferred as heat through the slab in time t. Based on experiment, the conduction rate, which is the amount of energy transferred per unit time, is calculated by: P cond Q t A T H L T C (Unit: W = J/s) is called the thermal conductivity; good thermal conductors (or poor thermal insulator) have high -values.
15 Thermal Material conductivity (Wm -1 K -1 ) Diamond 1000 Metals Silver 428 Copper 401 Gold 314 Aluminium 235 Brass 109 Iron 67 Steel 50 Lead 35 Stainless steel 14 Material Gases Thermal conductivity (Wm -1 K -1 ) Hydrogen 0.18 Helium 0.15 Air (dry) Building Materials Window glass 1.0 White pine 0.11 Fiberglass Roc wool Polyurethane form 0.024
16 Thermal Resistance to Conduction: A measure of a body s ability to prevent heat from flowing through it. R L ; L : the thicness of the slab Good thermal insulators (poor thermal conductors) have high R-values. Conduction through a Composite Slab: A composite slab consisting of two materials having thicnesses L1 and L2, and thermal conductivities 1 and 2. If the transfer is a steady-state process that is the temperature everywhere in the slab and the rate of energy transfer do not change with time. P cond 2 A(T L H 2 - T X ) 1 A(T L X 1 - T C )
17 P cond 2 A(T L H 2 - T X ) 1 A(T L X 1 - T C ) P cond A(TH - TC ) L / L / If the slab consists of n materials: P cond A(T n i1 H - T (L / i C i ) )
18 Checpoint 7: The figure shows the face and interface temperature of a composite slab consisting of four materials, of identical thicness, through which the heat transfer is steady. Ran the materials according to their thermal conductivities, greatest first. P cond x A T H T L x C The heat transfer is steady, therefore P a =P b =P c =P d. a (T1 - T2 ) b(t2 - T3 ) c(t3 - T4 ) d (T4 - T5 ) b d a c
19 Convection: Energy is transferred through fluid motion (gases, liquids). Physical mechanism: When a fluid comes in contact with an object whose temperature is higher than that of the fluid. The part of the fluid in contact with the hot object has a temperature higher than that of the surrounding cooler fluid, hence that fluid becomes less dense; buoyant forces cause it rise. The cooler fluid flows to tae the place of the rising warmer fluid, producing fluid motion.
20 Examples: convection in the Earth s atmosphere; in the oceans, in the Sun. Hurricane Felix (NASA)
21 Radiation: Thermal energy is transferred via electromagnetic waves. Physical mechanism: Thermal radiation is generated when heat from the movement of charged particles within atoms and molecules is converted to electromagnetic radiation. Properties: + Every object whose temperature above 0 K emits thermal radiation via electromagnetic waves. + No medium is required for heat transfer via radiation. + The rate of emitting energy of an object is given by: P rad σεat 4 σ W m K : the Stefan - Boltzmann constant ε is the emissivity of the object's surface (values from 0 to1) ε 1: an idealized blacbody radiator will absorp all the radiated energy it intercepts A is the object's surface area T is the object's surface temperature
22
23 Reference: e-education.psu.edu
24 Sample Problem (p. 497) L d = 2 L a (thicness) White pine d = 5 a (conductivity) The heat transfer is steady T 1 =25 0 C; T 2 =20 0 C; T 5 =-10 0 C T 4? a T1 T A L T a L 2 d T4 T A L a d 4 (T1 T2 ) T5 dla d 5 Unnown material bric L d 2L a and d 5 a T 4 a (2L a ) (5 )L a a (25 20) 10-8( 0 C)
25 Homewor: 43, 44, 46, 47, 48, 49, 50, 51, 54, 59, 60 (pages )
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