Chapter 13 Heat Reference:chap.14 Alan Giambattista, Betty McCarthy Richardson and Robert C. Richardson, 2009, “College Physics”, 3rd ed., McGraw-Hill. Only for education •Internal Energy •Heat •Heat Capacity •Specific Heat •Phase Transitions •Thermal Conduction •Thermal Convection •Thermal Radiation 普物AB—海大輪機系 1 張宏宜 Internal Energy The internal energy of a system is the sum total of all the energy of all the molecules in the system. It does not include macroscopic kinetic energy nor energy due to external interactions (potential energy). • Energy of all molecules including – Random motion of individual molecules • <Ktr > = 3/2 k T for ideal gas • Vibrational energy of molecules and atoms – Chemical energy in bonds and interactions • DOES NOT INCLUDE – Macroscopic motion of object – Potential energy due to interactions w/ other objects 普物AB—海大輪機系 2 張宏宜 Internal Energy Internal energy includes: •Translational and rotational kinetic energy of the particles due to their individual motions. •Vibrational kinetic and potential energy •Potential energy due to interactions between particles in the system. •Chemical and nuclear energy (binding energies) Internal energy does not include: •The kinetic energy of the molecules due to translations, rotations, and vibrations of the whole or large fraction of the system. •Potential energy due to the interactions of the molecules of the system with bodies outside of the system (external interactions). 普物AB—海大輪機系 3 張宏宜 Example (text problem 14.5) A child of mass 15 kg climbs to the top of a slide that is 1.7 m above a horizontal run that extends for 0.5 m at the base of the slide. After sliding down, the child comes to rest just before reaching the very end of the horizontal portion of the slide. How much internal energy was generated during this process? The change in mechanical energy of the child is U = mgh ΔE= Ef –Ei = -mgh = -250 J. KE = 0 This is the increase in internal energy and is distributed between the child, the slide, and 1.7 m the air. U = 0 普物AB—海大輪機系 KE = 0 4 張宏宜 Heat Heat is energy in transit between two systems at different temperatures. Heat flows from the system at high temperature to the system at low temperature. An experiment by Joule showed that the quantity of work done on a system or the same quantity of heat flowing into a system causes the same increase in the system’s internal energy. Heat is measured in joules or calories. 1 cal = 4.186 J (the mechanical equivalent of heat); 1 calorie (used on food packaging) = 1 kcal. 普物AB—海大輪機系 5 張宏宜 Heat Capacity and Specific Heat For many substances, under normal circumstances ΔT∝Q. Or Q = CΔT where C is the heat capacity. The specific heat capacity, or just specific heat, of a substance is the heat capacity per unit mass. C Q c= = or Q= mcΔ T m mΔ T Q = mcΔT: “Cause” = “inertia” x “effect” (just like F=ma) cause = Q effect = ΔT inertia = mc (mass x specific heat capacity) 普物AB—海大輪機系 6 張宏宜 Example (text problem 14.12) If 125.6 kJ of heat are supplied to 5.00×102 g of water at 22 °C, what is the final temperature of the water? Q mc= Δ T= ( mcf− i) T T Q TT= + f i mc 125.6 kJ =22 ° C + =82 ° C 0 . 5 (kg 4 . () 186 ° )kJ/kg C 普物AB—海大輪機系 7 張宏宜 Example (text problem 14.19) A 0.400 kg aluminum teakettle contains 2.00 kg of water at 15.0 °C. How much heat is required to raise the temperature of the water (and kettle) to 100 °C? The heat needed to raise the temperature of the water to Tf is 2 Qkgw= 4 m w . wcΔ 186= w T( )( kJ/kg° )( C ° ) = 85 C 712 kJ. The heat needed to raise the temperature of the aluminum to Tf is 0 . 4Q kgAl =Al m 0 Al . cΔ 900= Al T( )( kJ/kg°) ( C ° ) = 85 C 30 . 6 kJ. Then Qtotal = Qw + QAl = 742.6 kJ. 普物AB—海大輪機系 8 張宏宜 Specific Heat of Ideal Gases The average kinetic energy of a molecule in an ideal gas is 3 K= kT. tr 2 And the total kinetic energy of the gas is 3 3 K= NkT = nRT. tr 2 2 Define the molar specific heat at constant volume; this is the heat capacity per mole. Q C = V nΔ T 普物AB—海大輪機系 9 張宏宜 Specific Heat of Ideal Gases Heat is allowed to flow into a gas, but the gas is not allowed to expand. If the gas is ideal and monatomic, the heat goes into increasing the average kinetic energy of the particles. All energy is translational Kinetic At constant Volume work = 0 3 The added heat KΔis = Q = nR. Δ T tr 2 3 nRΔ T Q 3 C = = 2 12= .R 5 = J/K/mol V nΔ T nΔ T 2 If the gas is diatomic (two atoms) Can also rotate 5 CR20V= . 8 = J/K/mol 普物AB—海大輪機系 2 10 張宏宜 Specific Heat of Ideal Gases Internal energy will be distributed equally among all possible degrees of freedom (equipartition of energy). Each degree of freedom contributes ½kT of energy per molecule and ½R to the molar specific heat at constant volume. Rotational motions of a 2-atom molecule: 普物AB—海大輪機系 11 張宏宜 Example (text problem 14.26) A container of nitrogen gas (N2 ) at 23 °C contains 425 L at a pressure of of heat are adde3.5 atm. If 26.6 kJ d to the container, what will be the new temperature of the gas? For a diatomic gas, Q= nCVΔ . T PV The number of moles n is given by the ideal gas law n = i i . RTi The change in temperature is ⎛ Q ⎞⎛ RTi ⎞ ΔT =⎜ ⎟⎜ ⎟ ⎝ CV ⎠PV⎝ i i ⎠ 26⎛ . 6× 3 10⎞ J 296R() K = ⎜ ⎟ ⎜ ⎟ 5 2 −3 3 3 . 5 atm⎝ 12 . .R 5 013⎠ () 10() × N/m() /atm() 425 L 10 m /L =21K The final temperature of the gas is Tf = Ti + ΔT = 317 K = 44 °C. 普物AB—海大輪機系 12 張宏宜 Phase Transitions A phase transition occurs whenever a substance changes from one phase (solid, liquid, or gas) to another. steam water water temp changes T temp to steam rises rises (boils) 100oC Latent Heat Q added to water Latent heat is the amount of heat per unit mass required to change the phase of a substance. The latent heat of fusion (Lf) is the heat per unit mass needed to produce the solid-liquid phase transition. The latent heat of vaporization (Lv) is the heat per unit mass needed to produce the liquid-gas phase transition. 普物AB—海大輪機系 13 張宏宜 Example (R&S 6) A 75 g cube of ice at -10.0 °C is placed in 0.500 kg of water at 50.0 °C in an insulating container so that no heat is lost to the environment. Will the ice melt completely? What will be the final temperature of this m?syste The heat required to completely melt the ice is Qice m=ice c iceΔ T+ ice ice m f L 0 . 075( kg ) 2 . 1 ()= kJ/kg C° ) 10 ° C + ( 0 . 075 ( kg ) 333 () . 7 kJ/kg 27= kJ The heat required to cool the water to the freezing point is Qw= m w wcΔ w T 0 . 5 kg( 4= . ) 186 ( °kJ/kg () ° C ) 50 C 105= kJ 普物AB—海大輪機系 14 張宏宜 Example continued: Since Qice < Qwater the ice will completely melt. To find the final temperature of the system, note that no heat is lost to the environment; the heat lost by the water is gained by the ice. 0 =QQice + w mcTmLmcTT=0 iceice Δ ice f + ice w + ) f− ice, ( + i () w mcTT w − f w, i mcTmL=0 ice Δice ice + f ice m+() w +mcTmcT w f − w w i 0= 27 m +icekJ() m + w c w105 T f − kJ T32f= . 4 ° C 普物AB—海大輪機系 15 張宏宜 Example (text problem 14.45): Compute the heat of fusion of a subsktance from these data: 31.15 J will change 0.500 kg of the solid at 21 °C to liquid at 327 °C, the melting point. The specific heat of the solid is 0.129 kJ/kg K. Q mc= Δ T+ f mL Q− mc Δ T L = 22 .= 8 kJ/kg f m 普物AB—海大輪機系 16 張宏宜 Phase diagram On a phase diagram, the triple point is the set of P and T where all three phases can coexist in equilibrium. Sublimation is the process by which a solid phase transitions into a gas (and gas → solid). The critical point marks the end of the vapor pressure curve. A path around this point (i.e. the path does not cross the curve) does not result in a phase transition. Past the critical point it is not possible to distinguish between the liquid and gas phases. 普物AB—海大輪機系 17 張宏宜 Thermal Conduction Through direct contact, heat can be conducted from regions of high temperature to regions of low temperature. Energy is transferred by collisions between neighboring atoms or molecules. The rate of energy transfer by conduction is ΔT PA= κ d where κ is the thermal conductivity, A is the cross- sectional area, and ΔT/d is the temperature gradient. 普物AB—海大輪機系 18 張宏宜 Thermal Conduction • Hot molecules have more KE than cold molecules • High-speed molecules on left collide with low- speed molecules on right – energy transferred to lower-speed molecules – heat transfers from hot to cold • I = rate of heat transfer = Q/t [J/s=W] – I = κ A (TH -TC )/L • Q/t = κ A ΔT/Δx L = Δx TH TC = thermal conductivity κ “ ” Hot Cold • Units: W/mK Area A • good thermal conductors…high κ • good thermal insulators … low κ – R = L/(Aκ) = thermal resistance 普物AB—海大輪機系 19 張宏宜 Thermal Conduction d AlsoΔTP =PR = κA where R is the thermal resistance.