Temperature & Heat Energy Lecture Notes for General Physics I by Dr. Donald Luttermoser, Study notes of Physics

Download Temperature & Heat Energy Lecture Notes for General Physics I by Dr. Donald Luttermoser and more Study notes Physics in PDF only on Docsity! PHYS-2010: General Physics I Course Lecture Notes Section XII Dr. Donald G. Luttermoser East Tennessee State University Edition 2.3 Abstract These class notes are designed for use of the instructor and students of the course PHYS-2010: General Physics I taught by Dr. Donald Luttermoser at East Tennessee State University. These notes make reference to the College Physics, 7th Edition (2005) textbook by Serway and Faughn. Donald G. Luttermoser, ETSU XII–3 Solution (b): From the same argument above, Eq. (XII-2) gives T2 − T1 = [(TC)2 + 273.15] − [(TC)1 + 273.15] T2 − T1 = (TC)2 − (TC)1 ∆T = ∆TC = 450 K . B. Thermal Expansion. 1. Heat energy added to matter causes the particles that make up the matter to speed up. a) Increased velocity of particles increases pressure. b) Increased pressure causes matter to expand due to the increased internal force. i) Liquids and gases fill a larger volume. ii) Solids get longer. c) A loss of heat energy causes objects to shrink in size. 2. For solids, the change in heat (as measured by temperature) dic- tates a change in linear size ∆L via ∆L = α L◦ ∆T . (XII-3) a) L◦ ≡ initial length [m]. b) ∆T ≡ change in temp. (usually measured in ◦C in Eq. XII-3). XII–4 PHYS-2010: General Physics I c) α ≡ coefficient of linear expansion (see Table 10.1 in text- book). 3. Surface areas also change in size ∆A via: ∆A = γ A◦ ∆T . (XII-4) a) A◦ ≡ initial area [m2]. b) γ ≡ coefficient of area expansion = 2α. 4. Volumes increase (or decrease), ∆V , as well when heat is added (or taken away) via: ∆V = β V◦ ∆T . (XII-5) a) V◦ ≡ initial volume [m3]. b) β ≡ coefficient of volume expansion = 3α. Example XII–2. Problem 10.11 (Page 346) from the Serway & Faughn textbook: The New River Gorge bridge in West Virginia is a 518- m long steel arch. How much will its length change between temperature extremes of −20◦C and 35◦C? Solution: Use Eq. (XII-3) here. ∆T = 35◦C – (–20◦C) = 55◦C, L◦ = 518 m, and α = 11 × 10−6 ◦C−1 from Table 10.1 on page 313 in the textbook. This gives a length change of ∆L = α L◦ ∆T = (11 × 10−6 ◦C−1)(518 m)(55◦C) = 0.31 m = 31 cm . Donald G. Luttermoser, ETSU XII–5 C. Heat and Internal Energy. 1. Internal energy, U , is the energy associated with the micro- scopic components of a system — atoms and molecules. It is given by U = KEtrans + KErot + KEvib + PEmol . (XII-6) a) KEtrans ≡ average translational (linear) kinetic en- ergy of the atoms and molecules. b) KErot ≡ average rotational kinetic energy =⇒ the ro- tation of a molecule about an axis. c) KEvib ≡ average vibrational kinetic energy =⇒ a con- tinuous change of distance between the atoms that make up the molecule. d) PEmol ≡ average intermolecular potential energy be- tween the atoms that make up the molecules =⇒ the bond energy. 2. Heat or thermal energy, Q, is a mechanism by which energy is transferred between a system and its environment because of a temperature difference between them. a) Heat is essentially related to kinetic energy =⇒ energy due to the motion of particles in matter. b) Heat is measured in calories in the cgs unit system. =⇒ one calorie of heat is required to raise 1 gram of water by a temperature of 1◦C. c) Heat is measured in kilocalories in the SI system. i) One kilocalorie (= 103 calories) of heat is required XII–8 PHYS-2010: General Physics I Example XII–3. Problem 11.9 (Page 379) from the Serway & Faughn textbook: A 5.00-gm lead bullet traveling at 300 m/s is stopped by a large tree. If half the kinetic energy of the bullet is transformed into internal energy and remains with the bullet while the other half is transmitted to the tree, what is the increase in temperature of the bullet? Solution: The mechanical energy transformed into internal energy of the bullet is Q = 1 2 (KEi) = 1 2 ( 1 2 mv2i ) = 1 4 mv2i , where m is the mass of the bullet and vi = 300 m/s is the initial speed of the bullet. From Table 11.1 on page 335 in the textbook, the specific heat of lead is 128 J/kg·◦C. Using Eq. (XII-8), the change in temperature of the bullet is ∆T = Q mc = 1 4mv 2 i mc = v2i 4c = (300 m/s)2 4 (128 J/kg ·◦ C) = 176◦C . Example XII–4. A 0.40-kg iron horseshoe that is initially at 500◦C is dropped into a bucket 20 kg of water at 22◦C. What is the final equilibrium temperature? Neglect any energy transfer to or from the surroundings. Solution: From Table 11.1 in the textbook, the specific heat of iron (Fe) is 448 J/kg·◦C and water is 4186 J/kg·◦C. Using Eq. (XII-11) we get an equi- librium temperature T of mwcw (T − Tw) = mFecFe (TFe − T ) mwcwT − mwcwTw = mFecFeTFe − mFecFeT mwcwT + mFecFeT = mwcwTw + mFecFeTFe (mwcw + mFecFe) T = mwcwTw + mFecFeTFe T = mwcwTw + mFecFeTFe mwcw + mFecFe Donald G. Luttermoser, ETSU XII–9 with mwcwTw = (20 kg)(4186 J/kg ·◦C)(22◦C) = 1.84 × 106 J , mFecFeTFe = (0.40 kg)(448 J/kg ·◦C)(500◦C) = 8.96 × 104 J , and mwcw + mFecFe = (20 kg)(4186 J/kg ·◦C) + (0.40 kg)(448 J/kg ·◦C) = 8.39 × 104 J/◦C . Plugging this into out equilibrium temperature equation above, we get T = 1.84 × 106 J + 8.96 × 104 J 8.39 × 104 J/◦C = 23◦C . E. Latent Heat & Phase Changes. 1. When a substance undergoes a physical alteration from one form to another (i.e., liquid to gas), it is referred to as a phase change. a) Under such circumstances, T does not change when heat is added or taken away from the system =⇒ the energy goes into changing the phase. b) Mathematically: Q = ±m L . (XII-12) i) Q ≡ heat gained or lost to the system. ii) m ≡ mass of the system. XII–10 PHYS-2010: General Physics I iii) L ≡ latent heat. iv) You use the ‘+’ sign when energy is being added during a phase change (i.e., melting or boiling). v) You use the ‘–’ sign when energy is being taken away during a phase change (i.e., freezing or con- densation). c) When the phase change is from gas to liquid (or vise- versa), L is called the latent heat of vaporization, Lv. i) Changing from gas to liquid is called condensa- tion. ii) Changing from liquid to gas is called boiling. d) When the phase change is from liquid to solid (or vise- versa), L is called the latent heat of fusion, Lf. i) Changing from liquid to solid is called freezing. ii) Changing from solid to liquid is called melting. e) Table 11.2 of the textbook shows latent heats of fusion and vaporization for some common substances. 2. We can deduce temperature changes in systems that undergo phase changes by using the conservation of energy for each state of the system. Donald G. Luttermoser, ETSU XII–13 will go towards heating the block of ice. Also, the temperature of the bullet will heat the ice block. Let mmelt be the amount of mass in the ice block that melts, mb = 3.00 gm = 3.00 × 10−3 kg be the mass of the bullet, Tb = 30.0 ◦C be the initial temperature of the bullet, vb = 240 m/s be the initial speed of the bullet, and T = 0◦C be the final temperature of the bullet (since it will cool to the same temperature of the block of ice). Using the tables from the textbook, the latent heat of fusion of water is Lf = 3.33 × 105 J/kg and the specific heat for lead is cb = 128 J/kg·◦C. Using the conservation of energy (and noting that J/kg = m2/s2), we get Qgain = Qloss Qmelt = KEb + Qb-loss mmeltLf = 1 2 mbv 2 b + mbcb (Tb − T ) mmelt = mb  v 2 b/2 + cb (Tb − T ) Lf   = (3.00 gm)  (240 m/s) 2/2 + (128 J/kg ·◦C)(30◦C − 0◦C) 3.33 × 105 J/kg   = 0.294 gm . F. Transport of Heat Energy. 1. Thermal energy (i.e., heat) can only flow by one of three dif- ferent mechanisms: conduction, convection, and radiation transport. 2. Heat Transfer by Conduction. a) Conduction is the process by which heat is transferred via collisions of internal particles that make up the object =⇒ individual (mass) particle transport. XII–14 PHYS-2010: General Physics I i) Heat causes the molecules and atoms to move faster in an object. ii) The hotter molecules (those moving faster) col- lide with cooler molecules (those moving slower), which in turn, speeds up the cooler molecules mak- ing them warm. iii) This continues on down the line until the object reaches equilibrium. b) The amount of heat transferred ∆Q from one location to another over a time interval ∆t is ∆Q = P ∆t . (XII-14) i) P ≡ heat transfer rate. ii) P is measured in watts when Q is measured in Joules and ∆t in seconds. iii) As such, P is the same thing as power since they are both measured in the same units. c) Heat will only flow if a temperature difference exists be- tween 2 points in an object. i) For a slab of material of thickness L and surface area A, the heat transfer rate for conduction is Pcond = ∆Q ∆t = KA ( Th − Tc L ) . (XII-15) Donald G. Luttermoser, ETSU XII–15 A Tc Th (on back side) Heat Flow (if Th > Tc) L ii) Th is the temperature of the hotter side and Tc is the temperature of the cooler side. iii) K ≡ thermal conductivity of the material (see Table 11.3 in textbook) =⇒ [K] = J/(s m ◦C) = W/m/◦C. iv) The larger K is, the better the material is in conducting heat. v) The smaller K is, the better the material is as a thermal insulator. vi) The effectiveness of thermal insulation is rated by another quantity =⇒ thermal resistance, R, where R ≡ L K , (XII-16) where in the U.S., [R] = ft2 h ◦F/BTU (‘h’ = hour), and elsewhere, [R] = m2 ◦C/W (see Table 11.4 in the textbook). XII–18 PHYS-2010: General Physics I iii) A ≡ surface area of the object [m2]. iv) e ≡ emissivity [unitless] (e = 1 for a perfect ab- sorber or emitter). v) T ≡ temperature [K]. c) A body also can absorb radiation. If a body absorbs a power of radiation Pabs, it will change its temperature to T◦. i) The net power radiated by the system is then Prad = Pnet = Pem − Pabs (XII-18) or Prad = σAeT 4 − σAeT 4◦ , Prad = σAe ( T 4 − T 4◦ ) . (XII-19) ii) In astronomy, the total power radiated by an ob- ject over its entire surface is called the luminosity, L = Prad, of the object. Since the amount of energy falling on the surface of the Sun (or any isolated star) from interstellar space is negligible to that of the power radiated, L = Pem for isolated stars. iii) If an object is in equilibrium with its surround- ings, it radiates and absorbs energy at the same rate =⇒ its temperature remains constant =⇒ this ra- diative equilibrium results in the object being in thermal equilibrium: Prad = 0 =⇒ T = T◦ , where T◦ is the temperature of the surroundings. Donald G. Luttermoser, ETSU XII–19 d) An ideal absorber is defined as an object that absorbs all of the energy incident upon it. i) In this case, emissivity (e) = 1. ii) Such an object is called a blackbody: Pbb = σAT 4 . (XII-20) iii) Note that a blackbody radiator can be any color (depending on its temperature =⇒ red blackbod- ies are cooler than blue blackbodies), it does not appear “black” (unless it is very cold). iv) The energy flux of such a radiator is Fbb = Pbb A = σ T 4 . (XII-21) v) This radiative flux results from the condition that in order to be in thermal equilibrium, the heat gained by absorbing radiation must be (virtually immediately) radiated away by the object. vi) This is not the same thing as reflecting the radi- ation off of the surface (which does not happen in a blackbody). The incident radiation does get “ab- sorbed” by the atoms of the object and deposited in the thermal “pool.” It is just that this radiation immediately gets re-emitted just after absorption. XII–20 PHYS-2010: General Physics I Example XII–7. Problem 11.42 (Page 382) from the Serway & Faughn textbook: The surface temperature of the Sun is about 5800 K. Taking the Sun’s radius to be 6.96×108 m, calculate the total energy radiated by the Sun each second. (Assume e = 0.965.) Solution: We will assume the Sun’s shape to be a sphere, where the surface area of a sphere is A = 4πR2 = 4π(6.96 × 108 m)2 = 6.09 × 1018 m2 . Making use of Eq. (XII-17), the total power (energy per second) that the Sun radiates to space is Pem = σAeT 4 = ( 5.67 × 10−8 W m2 · K4 ) (6.09 × 1018 m2)(0.965)(5800 K)4 = 3.77 × 1026 W . This is the value of the Sun’s luminosity: L = Pem = 3.77 × 1026 W .