Chapter 3 - Physics PDF

Chapter 3  Temperature and Heat * Temperature T: The degree of hotness or coldness of a body or environment. * Heat Q: is the energy that is transferred between objects because of a temperature difference. Zeroth law of thermodynamics “If object A is in thermal equilibrium with objec...

Chapter 3  Temperature and Heat * Temperature T: The degree of hotness or coldness of a body or environment. * Heat Q: is the energy that is transferred between objects because of a temperature difference. Zeroth law of thermodynamics “If object A is in thermal equilibrium with object B, and object C is also in thermal equilibrium with object B, then objects A and C will be in thermal equilibrium if brought into thermal contact”.  Temperature Scales The Celsius The Fahrenheit The Kelvin Scale TC Scale TF Scale TK Water freezes at 0 oC. Water freezes at 32 oF. Water freezes at 273 K. Water boils at 100 oC. Water boils at 212 oF. Water boils at 373 K. 9 A temperature T of five T F = TC + 32 T K = TC + 273 degrees is 5 oC. 5 A temperature change DT of five degree is 5 Co. 5 TC = (T F - 32 ) DT K = DTC 9 Absolute Zero: It is the lowest temperature below which it is impossible to cool an object.  Heat and Mechanical Work * What happen when you rub your hand against each other? Mechanical work is converted to heat. * What are the units of heat? 1. One kilocalorie [kcal or Calorie or C]: is the amount of heat needed to raise the temperature of 1 kg of water from 14.5 oC to 15.5 oC. 1 kcal = 1 Calorie = 1 C = 4186 J * The mechanical equivalent of heat 1 cal = 4.186 J Example: 250 C candy bar = 250 000 cal 2. Btu [Biritch thermal unit]: is the energy required to heat 1 lb of water from 63 oF to 64 oF 1 Btu = 0.252 kcal = 1055 J  Example 2-1 * A 74.0-kg person drinks a thick, rich, 305-C milkshake. How many stairs must this person climb to work off the shake? Let the height of a stair be 20.0 cm. Solution: 1. Convert the energy of milkshake to joules: Q = 305 C = 305 000 cal = 305 000 cal × 4.186 J/cal = 1.28 × 106 J 2. Energy of milkshake = work done against gravity Q = mgh Or 6 Q 1.28¥ 10 h= = = 1760 m mg 74 ¥ 9.8 The number of stairs = 1760/0.20 = 8800 stairs  Heat Capacity C “is the amount of heat (Q) necessary for a given temperature change (∆T)” Q C = [Unit: J/K or J/Co] Or DT Q = C DT * Q is positive if DT is positive (means: heat is added to the system). * Q is negative if DT is negative (means: heat is removed from the system). * For given DT, Q increases with mass m and so C depend on mass.  Specific Heat c “is the amount of heat required to increase the temperature of one kilogram of substance by one degree Celsius.” Q c= m DT [Unit: J/kg.K or J/kg.Co] Q = m c DT * For given DT, Q increases with mass m but c remains constant. So c does not depend on mass. * c depends on the type of material.  Latent heat L “It is the heat that must be added to or removed from one kilogram of a substance to convert it from one phase at same temperature“ Q L = SI unit is J/kg m The latent heat required to melt (or fuse) a substance is called the latent heat of fusion, Lf. The latent heat required to convert a liquid to gas is called the latent heat of vaporization, Lv. The latent heat required to convert a solid to gas is called the latent heat of sublimation, Ls. Adding salt to icy water increases the efficiency of cooling. Why? Salt prevent water to be crystalline at 0oC and so decreases the freezing point of water, making ice to melt at 0oC by drawing the required melting latent heat from the surrounding (e.g., cans, meat, fish, ……) which makes their temperature very low.  Conduction, Convection, and Radiation * Heat can be changed in a Varity of ways, i. Conduction. ii. Convection. iii. Radiation. (i) Conduction: It is the flow of the heat through a physical material and depends on the type of material. * Heat flow by conduction Q through a rod depends on: 1. Type of its material. 2. Its cross sectional area (A). 3. Temperature difference (∆T) between its ends. 4. Length of the conductor (L). 5. Time of heat flow (t). æ DT ö Q = kA ç ÷t è L ø L Where k is called the thermal conductivity (depends on type of material). What is its unit?  Application of heat Conduction 1) Insulated window: constructed from two panes of glass separated by air-filled gap as air is a poor heat conductor. 2) Countercurrent exchange: it means the exchange of heat between warm blood in arteries and the returning cold blood in venous from the leg to the body. For a bird standing in cool water, this serve to maintain the core body temperature of the bird, while at the same time keeping its legs and feet at much cooler temperatures to reduce the amount of heat lost to the water.  Convection “It is the physical flow of matter that carries heat due to a temperature difference.” * Warm portions of fluid rise because of their lower density. * Cool portions of fluid sink because of their higher density. * Examples of heat flow by convection are: * See breeze: During the day, warm air over the land become less dense and rises up while cooler air above water flows in to take its place. * Land breeze: at night, warm air over the water become less dense and rises up and replaced by cooler air from over the land.  Radiation * It is the energy transfer as an object emit or absorb radiation in the form of electromagnetic waves such as visible, infrared and ultraviolet radiation. * Unlike conduction and convection, radiation has no need for a physical material to mediate the energy transfer. Radiation transfer through empty space (vacuum).  Stefan-Boltzmann law “The radiated power (P) is proportional to the fourth power of the absolute temperature T.” P = es AT 4 [SI unit: J/sec or Watt] * 𝛔 = Stefan-Boltzmann constant = 5.6696 ×10-8 W/m2.K4 * e = emissivity [varies from 0 for poor radiator to 1 for perfect radiator]. * A dark-colored object has e ≈ 1, and light-colored object has e closer to 0. * A black body is a perfect emitter or a perfect absorber. * The opposite of blackbody is an ideal reflector which absorbs or emits no radiation.  Applications on radiation * Thermos bottle is designed to reduce the rate of heat transfer. * The inside of a Thermos bottle is highly reflective to reduce heat losses by radiation. * A Thermos bottle has a vacuum between the inner and the outer walls to reduce heat losses by convection and conduction and permits the flow of heat by radiation only.  Net radiated power * The net radiated power is Pnet = es A (T 4 -T s4 ) T = temperature of an object Ts = temperature of its surroundings. * The net radiated power is positive if the object’s temperature is greater than its surroundings and radiates more energy. * The net radiated power is negative if the object’s temperature is less than its surroundings and absorbs more energy. * The net radiated power is zero if the object has the same temperature as its surroundings.  Work and Energy Work is the component of force in the direction of displacement times the magnitude of the displacement. When a constant force (F) in the direction of displacement (d) the work is W = Fd [SI unit: N.m or Joule] When a constant force (F) act at an angle (θ) of the displacement (d) the work is W = (F cosθ)d  Work is a scalar quantity Work depends on the angle between force and the displacement If 270° < q < 90° there is a force component in the same direction of displacement and work is positive. If 90° < q < 270° there is a force component in the opposite direction to displacement and work is negative. If q = 90 or 270 there is a force perpendicular to the direction of displacement and work is zero.  The power (P) is a measure of how quickly work is done. Or the rate at which work is done. W P= t SI unit: watt= J/s or horsepower] 1 horsepower = 1 hp = 746 watt  The Mechanical energy (E): “ is the sum of the potential energy U and kinetic energy K” The kinetic energy (K): “the energy due to motion” m = mass (kg), v = speed (m/s) 1 SI unit: kg.m2/s2 = joule K = mv 2 2 The potential energy (U): “the energy due to position” “The work done by a conservative force is equal to the negative of the change in potential energy” W c = -DU = - (U f - U i ) DU = -W = - mg ( y f - y i ) cos180o = mgh m = mass (kg), g = gravitational acceleration (m/s2), y = height = altitude (m).  The First Law of Thermodynamics “The change in a system’s internal energy (∆U) is related to the heat (Q) and the work (W) as follows: DU = Q -W Q is positive when heat is added to the system and negative when heat is removed from the system. W is positive when work is done by the system and negative when work is done on the system. DU is positive when heat is added to the system or work is done on the system. DU is negative when heat is removed from the system or work is done by the system.  Example 2-2 (a) Jogging along the beach one day, you do 4.3 × 105 J of work and give off 3.8 × 105 J of heat. What is the change in your internal energy? (b) Switching over to waking, you give off 1.2 × 105 J of heat and your internal energy decreases by 2.6 × 105 J. How much work have you done while walking? Solution: (a) W = 4.3 × 105 J [(+) means you do work] Q = - 3.8 × 105 J [(-) means you give off heat (lose heat)] DU = Q -W = -3.8 ´ 105 - 4.3 ´ 105 = -8.1 ´ 105 J (b) Q = - 1.2 × 105 J [(-) means you give off heat (lose heat)] ∆U =- 2.6 × 105 J [(-) means the internal energy decrease] W = Q - DU = -1.2 ´ 105 - ( -2.6 ´ 105 ) = 1.4 ´ 105 J First Law and Human Metabolism Suppose in a time ∆t, a person does an amount of mechanical work ∆W. Heat will usually leave the body, so ∆Q (heat change) will be negative. From the first law of thermodynamics, DU = Q -W The rate of potential energy change, Du = D Q - D W Du DQ DW = - Dt Dt Dt Ø Metabolic rate: Du “It is the rate of change of the internal energy” or M.R. = [unit: J/s or Watt] Dt 1 Du The metabolic rate = - [Unit: W/kg or J/kg.s] Per unit mass m Dt Ø The energy Equivalent of Oxygen: “It is the ratio of the energy released to the oxygen consumed” E E eq = [Unit: kJ/L] V Ø The Energy Content per unit mass: “It is the ratio of the energy released divided by the mass” E Ec = [Unit: kJ/g] m Ø The efficiency of food utilization (e): The efficiency would be 100 % if all the additional energy was converted to mechanical work DW e= Dt ´ 100 éæ Du ö æ Du ö ù êç Dt ÷ - ç Dt ÷ ú ëè ø è øbasal û