Chapter 1. Temperature and Heat

Transcription

1 Chapter 1 Temperature and Heat Temperature and the Zeroth Law of Thermodynamis. Temperature - Temperature is a physical property of all bodies and is a relative measure, or indication, of hotness and coldness. - Temperature is a scalar quantity. Heat : is the net energy transferred from one object to another because of a temperature difference. Thermometer : Devices used to measure the temperature of a system. -Common thermometer liquid-in-glass-type -> based on the thermal expansion of a liquid (mercury/alcohol) Thermal contact : - When heat is transferred between two objects, whether or not they are physically touching, they are said to be in thermal contact. Thermal Equilibrium : When there is no longer a net heat transfer between objects in thermal contact, they are at the same temperature and are said to be in thermal equilibrium. Zeroth Law of Thermodynamics ( The Law of Equilibrium) states that: If bodies A and B are separately in thermal equilibrium with a third body C, then A and B will be in thermal equilibrium with each other if placed in thermal contact. Two objects in thermal equilibrium with each other are at the same temperature. Thermometers and Temperature Scales. Thermometer : - Devices used to measure the temperature of a system. 1

2 Three common scales of a temperature measurement : 1. Celsius 2. Fahrenheit. Kelvin They are defined according to three physical states of matter: - absolute zero, - freezing point of water, and - boiling point of water. The constant-volume Gas Thermometer and The Kelvin Scale. Gas thermometer is a standard instrument because its measurement is independent from any type of gas. The physical property used in a constant volume gas thermometer is the pressure variation with temperature of a fixed volume of gas. By international agreement, the standard thermometric property is the pressure of gas at constant volume. Figure 1.1 : A Constant-Volume gas thermometer. - The triple point of water, which is the single temperature and pressure at which water, water vapour, and ice can coexist in equilibrium, is set at an absolute temperature of Kelvin (=.1 C and pressure of 4.58mm of mercury) SI unit of the Kelvin is defined as 1/27.16 of the temperature of the triple point of water. 2

3 The Celsius, Kelvin and Fahrenheit Temperature Scales. F C K Boiling point (water) Freezing point (water) Absolute zero Figure 1.2 : Comparison of the Kelvin Celsius, and Fahrenheit temperature scales. Physical state C F K Absolute zero Freezing point of water Boling point of water The temperature scales are related according to: i) Celsius-to-Fahrenheit conversion: T F = 5 9 T c + 2 = 1.8 T c + 2 (1.1) ii) Fahrenheit-to-Celsius conversion: T C = 9 5 (TF - 2) (1.2) iii) Celsius-to-Kelvin conversion: T K = T c (1.) Example 1.1 Calculate: (a) the typical room temperature of 2 C and a cold temperature of -18 C on the Fahrenheit scale; and (b) another cold temperature of -1 F and normal body temperature, 98.6 F, on the Celsius scale?

4 c) The absolute zero on the Fahrenheit scale? Solution: a) Celsius-to-Fahrenheit conversion 2 C: T F = 5 9 Tc + 2 = 5 9 (2) + 2 = 68 F -18 C: T F = 5 9 Tc + 2 = 5 9 (-18) + 2 = -.4 F b) Fahrenheit-to-Celsius conversion: -1 F: T C = 9 5 (TF - 2) = 9 5 (-1-2) = -2. C 98.6 F: T C = 9 5 (TF - 2) = 9 5 (98.6-2) = 7. C c) Since there is no direct relation between Kelvin and Fahrenheit, so first convert K to a Celsius value: T c = T K = = C Then, converting to Fahrenheit : T F = 5 9 Tc + 2 = 5 9 (-27.15) + 2 = F Thermal Expansion of Solids and Liquids. When the temperature of a body increases the average distance between the atoms usually increases and so the volume of the body increases. Thermal expansion is a physical phenomenon in which increases in temperature can cause substance in the solid, liquid, and gaseous states to expand. The thermal expansion of a material is characterized by its coefficient of expansion. 4

5 Figure 1.: Thermal expansion of a homogeneous metal washer. As the washer is heated, all dimensions increase. Figure 1.4 A bimetallic strip bends as the T changes because the two metals have different expansion coefficients. (Fig. b use to break or make electrical contact) - Thermal expansion of solids: Linear : A solid subjected to an increase in temperature T experiences an increase in length L that is proportional to the original length L of the solid. T L T=T + L 5

6 - The change in length of the rod, L L o T where the constant of proportionality depends on the substance from which the rod is made. - The relation for the linear expansion of solid is L L LT or T or L L (1 T ) L where is the average coefficient of linear expansion (1.4) Example 1.2: The main span of the Penang bridge is 1275 m long when at its coldest. The bridge is subjected to temperature extremes ranging from 15 C to 4 C. What is its change in length between these two temperature? Solution : 6 L LT (12x1 )(1275m)[4 ( 15)]. 84m Example 1. The Sultan Alaudin Bridge in Klang Valley is a 518 m long steel ach. How much will its length change between temperature extremes 2 C and 5 C? Solution: Use L = Lo(1 + (T - To)): L-2 = Lo + Lo(-2. C) - LoTo, and L5 = Lo + Lo (5. C) - LoTo. L = L5 - L-2 = Lo (55 C), or L = (11 x 1-6 / C)(518 m)(55 C) =.1 m = 1 cm. Area: Since the length of an object changes with temperature, it follows that its area changes as well. A 6

7 From L L T L L L T L L (1 T ) Since area (A) is length squared (L 2 ) for a square, we get A L L 1 T ) A (1 2 T) or A A 2T ( (1.5) Thus, the thermal coefficient or area expansion is twice as large as the coefficient of linear expansion. Volume: As the system is heated there will be an expansion of the container. V - A liquid subjected to an increase in temperature T experiences an increase in volume V that is proportional to the original length V of the liquid. - The relation for the linear expansion of liquids is V V V V T or T or V V (1 T ) (1.6) Thermal Volume Expansion of Fluids Volume Expansion V V T (1.7) where is the coefficient of volumetric expansion and is equal to. 7

8 TABLE 1.1 Average coefficient of Expansion for Some Materials Near Room Temperature Material Material Aluminum 24 x 1-6 Ethyl alcohol 1.12 x 1-4 Steel 11 x 1-6 Mercury 1.82 x 1-4 Example 1.4: A copper flask with a volume of 15 cm is filled to the brim with olive oil. If the temperature of the system is increased from 6 C to 1 C, how much oil spills from the flask? Solution: 1. Calculate the change in volume of the olive oil : 1 VT (.68x1 K )(15cm )(25K) 2.6cm V oil 2. Calculate the change in volume of the flask : 6 1 VT (17x1 K )(15cm )(25K).19cm V flask. Find the difference in volume expansions. This is the volume of oil that spills out. V V 2.6cm.19cm 2.4cm oil flask (if the system were cooled, the oil would lose volume more rapidly than the flask. This would result in a drop in oil level. HEAT Heat (or thermal energy) is the net energy transferred from one object to another (between a system and its surroundings) because of a temperature difference. SI Unit : joule (J) Other units: calorie (cal) / kilocalorie (kcal) [ 1 kcal=1cal] 1 kcal is defined as the amount of heat needed to raise the temperature of 1 kg of water by 1 C o (from 14.5 to 15.5 o C). 1 cal is defined as the amount of heat needed to raise the temperature of 1 g of water by 1 C o. 8

9 The Mechanical Equivalent of Heat Relationship between unit of heat and Joule (standard unit) is called the mechanical equivalent of heat 1 kcal =4186 J = kj or 1 cal=4.186j Specific Heat (c) -Specific heat is the amount of heat energy required to change the temperature of 1 kg of substance by 1 C o. -Specific heat C is depends on the substance, its phase and T Q mct (1.8) Unit : J/(kg.K) or J/(kg. o C) -The heat needed to change the temperature of a mass m by T is Q mct (1.9) where Q is heat, c is the specific heat T = T f - T i -The amount of heat Q required to cause a temperature change depends on 1. Size of the temperature change 2. The mass of the system. The substance and phase involved Table 1.2: Specific Heat of some materials Material c (J / kg. o C) c (cal/ g. o C) Aluminum Copper Gold Water

10 -When the object gains heat ( T > ), Q > (+Q) : Energy being added to a system -If the object loses heat ( T < ), Q < (-Q) : Energy is removed from a system Example 1.5 How many joules of energy are required to raise the temperature of 1g of gold from 2. C to 1 C? Solution: Q = mc T = (.1 kg)(129 J/kg C)(8. C) = 12 J Example 1.6 A 5.g piece of cadmium is at 2 C. If 4 cal of heat is added to the cadmium, what is its final temperature? Solution: Q = 4 cal = 1674 J, so T = Q mc = 1674 J (5. x 1-2 kg)(2 J/kg C) = 146 C, and Tf = 166 C Conservation of Energy: Calorimetry -Calorimetry is the study of the transfer of the thermal energy between a system and (solid or liquid) and its environment. -Calorimeter is a device where the heat transfer occurs. -To determine the specific heat, c x, of a system we conduct the following steps: 1. heat the system to a known temperature 2. put it in a bowl of water of known mass and temperature,m w and T w respectively. -Because the law of conservation of energy implies that leaves the warmer system equals the heat that enters the water, the (system+water) will have the same temperature, T, when equilibrium is reached. -Within an isolated system, the heat lost by an object will be gained by other objects within the system. -According to the conservation of energy, Q (1.1) absolute value of heat lost = heat gained 1

11 -Q lost = Q gained (1.11) -The law of conservation is then written as follows : m w. c w.(t-t w ) = m x. c x.(t x -T) or C x mw. c mx w T T T x T w (1.12) where, m x : mass of the system (solid, liquid) c x : specific heat of the system T x : initial temperature T : equilibrium temperature Example 1.7 A.25 kg block of a pure material is heated from 2. C to 65. o C by the addition of 1.4 kcal of heat. Calculate its specific heat, and identify the substance of which it is most likely composed. Solution : Q = mc T c Q mt It is copper x45 o.924kcal / kg C Example 1.8: A half litre of water at o C is cooled, with the removal of 6 kj of heat. What is the final temperature of the water? Solution : From Q = mc T Q 6kJ o o o Tf T1 C C C mc (.5kg)[4.186kJ /( kg. C)] O C Example 1.9: A.5 kg mass of metal is heated to 2 o C then put in a bowl containing.4kg of water that is initially at 2 o C. If the final equilibrium temperature is 22.4 o C, find the specific heat of this metal. 11

12 Solution: Because the heat lost by the piece of metal equals the heat gained by the water, we have m x. c x.(t ix -T fx ) = m w. c w.(t fw -T iw ) (.5Kg). (c x ). (2 o C o C) = (.4Kg). (4186 J/Kg. o C).(22.4 o C-2 o C) thus, c x = 45 J/Kg. o C Latent Heat and Phase Changes -Matter normally exists in three phases: solid, liquid, and gas. -When a heat is transferred between a substance and its surroundings, it leads to a physical change or alteration of the substance ( from one form to another). We call this change phase change. A phase change from solid to liquid is called melting. A phase change from liquid to gas is called boiling. Solid melting Liquid boiling freezing Solid to gas phase S condensati on Gas Solid D gas S = Sublimation D = Deposition Latent heat (or hidden heat ) is the heat involved in a phase change such as from solid to liquid or from liquid to gas, and does not go into changing the temperature, but in breaking or forming molecular bonds. Q ml (1.1) where L= latent heat m = mass of the substance 12

13 Unit for L : J/kg or kcal/kg LH for a solid-liquid phase change latent heat of fusion(l f ) Q= ml f (melting-freezing) LH for a liquid-gas phase change latent heat of vaporization (L v ) Q = ml v (vaporization-condensation) A plot of T vs heat energy for a quantity of water (Refer Fig.11.6) Figure 1.5 : A plot of temperature versus heat added when 1. g of ice, initially at -. C, is converted to steam at 12 C. Part A : Ice :T=-. C to. C Q = mc T Part B : Ice-water mixture:t= to (phase change) Q ml Part C : Water : T= C to 1 C (no phase change) Q = mc T Part D : Water-steam mixture :T=1 C to 1 C (phase change) Q ml Part E : Steam Q = mc T Total amount of heat = Q A + Q B + Q C + Q D + Q E 1

14 Problem Solving Strategy _ Calorimetry Problems 1. Always be sure of the units used. If the unit of the specific heat is cal/ g. o C, the masses must be in g and the temperature in o C. 2. Losses and gains in heat are found by using the equation : Q = m c T (if no phase change occur) Q = m. L v or Q = m.l f. ( if a phase change takes place). The sign of T must be kept positive ( T positive number) Example 1.1 : What mass of the steam that is initially at 1 o C is required to warm 2g of water in 1g glass mug from 2 o C to 5 o C? Solution: Because we have a heat transfer problem, the heat lost by the steam equals the heat gained by the water and mug. The steam losses heat in three steps: Step 1: Cooling to 1 o C: the heat liberated is: Q 1 = m x c s T = m x. (2.1 1 J/Kg. o C). ( o C) = m x.( J/Kg) Step 2: Conversion to water: Q 2 = m x. L v. = m x.( J/Kg) Step : Decrease of the temperature of the water to 5 o C: Q = m x c w T = m x.( J/Kg. o C)(5 o C) = m x.( J/Kg) If the heat lost by the steam equals the heat gained by the water, we will have: m x.(6.1 4 J/Kg)+m x.( J/Kg)+m x.( J/Kg) = (.2Kg)(4.191 J/Kg)( o C)+(.1kg)(87 J/Kg. o C)( o C) So, m x = kg Example 1.11: How much heat energy is needed to change 1kg of ice at 2 o C to steam at 12 C? 14

15 Solution : Use the following subscript : i=ice, w=water, and s=steam. Ice at 2 o C firsts need to be raised to o C, and then melted to o C liquid water. The water at o C needs to be raised to 1 o C, where it is vaporized to steam. Finally, the temperature of steam is raised to 12 o C. 1. The heat required to raise the temperature of ice from -2 o C to o C is Q 1 = c i m i T i = [21J/kg.C o ][1kg][ o C-(-2 o C)] = 4.2 x 1 5 J 2. The heat required to melt ice at o C to water at o C is Q 2 = ml f = (1kg)(. x 1 5 J/kg) =. x 1 6 J. The heat required to raise the temperature of water from - o C to 1 o C is Q = c w m w T w = [4186 J/kg.C o ][1kg][ 1 o C- o C)] = x 1 6 J 4. The heat required to vaporize water at 1 o C to steam at 1 o C is Q 4 = ml v = (1kg)(22.6 x 1 5 J/kg) = 2.26 x 1 6 J 5. The heat required to raise the temperature of water from 1 o C to 12 o C is Q 5 = c s m s T s = [21J/kg.C o ][1kg][ 12 o C--1 o C] = 4.2 x 1 5 J So, the total heat required is Q total = Q i =Q 1 + Q 2 + Q + Q 4 + Q 5 =.1 x 1 7 J 15