Chapter 2 (Heat & Temperature) PDF

1 CHAPTER 2 Heat & Temperature  Contents 1/ Heat and Temperature 2/ Methods of Temperature Measurement 3/ Thermal expansion 4/Quantity of Heat and Specific Heat 5/ Methods of Heat Transfer.  Cup gets cooler while 2.1 Heat hand gets wa...

1 CHAPTER 2 Heat & Temperature  Contents 1/ Heat and Temperature 2/ Methods of Temperature Measurement 3/ Thermal expansion 4/Quantity of Heat and Specific Heat 5/ Methods of Heat Transfer.  Cup gets cooler while 2.1 Heat hand gets warmer Heat The flow of thermal energy from one object to another.. Heat always flows from warmer to cooler objects. Ice gets warmer while hand gets cooler  2.2 Temperature:  Temperature is one of the seven SI base quantities.  The SI unit of temperature is Kelvins (K).  The temperature of a body is its degree of hotness (or coldness).  Thermometers are devices used to measure the temperature of a system.  Thermal equilibrium is a situation in which two objects would not exchange energy if they were placed in thermal contact  Temperature is the property that determines whether an object is in thermal equilibrium with other objects.  2.3 Thermal Equilibrium The zeroth law of thermodynamics : If bodies A and B are each in thermal equilibrium with a third body C , then A and B are in thermal equilibrium with each other.  2.4 Methods of Temperature Measurement There are several temperature scales. The Three most famous scales are:  Celsius (℃ )  Kelvin ( K)  Fahrenheit (℉ ) In order to establish a temperature scale it is necessary to make use of fixed points. A fixed point is a single temperature at which certain physical property always occurs.  2.4 Methods of Temperature Measurement 2.4.1 Fahrenheit temperature The Fahrenheit temperature scale, 32°F is the freezing point of pure water and 212°F is its boiling point. To convert from Celsius to Fahrenheit: To convert from Fahrenheit to Celsius:  2.4 Methods of Temperature Measurement 2.4.2 Kelvin (or absolute) On the Kelvin (or absolute) temperature scale, 0 K is the extrapolated temperature at which a gas would exert no pressure. To convert from Celsius to Kelvin:  2.4 Methods of Temperature Measurement 2.4.3 Temperature Conversions Below are relationships between the Kelvin (K) and Celsius (C) temperature scales. Temperatures have been rounded off to the nearest degree.  Example (2.1) (A) A small piece of Iron has a temperature T= 90 °F. Express the temperature in absolute. Solution: (B) A small piece of Iron has a temperature T= 300 K. Express the temperature in Fahrenheit. Solution: TC=TK - 273.15 TC=300 - 273.15 =26.850C  1 1 Example (2.2) (A) On a day when the temperature reaches 𝟓𝟎 ℉, what is the temperature in degrees Celsius and in Kelvins. Solution: 1. Celsius (°C) = (°F - 32) x 5/9 2. Kelvin (K) = °C + 273.15 1. Celsius (°C) = (50 - 32) x 5/9 = 18 x 5/9 = 10 °C 2. Kelvin (K) = 10 °C + 273.15 = 283.15 K Thus, the temperature is roughly 10°C and 283.15 K when it hits 50°F. (B) Convert the following temperatures to their values on the Fahrenheit and Kelvin scales: B1/ the sublimation point of dry ice, - 78.5°C Solution: TF = 9/5 ( Tc ) + 32 = 9/5 (- 78.5 )+ 32 = –109° F T = Tc + 273.15 = ( –78.5 + 273.15 ) = 195 K B2/ human body temperature, 37.00°C Solution: TF = 9/5 ( Tc ) + 32 = 9/5 ( 37.0 )+ 32 = 98.6° F T = Tc + 273.15 = ( 37.0 + 273.15 ) = 310 K  2.5 THERMAL EXPANSION Thermal expansion is the tendency of matter to change in length, area, and volume in response to a change in temperature through heat transfer Thermal expansion Linear Area Volume  2.5.1 LINEAR THERMAL EXPANSION 𝑻𝒊 𝑳𝒊 𝒇𝑻 𝑳𝒇 Linear Expansion 𝑳𝒇 − 𝑳𝒊 = 𝜶𝑳𝒊 𝑻𝒇 − 𝑻𝒊 ⟺ 𝚫𝑳 = 𝜶𝑳𝒊𝚫𝐓 𝜶 is the coefficient of linear expansion  2.5.1 LINEAR THERMAL EXPANSION Linear Expansion The strip bends as shown at temperatures above this reference temperature. Below the reference temperature the strip bends the other way. Many thermo- stats operate on this principle, making and breaking an electrical contact as the temperature rises and falls.  Example (2.3) (A) A copper wire 15 m long is cooled from 40 to -9ºC. How much change in length will it experience? Solution: Answer: For Cu   16.5x10 6 (  C) 1 L   Li T  [16.5 x 10 6 (1 / C)](15 m)[40C  (9C)] L  0.012 m  12 mm (B) A segment of steel railroad track has a length of 30.000 m when the temperature is 0.0°C. What is its length when the temperature is 40.0°C? Solution: - length is 30.000 m when the temperature is 0.0°C - Then the length when the temperature is 40.0°C equal 30.0144 m  2.5.2 AREA THERMAL EXPANSION 𝑨𝒇 𝑨𝒊 Area Expansion 𝑨𝒇 − 𝑨𝒊 = 𝟐𝜶𝑨𝒊 𝑻𝒇 − 𝑻𝒊 ⟺ 𝚫𝑨 = 𝟐𝜶𝑨𝒊𝚫𝐓 𝛂 is the coefficient of linear expansion  2.5.3 VOLUME THERMAL EXPANSION 𝑽𝒇 𝑽𝒊 Volume Expansion 𝑽𝒇 − 𝑽𝒊 = 𝜷𝑽𝒊 𝑻𝒇 − 𝑻𝒊 ⟺ 𝚫𝑽 = 𝜷𝑽𝒊𝚫𝐓 𝜷 is the coefficient of Volume expansion 𝜷 = 𝟑𝜶 Table (a) : Coefficients of Linear Expansion Table (b) : Coefficients of Volume Expansion  Example (2.4) On a hot day in Las Vegas, a n oil trucker loaded 37000 L of diesel fuel. He encountered cold weather on the way to Payson, Utah, where the temperature was 23 K lower than in Las Vegas, and where he delivered his entire load. How many litters did he deliver? (𝛽𝑑𝑖𝑒𝑠𝑒𝑙 = 9,5 × 10−4°𝐶−1). Neglect the steel tank expansion. Solution: the temperature decreased, the volume of the fuel did also , as given by (ΔV=βVΔT) Calculations : ΔV=(9.50×10-4)(37000)(-23.0)= –808.45 L Thus, the amount delivered was V=V+ΔV=37000L – 808.45L = 36,191.55L. 19  Example (2.5) The volume of a solid aluminum ball with initial radius 20 cm increases by 347 𝑐𝑚3when the ball is heated. What is the temperature change. 𝛼 = 23𝑋10−6/K. Solution: ΔV=VβΔT , β = 3a V = 4/3 R3π ΔT = ΔV/Vβ ΔT = ΔV / ( 4/3. R3. π. 3. 𝛼) ΔT = ΔV / ( 4. R3. π. 𝛼) ΔT = 347 / ( 4. 203. π. 23x10-6) ΔT = 150.07 Co  2.6 Quantity of Heat and Specific Heat 2.6.1 Quantity of Heat Q Quantity of Heat Q is energy that is transferred between a system and its environment because of a temperature difference between them. Quantity of Heat Q can be measured in joules (J), calories (cal), kilocalories (Cal or kcal), 𝟏 𝒄𝒂𝒍 = 𝟒. 𝟏𝟖𝟔 𝑱 𝟏𝑲𝒄𝒂𝒍 = 𝟏𝟎𝟎𝟎 𝒄𝒂𝒍 Q = mC ΔT Q = heat transferred (in joules) m = mass (in grams) C = specific heat capacity Δ T = change in temperature ( ΔT = Tfinal – Tinitial ) 2 2  2.6.2 Specific Heat Specific Heat : The quantity of heat required to raise the temperature of 1 gram of a substance by one degree Celsius (or Kelvin) The unit is J/g(oC) Q = mC ΔT Q = heat transferred (in joules) m = mass (in grams) C = specific heat capacity Δ T = change in temperature ( ΔT = Tfinal – Tinitial )  Example (2.6) What is the specific heat of silver if the temperature of a 15.4 g sample of silver is increased from 20.0oC to 31.2oC when 40.5 J of heat is added? Solution: Givens: m = 15.4 g , Ti = 20.0 oC , Tf = 31.2oC , Q = 40.5 J Q = mC∆T » 40.5=15.4(C)(31.2-20.0) 40.5=15.4(C)(11.2) » 40.5=172.48(C) C = 0.235 J/g(oC)  Example (2.7) What is the final temp of silver if the temperature of a 5.8 g sample of silver starts out at 30.0oC and 40.5 J of heat is added? The specific heat of silver is 0.235 J/g(oC). Solution: Givens: m = 5.8 g , Ti = 30.0oC , Q = 40.5 J, C = 0.235 , Tf = ??? Q = mC∆T » 40.5=5.8(0.235)(Tf -30.0) 40.5=1.363(Tf-30.0) »40.5=1.363Tf – 40.89 81.39=1.363Tf » Tf = 59.7139 » Tf = 60.oC  Example (2.8) What quantity of heat is required to raise the temperature of 100 mL of water from 45.6C to 52.8C? The specific heat of water is 4.184 J/g(C) and water has a density of 1.00 grams/mL. Solution: Givens: Q = ? , V = 100. mL , Ti = 45.6oC , Tf = 52.8oC , C = 4.184 , dwater=1.00 g/mL Q = mC∆T » Q =100(4.184)(52.8-45.6) » Q =3012.48 Q =3010 J  2.6.3 Heat Capacity (C) The heat capacity (C) of a particular sample is defined as the amount of energy needed to raise the temperature of that sample by 1°C. 𝑸 = 𝑪 ∆𝑻 = 𝑪(𝒕𝒇 − 𝒕𝒊) 𝑪 = 𝒎𝒄 𝑸 = 𝒎𝒄 ∆𝑻 (c is the specific heat)  2.7 Heat transfer mechanisms 2.7.1 Heat Transfer Processes 2.7.1.1 Conduction – transfer of heat from a region of higher temperature to a region of lower temperature by increased kinetic energy moving from molecule to molecule through collisions between molecules. Occurs in solids. 2.7.1.2 Convection – transfer of heat from a region of higher temperature to a region of lower temperature by the flow of higher energy molecules. Occurs in gases and liquids. 2.7.1.3 Radiation – transfer of heat by emission and absorption of radiant energy (energy that can travel through space as electromagnetic radiation, like visible light).  2.7 Heat transfer mechanisms 2.7.2 Thermal Conduction: The rate 𝑃𝐶𝑜𝑛𝑑 at which energy is conducted through a slab A : the area of the slap. L : the distance between the faces. k : is the thermal conductivity of the material.  Example (2.9) A Styrofoam box used to keep drinks cold at a picnic has total wall area (including the lid) of 0.8 m? and wall thickness of 2 cm. It is filled with ice, water, and cans of cola at 0 °C. What is the rate of heat flow into the box if the temperature of the outside wall is 30 °C?. Solution: A = 0.80 m2 , L = 2 cm , Tc = 0°C , TH=30°C , k=0.01 W/m. K R= KA ( Th– Tc / L ) 0.01x0.8 (30-0 / 2x10-3 = 12W Q = H x t = 12 x 1 × 24 x 3600 = 1.04 x 106 J  2.7 Heat transfer mechanisms 2.7.3 Convection occurs when temperature differences cause an energy transfer by motion within a fluid. 2.7.4 Thermal Radiation via the emission of Radiation is an energy transfer electromagnetic energy. The rate 𝑃𝑟𝑎𝑑 is given by 𝑷𝒓𝒂𝒅 = 𝝈𝜺𝑨𝑻𝟒 𝜎 = 5.6704 𝑥 10−8 𝑊/𝑚2𝐾4is the Stefan– Boltzmann constant 𝜀 𝑖𝑠 𝑡ℎ𝑒 𝑒𝑚𝑖𝑠𝑠𝑖𝑣𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑏𝑗𝑒𝑐𝑡’𝑠 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝐴 𝑖𝑠 𝑖𝑡𝑠 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎. 𝑇 𝑖𝑠 𝑖𝑡𝑠 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 (𝑖𝑛 𝑘𝑒𝑙𝑣𝑖𝑛𝑠).  Example (2.10) A thin square steel plate, 10 cm on a side, is heated in a blacksmith's forge to a temperature 800 °C. If the emissivity is 0.6, what is the total rate of radiation of the energy? Where, Stefan's Constant (s) = 5.67 × 10 -8 W/ m2. K4 Solution: T = 800 + 273 = 1073 K, σ = 0.60 P=σ⋅E⋅A⋅T4 , P = ( 5.67×10−8)⋅(0.60)⋅(0.01)⋅(1073.15)4 T4=(1073.15)4≈1.307×1012K4 Now substituting T4 back into the equation: P= ( 5.67×10-8)⋅( 0.60)⋅( 0.01)⋅( 1.307×1012) Calculating this gives: P≈ 445.2W  Example with key answer for solution (1) You place a small piece of ice in your mouth. Eventually, the water all converts from ice at T1 =32 °F to body temperature, T2 =98.6 °F. express these temperatures in both centigrade and absolute. Answer (1) (TC1= 0 0C , TK1= 273.15K) , (TC2= 37 0C , TK2= 310.15K) (2) A hot-water bottle contains 750 g of water at 65 °C. Water has a specific heat of 1.00 cal/g °C. If the water cools to body temperature (37 °C), how many calories of heat could be transferred to sore muscles? Answer (2) (energy = 21,000 calories) (3) What is the specific heat of a metal if 24.8 g absorbs 65.7 cal of energy and the temperature rises from 20.2 C to 24.5 C? Answer (3) c = 0.62 cal/g °C) (4) How many kilojoules are needed to raise the temperature of 325 g of water from 15.0 °C to 77.0 °C? Answer (4) Q = 84.3 kJ  (5) A surveyor uses a steel measuring tape that is exactly 50.000 m long at a temperature of 20°C. The markings on the tape are calibrated for this temperature. What is the length of the tape when the temperature is 35°C? Answer (5) L = 50.009 m (6) Suppose that the temperature of a block of copper is increased from 127°C to 137°C for 50m3 copper. We wish to know what change in volume would be necessary to keep the pressure constant? Answer (6) ΔV = 0.0255 m3 Good luck