VIEWS: 0 PAGES: 15 POSTED ON: 7/24/2012 Public Domain
ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3 CHAPTER 3: TEMPERATURE AND HEAT 3.1 Temperature and Thermometers Temperature is a physical property of all bodies and is a relative measure or indication of hotness and coldness of objects. Thermometers are devices used to measure the temperature of a system. Three common scales of a temperature measurement: 1. Celsius 2. Fahrenheit 3. Kelvin Table shows a comparison of the Celsius, Fahrenheit and Kelvin temperature scale Celsius Fahrenheit Kelvin scale scale scale Steam point 100 C 212 F 373.15 K Ice point 0 C 32 F 273.15 K On the Celsius temperature scale, there are 100 equal divisions between the ice point (0C) and the steam point (100C). On the Fahrenheit temperature scale, there are 180 equal divisions between the ice point (32F) and the steam point (212F). The relationship between Celsius, TC and Fahrenheit, TF temperature scale is 9 TF TC 32 5 3.2 The Kelvin Temperature Scale For scientific work, the Kelvin temperature scale is the scale of choice. As shown in the table above, the ice point (0C) occurs at 273.15K on the Kelvin scale. Thus, the Kelvin temperature, T and the Celsius temperature are related by, T TC 273 .15 Thus, the Kelvin temperature, T and the Fahrenheit temperature are related by, TF 9 T 273.15 32.0 5 The number 273.15 in the equation is an experimental result, obtained in studies that utilize a gas-based thermometer. Gas thermometer is a standard instrument because its measurement is independent from any type of gas. One type of gas thermometer is the constant-volume gas thermometer. The physical property used in a constant-volume gas thermometer is the pressure variation with temperature of a fixed volume of gas. A constant-volume gas thermometer consists of a gas-filled bulb to which a pressure gauge is attached, as in figure below. DEPARTMENT OF ENGINEERING 41 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3 Figure shows a constant-volume gas thermometer The basic idea of this thermometer is that the volume of gas is held constant by raising or lowering the right column of the U tube manometer in order to keep the mercury level in the left column at the same reference level. With the mercury level that can be adjusted, the gas occupies a constant volume and its pressure is related to P2= P1 + gh where is the density of mercury P1 = 0 Pa (evacuated) P2 = Pgas As the temperature changes, the pressure changes. The temperature increases when a gas confined to a fixed volume is heated and decreases when the gas is cooled. If the results are plotted on a pressure versus temperature graph, a straight line is obtained as in figure below. Figure shows a plot of absolute pressure versus temperature for a low density gas at constant volume If the straight line is extended to lower temperatures where the pressure is zero, the line crosses the temperature axis at -273.15C. This is the location of absolute zero. DEPARTMENT OF ENGINEERING 42 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3 EXERCISE 1 (a) “Room temperature” is often taken to be 68°F. What is this on the Celsius scale? (b) The temperature of the filament in a lightbulb is about 1800°C. What is this on the Fahrenheit scale? SOLUTION (a) T C T F 32 68 32 20 C o 5 9 o 5 9 o (b) T F T C 32 1800 32 3300 F o 9 5 o 9 5 o EXERCISE 2 What’s your normal body temperature? It may not be 98.6F, the oft-quoted average that was determined in the nineteenth century. What is the difference between these averages expressed in Celsius degrees? SOLUTION The difference between these two averages, expressed in Fahrenheit degrees, is 98.6 F 98.2 F 0.4 F 9 Since 1 C° is equal to F°, we can make the following conversion 5 1 C (0.4 F) 0.2 C (9/5) F 3.2 Thermal Expansion When the temperature of a body increases, the average distance between the atoms increases and the volume of the body also increases. The phenomenon is known as thermal expansion. Thermal expansion is a physical phenomenon in which the increment in temperature can cause substances to expand. Linear Expansion Figure below illustrates the linear expansion of a rod whose length is Lo when the temperature is To. Figure shows when the temperature of a rod is raised by T, the length of the rod increases by L For linear expansion, an object of length Lo experiences a change L in length when the temperature changes by T L L0 T where is the coefficient of linear expansion OR L – Lo = Lo T OR L L0 (1 T ) (in terms of final length) DEPARTMENT OF ENGINEERING 43 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3 Thermal stress For an object held rigidly in place, a thermal stress can occur when the object attempt to expand or contract. Thus, the force / stress needed to prevent a solid object from expanding must be strong enough to counteract any change in length due to a change in temperature. F L L T Stress Y Y 0 YT A L0 L0 EXERCISE 3 An aluminum bar has the desired length when at 15°C. How much stress is required to keep it at this length if the temperature increases to 35°C? SOLUTION The thermal stress must compensate for the thermal expansion. Y is Young’s modulus for the aluminum. Stress F A Y T 25 106 Co 70 10 9 N m2 35 C 15 C 3.5 10 o o 7 N m2 EXERCISE 4 A steel beam is used in the roadbed of a bridge. The beam is mounted between two concrete supports when the temperature is 23C, with no room provided for thermal expansion. What compressional stress must the concrete supports apply to each end of the beam, if they are to keep the beam from expanding when the temperature rises to 42C? SOLUTION F L Lo T Y Y Y T Stress = A Lo Lo YT (2 x1011 )(12 x106 )(19) 4.6 x107 N / m2 The Bimetallic Strip Each substance has its own characteristic coefficient of expansion. For example, brass rod expands more than the steel rod because brass has a larger coefficient of expansion than steel. A simple device that uses this principle is a bimetallic strip. A bimetallic strip is made from two thin strips of metal that have different coefficients of linear expansion as shown in figure below Figure shows: (a) A bimetallic strip and how it behaves when (b) heated and (c) cooled DEPARTMENT OF ENGINEERING 44 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3 APPLICATION Bimetallic strips are frequently used as adjustable automatic switches in electrical appliances. Refer figure below. In part (a), the electricity passes through the heating coil that heat the water. The electricity can flow because bimetallic strip touches the contact on the strength adjustment knob, thus providing a continuous path of the electricity. In part (b), when the bimetallic strip gets hot enough to bend away, the contacts separate and the electricity stops because it is no longer has a continuous path along which to flow and the brewing cycle is shut off. NOTE: Expansion of holes-Let’s say we have a hole in piece of solid material. When heated, the material will expand, but what about the hole??? The hole also expands at the same rate and at the same direction (outwards)-refer to the diagram. (a) Unheated (b) Heated – dotted line represents after heated Volume Thermal Expansion For volume expansion, the change V in the volume of an object of volume Vo is given by V 3V0 T or V V0 T where is the coefficient of volume expansion and is equal to 3 = 3 (liquids expands more than solids) DEPARTMENT OF ENGINEERING 45 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3 EXERCISE 5 A rod made from a particular alloy is heated from 25 0C to the boiling point of water. Its length increases by 8.47 10-4 m. The rod is then cooled from 250C to the freezing point of water. By how much does the rod shrink? SOLUTION Assuming that the rod expands linearly with heat, we first calculate the quantity L/ T using the data given in the problem. L 8.47 10–4 m –5 1.13 10 m/C T 100.0 C – 25.0 C Therefore, when the rod is cooled from 25.0 °C, it will shrink by L (1.13 10 –5 m/C) T (1.13 10 –5 m/C) (0.00 C – 25.0 C) –2.82 10 –4 m EXERCISE 6 A cylinder of diameter 1.00000cm at 30C is to be slid into a hole in a steel plate. The hole has a diameter of 0.99970cm at 30C. To what temperature must the plate be heated? For steel, = 1.2 x 10-5C-1 SOLUTION L= LoT L T= Lo L= 1 10-2 – 9.997 10-3 = 3 10-6m = 1.2 10-5C-1 Lo= 9.997 10-3m 3 106 T= = 25C 1.2 105 (9.997 103 ) The temperature of the plate must be = 30 + 25 = 55C EXERCISE 7 The brass bar and the aluminum bar in the drawing are attached to an immovable wall. At 28C the air gap between the rods is 1.3 x 10 -3m. At what temperature will the gap be closed? SOLUTION L = L0T gives for the expansion of the aluminum LA = ALAT and for the expansion of the brass LB = BLBT Taking the coefficients of thermal expansion for aluminum and brass from table adding Equations (1) and (2), and solving for T give DEPARTMENT OF ENGINEERING 46 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3 L A L B 1.3 10 3 m T 21C A L A BL B 23 10 6 C 1 1.0 m 19 10 6 C 1 2.0 m The desired temperature is then T = 28 °C + 21 C° = 49 C EXERCISE 8 A lead object and a quartz object each have the same initial volume. The volume of each increases by the same amount because the temperature increases. If the temperature of the lead object increases by 4.0C., by how much does the temperature of the quartz increase? SOLUTION Recognizing that the lead and quartz objects experience the same change in volume and expressing that change with Equation 12.3, we have LeadV0TLead QuartzV0TQuartz VLead VQuartz In this result V0 is the initial volume of each object. Solving for TQuartz and taking values for the coefficients of volume expansion for lead and quartz from Table gives 1 Lead TLead 87 10 C 4.0 C 6 TQuartz 1 230 C Quartz 1.5 106 C EXERCISE 9 A glass is filled “to the mark” with 50.00cm3 of mercury at 18C. If the flask and its contents are heated to 38C, how much mercury will be above the mark? glass = 8.5 x 10-6C-1 and βmercury = 182 x 10-6C-1 SOLUTION Both of glass and mercury will expand Vmercury= βmercuryVoT = 182 10-6 (5 10-5) (38 – 18) = 1.82 10-7m3 Vglass= βglassVoT = 3glassVoT = (3 8.5 10-6) (5 10—5) (38 -18) = 2.55 10-8m3 Volume of mercury above mark = Vmercury - Vglass = 1.82 10-7 – 2.55 10-8 = 1.565 10-7m3 EXERCISE 10 Suppose that the steel gas tank in your car is completely filled when the temperature is 170C. How many gallons will spill out of the twenty gallon tank when the temperature rises to 35 0C? DEPARTMENT OF ENGINEERING 47 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3 SOLUTION Both the gasoline and the tank expand as the temperature increases. The coefficients of volumetric expansion g and s for gasoline and steel are available in Table. The volume expansion of the gasoline is Vg gV0 T 950 10 –6 C 1 (20.0 gal)(18 C) 0.34 gal While the volume of the steel tank expands by an amount V V T 10 C s s 0 36 –6 1 (20.0 gal)(18 C) 0.013 gal Vg Vs 0.33 gal The amount of gasoline which spills out is 3.5 Heat and Internal Energy Heat is defined as the energy that flows from a higher temperature object to a lower temperature object because of the difference in temperatures. (Refer figure) The SI unit is Joule , J Heat is sometimes measured with a unit called the kilocalorie (kcal) The conversion factor between kilocalories and joule is known as the mechanical equivalent of heat: 1 kcal = 4186 joule or 1 cal = 4.186 joule Internal energy of a substance is the sum of the kinetic energy (due to the random motion of the molecules), the potential energy (due to forces that act between the atoms of the molecule and between molecules) and other kinds of energy that the molecules of the substance have. Figure shows how heat as energy is transit from hot to cold. (a) Heat flows from the hotter coffee to the colder hand. (b) Heat flows from the warmer hand to the colder glass of ice water. DEPARTMENT OF ENGINEERING 48 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3 3.6 Heat and Temperature Change: Specific Heat Capacity Solids and Liquids Greater amounts of heat are needed to raise the temperature of solids or liquids to higher values Similar concepts apply when the temperature is lowered, except that heat must be removed. The heat, Q that must be applied or removed to change the temperature of a substance of mass, m by an amount T is Q mcT where c is the specific heat capacity of the substance. Unit for specific heat capacity: J / (kg C) Calorimetry To determine the specific heat of a solid or liquid, we conduct the following steps: i. raise the temperature of the substance to some value ii. place it in a vessel containing cold water of known mass and temperature iii. measure the temperature of the combination after the equilibrium is reached. Vessels having this property are called calorimeters and analysis performed using such vessels is called calorimetry. The principle of conservation of energy for perfectly insulated container requires that heat lost by warmer materials equals heat gained by cooler materials. (That is if there is no heat loss to the external surroundings) From the principle of conservation of energy for the isolated system, heat loss by the hotter object = heat gained by the cooler object. Q 0 Q gained Qlost [mc (T )] gained [mc (T )] lost **When calculating heat contributions, always write any temperature changes as the higher minus the lower temperature. EXERCISE 11 A 0.4kg iron horseshoe that is initially at 500C is dropped into a bucket containing 20kg of water at 22C. What is the final equilibrium temperature? Neglect any energy transfer to or from the surroundings SOLUTION - Qhot = Qcold (mCT)iron = (mCT)water (0.4) (452) (500 – Tf) = (20) (4186) (Tf – 22) (90400 – 180.8Tf) = (837201Tf – 1341840) Tf= 39.3C DEPARTMENT OF ENGINEERING 49 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3 EXERCISE 12 An aluminum cup contains 225g of water and a 40g copper stirrer, all at 27C. A 400g sample of silver at an initial temperature of 87C is placed in the water. The stirrer is used to stir the mixture gently until it reaches its final equilibrium temperature of 32. Calculate the mass of the aluminum cup. SOLUTION The total energy absorbed by the cup, stirrer and water equals the energy given up by the silver sample. Thus, (mcup cal + mstirrer ccopper + mwater cwater) (Twater) = (mcT)silver Tsilver mcup = [ ( msilver csilver - mstirrerccopper - mwater ) ] Twater 1 87 32 mcup = [ (0.4) (235) - 0.04 (387) – 0.225 9 10 32 27 2 mcup = 0.085kg EXERCISE 13 An engineer wishes to determine the specific heat of a new alloy. A 0.15kg sample of the alloy is heated to 540C. It is then quickly placed in 400g of water at 10.0C, which is contained in a 200g aluminum calorimeter cup. The final temperature of the mixture is 30.5C. Calculate the specific heat capacity of the alloy. (The specific heats of water and aluminum are 4186J/kg C and 900J/kgC) SOLUTION (heat lost by sample) = (heat gained by water) + (heat gained by calorimeter cup) ms cs Ts mwcw Tw mcal ccal Tcal (0.15)cs (540 30.5) (0.4)(4186)(30.5 10) (0.2)(900)(30.5 10) 76.4cs 34300 3700 cs 500 J / kg C EXERCISE 14 A thermometer has a mass of 31.0 g, a specific heat capacity, c = 815 J/kgC and a temperature of 12.0C. It is immersed in 119 g of water and the final temperature of the water and thermometer is 41.5C. What was the temperature of the water before the insertion of the thermometer? SOLUTION Since there is no heat lost or gained by the system, the heat lost by the water in cooling down must be equal to the heat gained by the thermometer in warming up. The heat Q lost or gained by a substance is given by Equation as Q = cmT, where c is the specific heat capacity, m is the mass, and T is the change in temperature. Thus, we have that cH O mH O TH O ctherm mtherm Ttherm 2 2 2 Heat lost by water Heat gained by thermometer We can use this equation to find the temperature of the water before the insertion of the thermometer. Solving the equation above for TH 2 O and using the value of cH O from Table we 2 have DEPARTMENT OF ENGINEERING 50 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3 ctherm mtherm Ttherm TH O 2 cH O mH O 2 2 815 J/ kg C 31.0 g 41.5 C 12.0 C 1.50 C 4186 J/ kg C 119 g The temperature of the water before the insertion of the thermometer was T 41.5 C 1.50 C = 43.0 C 3.7 Heat and Phase Change: Latent Heat In some situations, the transfer of energy does not result in a change in temperature. This occurred when the substance change from one form to another, which referred to as a phase change. Figure below summarizes the three familiar phases of matter (solid, liquid and gas) and the phase changes that can occur between any two of them melting evaporating solid liquid gas freezing condensing subliming solid gas condensing The heat, Q that must be supplied or removed to change the phase of a mass, m of a substance is Q mL where L is the latent heat of the substance. (Unit: J/kg)SI unit of L=J / kg Heat of fusion: the heat required to change 1.0 kg of substance from the solid to the liquid state. Heat of vaporization: the heat required to change a substance from the liquid to the vapor phase. Figure below displays a graph that indicates what typically happens when heat is added to a material that changes phase. DEPARTMENT OF ENGINEERING 51 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3 The graph shows the way temperature of water changes as heat is added, starting with ice at -40C. The pressure is atmospheric pressure. a) From -40C to 0C: no phase change occurs. The temperature of the ice changes as heat is added. Amount of energy added is Q = mciceT. b) At 0C: no temperature change occurs. the ice-water mixture remains at this temperature even though energy is being added until all the ice melts (phase change occurs) to become water at 0C. The energy required to melt is Q = mLf c) From 0C to 100C: no phase change occurs. The energy added to the water is used to increase its temperature. The amount of energy necessary to increase the temperature from 0C to 100C is Q = mcwaterT. d) At 100C: another phase change occurs as the water changes to steam. The water-steam mixture remains at constant temperature, 100C even though energy is being added until all the liquid has been converted to steam. The energy required to convert water to steam at 100C is Q = mLv e) Above 100C: during this portion of the curve, no phase change occurs, so all the energy added is used to increase the temperature of the steam. The energy that must be added to raise the temperature of the steam is Q = mcsteam T. EXERCISE 15 Suppose that the amount of heat removed when 3.0 kg of water freezes at 00C were removed from ethyl alcohol at its freezing/melting point of -1140C. How many kilograms of ethyl alcohol would freeze? SOLUTION We want the same amount of heat removed from the water as from the ethyl alcohol, i.e., Qwater = Qalcohol or (mLf)water = (mLf)alcohol DEPARTMENT OF ENGINEERING 52 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3 Lf water 33.5 10 4 J/kg m alcohol m water 3.0 kg 9.3 kg L f alcohol 10.8 10 4 J/kg EXERCISE 16 A woman finds the front windshield of her car covered with ice at -120C. The ice has a thickness of 4.5 x 10-4 m, and the windshield has an area of 1.25 m 2. The density of ice is 917 kg/m3. How much heat is required to melt the ice? SOLUTION The heat required is Q = mLf + cmT, where m = V. Thus, Q = VLf + cVT –4 4 3 Q =(917)(4.50 10 )(1.25){33.5 10 +[2.00 10 ](12.0} 1.85 10 5 J EXERCISE 17 Find the mass of water that vaporizes when 2.1 kg of mercury at 205 C is added to 0.11 kg of water at 80.0C SOLUTION From the conservation of energy, the heat lost by the mercury is equal to the heat gained by the water. As the mercury loses heat, its temperature decreases; as the water gains heat, its temperature rises to its boiling point. Any remaining heat gained by the water will then be used to vaporize the water. According to Equation, the heat lost by the mercury is Qmercury (cmT )mercury . The heat required to vaporize the water is, from Equation, Qvap (m vap L v )water . Thus, the total amount of heat gained by the water is Qwater (cmT )water (mvap Lv ) water . Qlost by Qgained by mercury water (cmT ) mercury (cmT ) water ( mvap Lv ) water where Tmercury (205 C 100.0 C) and T water (100.0 C 80.0 C) The specific heats of mercury and water are given in Table, and the latent heat of vaporization of water is given in Table. Solving for the mass of the water that vaporizes gives T c m e r cur y m e r cur y m e r cur y c wa te rmwa te rTwa te r m m vap ( Lv ) wa te r [139 J/(kg C)](2.10 kg)(105 C) [4186 J/(kg C)](0.110 kg)(20.0 C ) 22.6 10 5 J/kg 9.49 10 –3 kg EXERCISE 18 A cube of ice is taken from the freezer at 8.5º C and placed in a 95-g aluminum calorimeter filled with 310 g of water at room temperature of 20.0°C. The final situation is observed to be all water at 17.0°C. What was the mass of the ice cube? DEPARTMENT OF ENGINEERING 53 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3 SOLUTION The heat lost by the aluminum and 310 g of liquid water must be equal to the heat gained by the ice in warming in the solid state, melting, and warming in the liquid state. mAl cAl Ti Al Teq mH2 O cH2 O Ti H2 O Teq mice cice Tmelt Ti ice Lfusion cH2 O Teq Tmelt 0.095 kg 900 J kg Cº 3.0 Cº 0.31 kg 4186 J kg Cº 3.0 Cº mice 9.90 10 3 kg 2100 J kg Cº 8.5Cº 3.3 10 5 J kg 4186 J kg Cº 17 Cº EXERCISE 19 A 0.2 kg piece of aluminum that has a temperature of -155C is added to 1.5 kg of water that has a temperature of 3.0C. At equilibrium the temperature is 0.0C. Ignoring the container and assuming that the heat exchanged with the surroundings is negligible, determine the mass of water that has been frozen into ice. SOLUTION Using the energy-conservation principle cAluminum mAluminum TAluminum cWater mWater TWater mIce Lf, Water Heat gained by aluminum Heat lost by water Heat lost by water that freezes Solving for mIce, taking values for the specific heat capacities from Table, and taking the latent heat for water from Table , we find that cAluminum mAluminum TAluminum cWater mWater TWater mIce Lf, Water 9.00 102 J/ kg C 0.200 kg 0.0 C 155 C 33.5 10 J/kg 4 4186J/ kg C 1.5 kg 3.0 C 0.0 C 0.027 kg 33.5 104 J/kg EXERCISE 20 An unknown material has a normal melting/freezing point of -25C and the liquid phase has specific heat capacity of 160 J/kgC. One –tenth of a kilogram of the solid at -25C is put into a 0.15kg aluminum calorimeter cup that contains 0.1kg of glycerin. The temperature of the cup and the glycerin is initially 27C. All the unknown material melts, and the final temperature at equilibrium is 20.0C. The calorimeter neither losses energy nor gains energy from the external environment. What is the latent heat of fusion of the unknown material? SOLUTION The system is comprised of the unknown material, the glycerin, and the aluminum calorimeter. From the principle of energy conservation, the heat gained by the unknown material is equal to the heat lost by the glycerin and the calorimeter. The heat gained by the unknown material is used to melt the material and then raise its temperature from the initial value of –25.0 °C to the final equilibrium temperature of Teq 20.0 C . DEPARTMENT OF ENGINEERING 54 ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3 Qgained by Qlost by Qlost by unknow n gly cerine calorim eter m u Lf c u mu Tu c gl m gl Tgl c alm al Tal Taking values for the specific heat capacities of glycerin and aluminum from Table, we have (0.10 kg)Lf [160 J/(kg C)](0.10 kg)(45.0 C) [2410 J/(kg C)](0.100 kg)(7.0 C ) [9.0 10 2 J/(kg C)](0.150 kg)(7.0 C ) Lf 1.9 10 4 J/kg Solving for Lf yields, DEPARTMENT OF ENGINEERING 55