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Physics 1. Thermal Physics Chapters 21-23. Links: http://homepage.mac.com/phyzman/phyz/BOP/2-06ADHT/index.html and http://www.colorado.edu/physics/2000/index.pl. What do you think? know?. Why does popcorn pop?
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Physics 1 Thermal Physics Chapters 21-23 Links: http://homepage.mac.com/phyzman/phyz/BOP/2-06ADHT/index.html and http://www.colorado.edu/physics/2000/index.pl
What do you think? know? • Why does popcorn pop? • On a camping trip, your friend tells you that fluffing up a down sleeping bag before you go to bed will keep you warmer than sleeping in the same bag when it is still crushed. Why?
3. Why is it difficult to build a fire with damp wood? 4. Why does steam at 100oC cause more severe burns than liquid water does at 100oC?
5. Until refrigerators were invented, many people stored fruits and vegetables in underground cellars. Why was this more effective than keeping them in the open air? 6. In the past, when a baby had a high fever, the doctor might have suggested gently sponging off the baby with rubbing alcohol. Why would this help?
7. Why does water expand when it freezes? 8. Why, during the final construction of the St. Louis arch, was water sprayed on the previous sections as the last section was put in place?
Objectives • 1. Describe thermal energy and compare it to potential and kinetic energies. • 2. Describe changes in temperatures of two objects reaching thermal equilibrium • 3. Identify various temperature scales, and convert between them
Objectives • 4. Explain heat as energy transferred between substances at different temperatures in one of 3 ways (conduction, convection, radiation) • 5. Relate heat and temperature • 6. Apply principle of energy conservation to calculate changes in potential, kinetic, & internal energy
Objectives • 7. Perform calculations with specific heat capacity • 8. Interpret the various sections of a heating curve
Objectives • 9. Recognize that a system can absorb or release energy as heat in order for work to be done on or by that system • 10. Compute work done during thermodynamic process • 11. Distinguish between isovolumetric, isothermal, and adiabatic thermodynamic processes
Objectives • 12. Illustrate how the first law of thermodynamics is a statement of energy conservation • 13. Calculate heat, work, and change in internal energy using lst law of T-D • 14. Apply 1st law of T-D to describe cyclic processes • 15. Recognize why 2nd law of T-D requires 2 bodies at different temps.
1.Relate temperature to the kinetic energy of atoms and molecules • Temperature scales • In the USA, the Fahrenheit temperature scale is used. Most of the rest of the world uses Celsius, and in science it is often most convenient to use the Kelvin scale. • The Celsius scale is based on the temperatures at which water freezes and boils. 0°C is the freezing point of water, and 100° C is the boiling point. Room temperature is about 20° C, a hot summer day might be 40° C, and a cold winter day would be around -20° C. • To convert between Fahrenheit and Celsius, use these equations:
21.1Temperature Temperature and Kinetic Energy Temperature is related to the random motions of the molecules in a substance. In the simplest case of an ideal gas, temperature is proportional to the average kinetic energy of molecular translational motion.
Convert 72 oF to oC • C = 5/9 (F-32) • C = 5/9 (72-32) = 22oC • Convert -10 oC to oF • F = 9/5 C + 32 • F = 9/5(-10) + 32 = 14oF
Objective 3: Temperature Scales Temperature degree scales comparison
Celsius to Kelvin: T = Tc + 273.15 Problem: 1. The lowest outdoor temperature ever recorded on Earth is -128.6 o F., recorded at Vostok Station, Antarctica, in 1983. What is this temperature on the Celsius and Kelvin scales? Answers: -89.22oC, 183.93 K
Obj. #2 - Hotter temperature means • – more heat present in a substance • – the faster the molecules of the substance move. • Obj #5 – Relate heat and temperature • Heat units: calorie or joule (amount of heat energy present in a substance) • Temperature units: degree (proportional to heat energy present in a substance)
21.2Heat If you touch a hot stove, energy enters your hand from the stove because the stove is warmer than your hand. If you touch ice, energy passes from your hand into the colder ice. The direction of spontaneous energy transfer is always from a warmer to a cooler substance. The energy that transfers from one object to another because of a temperature difference between them is called heat. Heat unit: calories or Joules (4.186 J/cal)
2.Describe changes in temperatures of two objects reaching thermal equilibrium • The temperature of the hotter substance will decrease. The temperature of the colder substance will increase. Each change will stop when the temperatures are the same – thermal equilibrium. In other words, thermal energy travels from hot to cold. • Obj. #4 - Heat energy can be transferred by • Convection (motion of fluid), conduction (touching), or radiation (electromagnetic waves)
Convection Heat transfer in fluids generally takes place via convection. Convection currents are set up in the fluid because the hotter part of the fluid is not as dense as the cooler part, so there is an upward buoyant force on the hotter fluid, making it rise while the cooler, denser, fluid sinks. Conduction When heat is transferred via conduction, the substance itself does not flow; rather, heat is transferred internally, by vibrations of atoms and molecules.
Radiation: energy is transferred in the form of electromagnetic waves.
22.1Conduction • Touch a piece of metal and a piece of wood in your immediate vicinity. Which one feels colder? Which is really colder? • If the materials are in the same vicinity, they should have the same temperature, room temperature. • The metal feels colder because it is a better conductor. • Heat easily moves out of your warmer hand into the cooler metal. • Wood, on the other hand, is a poor conductor. • Little heat moves out of your hand into the wood, so your hand does not sense that it is touching something cooler.
22.1Conduction The good insulating properties of materials such as wool, wood, straw, paper, cork, polystyrene, fur, and feathers are largely due to the air spaces they contain. Birds fluff their feathers to create air spaces for insulation. Snowflakes imprison a lot of air in their crystals and are good insulators. Snow is not a source of heat; it simply prevents any heat from escaping too rapidly.
22.1Conduction Strictly speaking, there is no “cold” that passes through a conductor or an insulator. Only heat is transferred. We don’t insulate a home to keep the cold out; we insulate to keep the heat in. No insulator can totally prevent heat from getting through it. Insulation slows down heat transfer.
22.2Convection Convection occurs in all fluids, liquid or gas. When the fluid is heated, it expands, becomes less dense, and rises. Cooler fluid then moves to the bottom, and the process continues. In this way, convection currents keep a fluid stirred up as it heats.
22.5Absorption of Radiant Energy A blacktop pavement and dark automobile body may remain hotter than their surroundings on a hot day. At nightfall these dark objects cool faster! Sooner or later, all objects in thermal contact come to thermal equilibrium. So a dark object that absorbs radiant energy well emits radiation equally well.
22.3Radiation • Radiant energy is any energy that is transmitted by radiation. • From the longest wavelength to the shortest, this includes: • radio waves, • microwaves, • infrared radiation, • visible light, • ultraviolet radiation, • X-rays, • and gamma rays.
22.3Radiation • Radio waves send signals through the air. • You feel infrared waves as heat. • A visible form of radiant energy is light waves.
22.3Radiation Most of the heat from a fireplace goes up the chimney by convection. The heat that warms us comes to us by radiation.
22.5Absorption of Radiant Energy Good emitters of radiant energy are also good absorbers; poor emitters are poor absorbers.
22.5Absorption of Radiant Energy • Because the sun shines on it, the book absorbs more energy than it radiates. • Its temperature increases. • As the book gets hotter, it radiates more energy. • Eventually it reaches a new thermal equilibrium and it radiates as much energy as it receives. • In the sunshine the book remains at this new higher temperature.
22.6Newton’s Law of Cooling An object hotter than its surroundings eventually cools to match the surrounding temperature. Its rate of cooling is how many degrees its temperature changes per unit of time. The rate of cooling of an object depends on how much hotter the object is than the surroundings.
Specific heat capacity The amount of energy that must be added to raise the temperature of a unit mass of a substance by one temperature unit. • Units: cal/goC • For Water: 4.186 J/goC • For Aluminum: 0.900 J/goC • Which one heats faster?
21.6Specific Heat Capacity think! Which has a higher specific heat capacity—water or sand? Explain.
21.6Specific Heat Capacity think! Which has a higher specific heat capacity—water or sand? Explain. Answer: Water has a greater heat capacity than sand. Water is much slower to warm in the hot sun and slower to cool at night. Sand’s low heat capacity, shown by how quickly it warms in the morning and how quickly it cools at night, affects local climates.
A table similar to this is on page 413 of the Hewitt book. Not shown: Clay (14,000), Olive Oil (19,700), and Steel (448)
Heat Transfer Q = mCΔT = mC (Tf – Ti) • Heat Transfer • Q = mcΔT = mc(Tf – Ti) • Q, quantity of heat in joule • m, mass of substance in g • c, specific heat for water in 4.186 J/goC • t, temperature in Celsius
See table 12-1 on page 318 Find the amount of heat needed to change the temperature of 5.0 g of liquid water from 8.0oC to 100oC. Q = mcDt = 5.0g(4.186 J/goC) (92oC) = 1.9 x 103 J Again, specific heat is the amount of heat necessary to change one g of a substance 1 degree Celsius or Kelvin.
12/3 When you turn on the hot water to wash dishes, the water pipes have to heat up. How much heat is absorbed by a copper water pipe with a mass of 2.3 kg when its temperature is raised from 20.0oC to 80.0oC? • Q = mcDt • Q = (2300g)(0.386J/goC)(60.0oC) • Q = 53268 J or 5.3x104 J
Q = mc Dt Dt = Q mc Dt = 836,000 J goC 20,000g (4.186 J) Dt = 9.98oC
Specific Heat Capacities Which one is greatest that you use everyday? OR J/kgoC
Emily is testing her baby’s bath water and finds that it is too cold so she adds some hot water. If she adds 2.00 kg of water at 80.0oC to 20.0kg of bath water at 27.0 oC, what is the final temperature of the bath water? Q lost = Q gain 2.00kg(1cal)(80oC-tf) = 20.0kg(1cal)(tf-27oC) goC goC 160 – 2tf = 20 tf – 540 -22 tf = -700 tf = 31.8oC
Latent Heat is energy transferred during phase changes Crystalline materials change phase -- melt and freeze or vaporize and condense -- at a single, fixed temperature. Energy is required for a change of phase of a substance. It is called latent heat because there is no change or difference in temperature. Latent heat of fusion Lf describes the heat necessary to melt (or freeze) a unit mass of a substance.Q = m Lf Latent heat of vaporization Lv describes the heat necessary to vaporize (or condense) a unit mass of a substance. Q = m Lv
Formulas • Temperature change use: Q = mc D t • Melting or freezing: Q = m Lf • Evaporation or Condensation: Q = mLv • Lf is latent heat of fusion, 80 cal/g • Lv is latent heat of vaporization, 540 cal/g • These values are for water.
How much heat must be gained by 0.100 kg of ice at 0 oC in order for all of it to melt? Q = m Lf Q = 100g (80cal) g Q = 8000 or 8.00 x 103 g
Q = mcDt + mLf Specific Heat of Ice = 0.49 cal g oC Q = 100g(0.49cal)(20.0 oC) + 100g(80cal) g oC g Q = 8980 = 8.98 x 103 cal
Problem • A jar of tea is placed in sunlight until it reaches an equilibrium temperature of 32oC. In an attempt to cool the liquid to 0oC, which has a mass of l80 g, how much ice at 0oC is needed? Assume the specific heat capacity of the tea to be that of pure liquid water.
m tea = 180g m ice = ? which is the mass of the water that has melted • c tea = c water = 4186 J/kgoC • H f = 3.33 x 105 J/kg and t final = 32oC • Q lost = Q gained tea loses and water gains only melting the ice • (mcDt)tea = (mHf)ice • m ice= (mcDt)tea • Hfice • m ice = (.180kg)(4186J/kgoC)(32oC) • 3.33x105 J/kg • = 7.2 x 10-2kg
1st Law of Thermodynamics • U = Q - W The change in thermal energy of an object is equal to the heat added to the object minus the work done by the object. See Fig 12-11 on Page 326