Topic 3: Thermal Physics Sub Topic 3.1: Thermal Concepts § Molecular theory of solids, liquids and gases § Temperature and absolute temperature § Internal energy § Specific heat capacity § The topic of thermal physics is a good example of the use of international systems of measurement that allow scientists to collaborate effectively § Phase change § Specific latent heat § Describing temperature change in terms of internal energy § Using Kelvin and Celsius temperature scales and converting between them § Applying the calorimetric techniques of specific heat capacity or specific latent heat experimentally § Describing phase change in terms of molecular behaviour § Sketching and interpreting phase change graphs § Calculating energy changes involving specific heat capacity and specific latent heat of fusion and vaporization “Thermal physics deftly demonstrates the links between the macroscopic measurements essential to many scientific models with the microscopic properties that underlie these models.” The molecular theory of solids, liquids, and gases Solids The molecules (atoms, or ions) are held closely together by strong forces, usually in regular patterns. • They can vibrate about their mean positions • Have fixed shape and volume The molecular theory of solids, liquids, and gases Liquids • Molecules still vibrate, but some molecules have enough kinetic energy to overcome the forces, allowing them to move a little • Little to no regularity in arrangement • Constantly changing arrangement • Fixed volume, but variable shape The molecular theory of solids, liquids, and gases Gases • Much further apart than solids & gases • Weak forces betwixt (usually negligible except collisions) • Molecules move independently in different directions/speeds • Constantly changing • No fixed shape or volume Temperature and Absolute Temperature Temperature: Scalar Quantity. Measured in degrees Celsius (oC) or Kelvin (K) K vs. oC Celsius (sometimes called Centigrade) • Based off of when pure water freezes (0o C) and boils (100o C) at normal atmospheric pressure • Values are chosen, not discovered. • 10o C is not “twice as hot” as 5o C. Kelvin (Absolute temperature) • Based off Celsius scale • Zero reset to Absolute Zero • Where molecular motion stops at -273o C Converting between Celsius and Kelvin: Temperature (K) = temperature (oC) + 273 • In data booklet under unit conversions, NOT with thermal equations • Because increments of both scales are equal, β³K = β³oC • When using β³T, you can use either. When using only one temperature, use K. • *There is no oK. Only K. Don’t write oK. Common Temperatures in both Celsius and Kelvin Internal Energy Thermal Energy: • Energy transferred from hotter to colder as a result of temperature difference • Sometimes called “heat” • “Heat” can also be used wrongly to describe the heat inside an object. This is actually called internal energy. Internal Energy: • The total of the intermolecular potential and random kinetic energy in a substance. • Kinetic can be vibrational, translational, or rotational. • In gas, potential is negligible, making the energy almost entirely kinetic. Internal energy visualized • All substances are composed of individual molecules that are in vibration. • As we heat up a substance its vibrations become more energetic. This is an increase in the kinetic energy of the molecules. • Simultaneously, as heat energy is being added the molecules are also moving farther apart. This is an increase in the potential energy of the substance. • The two energies together are called the internal energy of the substance. Thus EINT = EK + EP. • When thermal energy (heat) is added to a substance it is stored as internal energy. More on Internal Energy EINT = EK + EP. total internal energy Total Internal Energy Potential Energy (due to inter-molecular forces) Think of Hooke’s law F = -kx and phase change Kinetic Energy (due to vibration and translation The thermometer measures internal kinetic energy Thermal equilibrium and heat transfer Thermal Equilibrium • When there is no net flow of thermal energy between 2 or more objects. • Thermal energy will flow from hotter (emits energy) to colder (absorbs energy) until both temperatures are the same. Specific Heat Capacity Specific Heat • The energy transferred to 1 kg of the substance causing its temperature to increase by 1 K. Capacity (c) -1 -1 • Unit: J kg K • π= # $β& or more commonly: π = ππβπ • Q: amount of energy supplied to mass (J) • m: mass (kg) • βT: Change in temperature (K) • (*This is the same as βoC) • c: Specific heat capacity (J kg-1K-1) In other words… Objects with a higher heat capacity heat up (or cool down) slowly compared to an equal mass of substance with a lower specific heat capacity. An Iron Kiln: To watch a guy use primitive technology to create an iron kiln from scratch, check out this link. Practice Problem: A piece of iron of mass 0.133 kg is placed in a kiln until it reaches the temperature π of the kiln. The iron is then quickly transferred to 0.467 kg of water held in a thermally insulated container. The water is stirred until it reaches a steady temperature. The following data are available. Specific heat capacity of iron = 450 J kg-1 K-1 Specific heat capacity of water = 4.2 x 103 J kg-1 K-1 Initial temperature of the water = 16oC Final temperature of the water = 45oC The specific heat capacity of the container and the insulation is negligible. a) State an expression, in terms of π and the above data, for the energy transfer of the iron in cooling from the temperature of the kiln to the final temperature of the water. Energy Transfer: Q Mass of iron: 0.133 kg Initial Temp of Iron: π Final temp of water (and iron): 45oC Specific Heat Capacity of Iron: 450 J kg-1 K-1 π = ππβπ π = 0.133 × 450 × 45 − π πΈ = ππ ×(ππ − π½) b) Calculate the increase in internal energy of the water as the iron cools in the water. Increase in Internal Energy: Q Mass of water: 0.476 kg Initial Temp of Water: 16 oC Final temp of water: 45 oC Specific Heat Capacity of Iron: 4.2 x 103 J kg-1 K-1 π = ππβπ π = 0.476 × 4200 × 45 − 16 πΈ = π. π×πππ J (using 2 s.f.) c) Use your answers from b) and c) to determine π *Energy absorbed by the water (Q in part b) is equal to the energy transferred by the iron (Q in part a). The negative indicates that energy was emitted by the iron. π = 60 ×(45 − π) −5.8×10D = 60×(45 − π) −5.8×10D = 2700 − 60π −6.1×10D = −60π 1010 = π π = 1010oC to 3 s.f. EXAMPLE: Air has a density of about r = 1.2 kg m-3. How much heat, in joules, is needed to raise the temperature of the air in a 3.0 m by 4.0 m by 5.0 m room by 5°C? • • • From the previous table we see that c = 1000 J kg-1 K-1 . The change in temperature is given: DT = 5°C. We get the mass from r = m / V or m = r rV = (1.2)(3.0 x 4.0 x 5.0) = 72 kg. Q = mcDT = (72)(1000)(5) = 360,000 J or 360 kJ. Let’s try another example: After a long day of yelling at kids, Mr. Christofi needs a (boiling) hot cup of tea to relax during an afterschool meeting, which starts in 10 minutes. The teachers lounge kettle has a power rating of 500 W and Mr. Christofi’s tea cup holds 500g of water (22oC starting temperature). If it takes 1 minute to walk to the teachers lounge, and 2 minutes to walk to the meeting (slowly with a hot tea!), will Mr. Christofi be late to the meeting? P= 500 W m = .500 kg Ti = 22oC Tf = 100oC c = 4180 J kg-1 K-1 t=? π = ππ‘ π = ππβπ ππ‘ = ππβπ 500π‘ = .500(4180)(100 − 22) π‘ = 326.04 π π‘stsuv 326.04 π = +1+3= 60π 8.43 min Phase Changes Phase Changes PHASE CHANGE PROCESS EXAMPLE solid ¾® liquid melting ice to water liquid ¾® solid freezing water to ice liquid ¾® gas gas ¾® liquid boiling condensing water to steam steam to droplets solid ¾® gas gas ¾® solid sublimation deposition frost evaporation frost What is temperature? Temperature Measure of the average kinetic energy of the particles in a substance. As temperature is increased, particles move faster, which increases the kinetic energy. This causes the substance to expand. Thermometers Expansion causes the liquid in the bulb to expand, rising up into the tube. A Thermostat uses 2 strips of metal bound together. As the temperature rises, the metals expand at different rates, causing the strips to bend. Specific Latent Heat Specific Latent Heat (L) • The energy required to achieve a change of phase • “Latent” means “hidden” Q = mL unit: J kg-1 Q = Thermal Energy (J) m = Mass (kg) L = Specific Latent Heat (J kg-1) Specific Latent Heat Specific Latent Heat of Fusion (Lf) (melting) • Energy required to change the phase of 1 kg of substance from solid to liquid (without DT) Specific Latent Heat of vaporization (Lv) (boiling) • Energy required to change the phase of 1 kg of substance from liquid to gas (without DT) Phase Change Graphs Latent heat of vaporization Calculating Thermal Energy Temperature Change: Use Q = mcDT Latent heat of fusion Phase Change: Use Q = mL Assumptions and Info about the graph: • Assumes energy is supplied by constant power source. • “Energy” could be replaced with time axis and it would produce the same graph (Q=Pt) Additional Information: • The graph remains flat until ALL of the substance has been converted (to solid, liquid, or gas) • Notice that the flat part of the graph has no change in temperature, therefore, the Latent Heat equation also does not contain a temperature variable. • One substance will have different Specific Latent Heat values (L) for fusion and vaporization. Make sure you use the correct one. Example Problem: How much energy is needed in order to change a 3 kg block of ice (starting temperature -40oC into 3 kg of water vapor at 160oC? 2014 Physics HL Paper 2 Question (7 marks) This question is about energy. a) At its melting temperature, molten zinc is poured into an iron mould. The molten zinc becomes a solid without changing temperature.(a) Outline why a given mass of molten zinc has a greater internal energy than the same mass of solid zinc at the same temperature. (3 Marks) b) The zinc is allowed to cool in the mould. The temperature of the iron mould was 20°C before the molten zinc, at its melting temperature, was poured into it. The final temperature of the iron mould and the solidified zinc is 89°C. The following data are available. Mass of iron mould =12kg Mass of zinc =1.5kg Specific heat capacity of iron =440Jkg–1K–1 Specific latent heat of fusion of zinc =113 kJ kg–1 Melting temperature of zinc =420°C Using the data, determine the specific heat capacity of zinc. (4 marks) Essential Idea “Thermal physics deftly demonstrates the links between the macroscopic measurements essential to many scientific models with the microscopic properties that underlie these models.” At a macroscopic level, we experience temperature; we heat things up, and observe ice melting. We can measure the pressure of a gas and discover how it relates to its temperature. But, to explain all this, we need to invent a microscopic world, where unseen entities interact according to their own rules; where entities give rise to our everyday experiences through a series of correspondencies.
