Lecture 34 Heat and phase changes. Heat We saw that energy is transferred due to a temperature difference. There is no work involved here. This transferred energy is called heat. Units: SI: J (Joules) cal (calorie) 1 cal = heat required to raise the temperature of 1 g of water from 14.5°C to 15.5°C Specific heat How much heat is needed to change by ?T the temperature of a mass m of material X? =? X Q mc T = X specific heatc water: c = 1 cal//(g °C) = 4186 J/(kg K) iron: c = 470 J/(kg K) c X does have some temperature dependence, but very small (ie, negligible for 221) Definition of calorie! Water has a very high specific heat. It?s ?hard? to increase the temperature of water. Example: Kettle Your electric kettle is labeled 2000 W. How long will it take to boil 0.5 liter of water if the water comes out of the tap at 15°C? Assuming that all the heat produced by the kettle is used to warm up the water, water water water Q mc T= ?Heat needed to warm up the water: () 5 water 1 kg water J 0.5 liters 4186 100 15 C 1.78 10 J 1 liter water kg C Q ???? =?°=× ???? ° ???? 5 1.78 10 J 89 s 1.5 min 2000 W Q t P × == = = But of course in reality it will be a little longer. What are we neglecting? Heat produced by the internal resistance of the kettle also warms up: ? air (negligible if kettle has a lid and is well insulated) ? kettle (at least internal wall) Energy released in resistor Energy absorbed by water Energy absorbed by wall Energy absorbed by air Flow of energy What does specific heat depend on? Temperature = average kinetic energy of particles Degrees of freedom (= ways to move, ie, to increase kinetic energy): Example: A molecule of helium is made of one atom. It can basically just ?bounce around? in 3 directions (3 degrees of freedom) A molecule of hydrogen is made of two atoms. It can bounce around (3 dof) and it can also rotate (+ 2 dof, total 5 degrees of freedom) Molar mass Heavier molecules require more additional energy to increase their average speed. DEMO: Specific heats PH Phases or states of matter Three basic states of matter: ?Solid ?Liquid ?Gas A phase change involves a critical change in the microscopic structure of matter. Example: Ice to water: Lattice disappears, molecules are free to move around. Analysis: Phase changes for water Q T 1 kg of ice is placed on a pan on the stove. Plot of temperature versus heat supplied by stove. Ice Ice + Liquid Liquid Gas Liquid + Gas 334 kJ/kg (Latent) heat of fusion 2256 kJ/kg (Latent) heat of vaporization 100°C Boiling point 0°C Melting point Slide 6 PH2 Heat-24 Comparison of specific heats. Different cylinders on a thin paraffin cake Paula Herrera, 9/18/2008 ACT: Specific heats Q T Ice Liquid Gas Which water phase has the largest specific heat? A. Ice B. Liquid water C. Steam 1 slope dT QmcT dQ mc =? ? = Small slope, large c ice water JJ Indeed: 2100 4186 kg K kg K cc== Latent heat Water: L fusion = 334 kJ/kg L vaporization = 2256 kJ/kg This energy is not used to increase the kinetic energy of the particles (does not increase the temperature) but to change the structure of matter. During a phase change, two or more phases coexist in dynamic equilibrium. Examples: Vapor and liquid water exactly at 100°C Ice and liquid water exactly at 0°C Vapor/Liquid water/Ice at the triple point (273.16 K and 610 Pa) In-class example: Ice melting How much heat is needed to turn 10 g of ice at -5°C into liquid water at 20°C? total warm ice to 0 C melt ice warm water to 20 C Q QQQ °° =++ () ()() warm ice to 0 C J 0.010 kg 2100 0 5 C 105 J kg C Q ° ?? = ??° = ?? ° ?? () ?? == ?? ?? melt ice kJ 0.010 kg 334 3340 J kg Q () () warm water to 20 C J 0.010 kg 4186 20 C 0 837 J kg C Q ° ?? =°?= ?? ° ?? A. 105 J B. 420 J C. 837 J D. 3330 J E. 4272 J total 4272 JQ = (mostly from melting) Follow-up example: Iced coffee 10 g of ice at -5°C are added to 30 ml of hot coffee inside a thermos that is then tightly closed. After the system reaches equilibrium, the temperature of the mix is 20°C. What was the initial temperature of the coffee? Because the system is thermally isolated (closed thermos), the hot coffee is the only source of energy, so it must provide the necessary 4272 J of heat. ( ) cool coffee coffee water initial 20 CQmc T=°?4272 J= ? cool coffee initial coffee water 20 C Q T mc =°? () 3 4272 J 20 C 54°C J 30 10 kg 4186 kg°C ? ? =°? = ?? × ?? ?? For the coffee, this is a decrease in energy 10 g of ice at -5°C are added to 30 ml of hot coffee inside a thermos that is then tightly closed. After the system reaches equilibrium, the temperature of the mix is 20°C. What was the initial temperature of the coffee? Energy balance: warm ice to 0 C melt ice warm water to 20 C cool coffee 0Q QQ Q °° ++ + = + + + ? Energy is absorbed Energy is released Energy released by coffee Energy absorbed by ice (warming) Energy absorbed by liquid water (warming) Flow of energy: from hot object to cold object Energy absorbed by ice (melting) Phase transition temperatures Water at 1 atm: T melting = 0°C T boiling = 100°C These temperatures indicate when the kinetic energy of the molecules is enough to break the structure. Transition temperatures depend on pressure At the top of mount Everest, where p = 0.26 atm, water boils at 69°C It is easier for molecules to break free into air! pT diagram Critical point Sublimation curve (gas/solid transition) Melting curve (solid/liquid transition) solid liquid gas Triple point Vapor pressure curve (gas/liquid transition) pT diagram (water) For water, the pT diagram looks a little different. water other solid liquid gas pT diagram (water) 1 atm T p solid liquid gas Critical point 0°C 100°C DEMO: Boiling water with ice PH3 Triple point (water) At 610 Pa T melting = T boiling Liquid water does not exist for p < 610 Pa!! T p solid liquid gas Triple point Critical point Triple point for water: p = 610 Pa (0.006 atm), T = 273.16K DEMO: Triple point of N 2 PH4 Critical point (water) Critical point for water: 647K and 218 atm T p solid liquid gas Triple point Critical point Supercritical fluid Liquid and gas are indistinguishable beyond critical point At critical point, ? gas = ? liquid Slide 18 PH3 Heat 33: Boiling water by cooling its vapor Paula Herrera, 9/23/2008 Slide 19 PH4 Heat-42: Cooling by evaporation (or nitrogen "snow", or the triple point of nitrogen!) Paula Herrera, 9/23/2008 Example: Mixing oil 30 g of oil at 10°C and 150 g of oil at 80°C are placed in a thermos. The thermos is closed and the system is allowed to go to equilibrium. What is the final temperature of the oil? Flow of heat: hot cold Q H < 0 (energy loss) Q C > 0 (energy gain) + = H C 0Q Q Closed thermos is an insulated system: no energy exchange with rest of the universe: += H C 0Q Q 30 g of oil at 10°C and 150 g of oil at 80°C are placed in a thermos. The thermos is closed and the system is allowed to go to equilibrium. What is the final temperature of the oil? ( ) ( ) ( ) ( )?°+ ?°= oil oil 30 g 10 150 g 80 0cT C cT C ( ) ( )?+ ?=30 10 150 80 0TT =°68T C Molar heat capacity Let?s change the question: How much heat is needed to change by ?T the temperature of n moles of material X? =?= ? XX Q mc T nMc T =mnM n = number of moles M = mass of one mole (molar mass) = XX CMc =? X Q nC T = X molar heat capacityC Soeren Prell Microsoft PowerPoint - Lecture 34 - Heat and phase changes
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