Properties of Solids & Liquids

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Discusses macroscopic and microscopic properties of solids and liquids.
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Properties of Solids & Liquids

  1. 1. Copyright Sautter 2015
  2. 2. SOLIDS, LIQUIDS AND PHASE CHANGES • What is the difference between a solid, a liquid and a gas? • ENERGY STATE ! Solids are the lowest energy state, liquids are intermediate and gases are the highest energy state of matter. • The energy differences which separate each state are called the Heats of Phase Change. The Heat of Phase Change is the energy required to change the state of a substance without changing its temperature. • The Heats of Phase Change are defined by the phase change which occurs, for example, Heat of Melting (often called fusion) is the heat needed to change a solid into a liquid at its melting point. The Heat of Vaporization is the heat needed to change a liquid into a gas at its boiling point.2
  3. 3. SOLIDS, LIQUIDS AND PHASE CHANGES • When solids, liquids and gases are heated or cooled and do not undergo phase change, the heated added or removed from the substance can be calculated using the equation: q = m x c x T • The letter q represents heat is calories or joules. The mass of the substance in grams is m. The letter c stands for “specific heat” in calories per gram degree Celsius or joules per gram degree Celsius. Specific heat indicated how easy or difficult it is to change the temperature of a substance. Large specific heats mean that it requires a large amount of heat addition or removal to change temperature. Small values mean that the substance changes temperature readily. • Delta T (T) is the Celsius temperature change of the substance 3
  4. 4. HEAT IN JOULES OR CALORIES MASS IN GRAMS SPECIFIC HEAT IN JOULES / G x 0C OR CALORIES / G x 0C TEMPERATURE CHANGE IN 0C CALCULATING HEAT QUANTITIES (NO PHASE CHANGE OCCURS) 4
  5. 5. SOLIDS, LIQUIDS AND PHASE CHANGES • Problem: What is the temperature of 15.0 grams of iron when 250 joules of heat are removed from it? (specific of iron = 0.450 j / g x 0C) • Solution: q = m x c x T, q = 250 joules m = 15.0 grams, c = 0.450 j / g x 0C T = q / (m x c) = 250 / (15.0 x 0.450) = 37.0 0C Since heat is removed from the sample the temperature is lowered by 37.0 degrees, therefore the answer is – 37.0 0C 5
  6. 6. SOLIDS, LIQUIDS AND PHASE CHANGES • Problem: A 100 gram sample of aluminum at 100 0C is placed in 200 grams of water at 25 0C. What is the final temperature of the system? (c for Al = 0.900 j / g 0C and c for water = 4.18 j / g 0C) • Solution: Heat lost = Heat gained (Conservation of Energy) Al loses heat to the water since the Al is at the higher initial temperature. • mAl x cAl x TAl = mw x cw Tw, T = temperature of system at equilibrium 100g(0.900 j /g0C)(1000C – T) = 200g(4.18 j /g0C)(T-250C) • 9000 – 90T = 836T – 20900 926T = 29900, T = 32.3 0C 6
  7. 7. SOLIDS, LIQUIDS AND PHASE CHANGES • When a substance undergoes phase change (change in physical state) heat is added or removed from the material however the temperature of the material remains unchanged. All energy that is released or absorbed during the phase change is used to change the state of the substance. None is used to alter the temperature. • Only after the phase change is complete can the temperature of the system rise or fall. (Note the temperature plateaus on the next graph) • The heat of phase change for different processes is described by the physical change which is occurring for example: Heat of Melting (sometimes called fusion), Heat of Vaporization, Heat of Freezing, etc. 7
  8. 8. Time of Heating T E M P E R A T U R E 0C 110 100 0 -10 All ice Melting starts Melting completed Ice & water All water Boiling starts Water & steam Boiling completed All steam Heating Curve for Water (-10 to 110 0C) 8
  9. 9. CALCULATING HEAT EXCHANGES DURING PHASES CHANGES • During a phase change the temperature of a substance remains constant. Based on that observation, q = m c T cannot be used to find heat gained or lost since T = 0. • In order to find heat values during phase change we must multiply the heat of phase change by the quantity of substance involved. Heat of phase change is the quantity of heat which is gained or lost when a substance changes state without a temperature change. • The heat of phase change may be expressed in calories or joules per gram of substance or calories or joules per mole of substance. • Q = Heat of Phase Change x Amount of Substance 9
  10. 10. Heat of Melting Heat of Vaporization Heat of Condensation Heat of Freezing T E M P E R A T U R E Time of Heating HEATS OF PHASE CHANGE DURING HEATING & COOLING 10
  11. 11. HEAT REQUIRED FOR PHASE CHANGE HEAT OF MELTING (FUSION) HEAT OF FREEZING HEAT OF VAPORIZATION HEAT OF CONDENSATION HEAT OF SUBLIMATION (ALL IN JOULES / GRAM, CALORIES/ GRAM OR PER MOLE AMOUNT OF SUBSTANCE CHANGING PHASE (GRAMS OR MOLES) CALCULATING HEAT GAINED OR LOST DURING A PHASE CHANGE 11
  12. 12. CALCULATING HEAT EXCHANGES DURING PHASES CHANGES • Problem: How much heat is needed to melt 200 grams of ice at its melt point (0 0C) ? The heat of fusion for ice is 6019 joules per mole. • Solution: Since a phase change is occurring without a temperature change: • qmelting = Heat of Melting x moles of ice melted • Moles = grams / grams per mole, moles = 200 / 18 = 11.1 • qmelting = 6019 joules / mole x 11.1 moles = 66,878 joules or 66.9 kilojoules. 12
  13. 13. CALCULATING HEAT EXCHANGES DURING PHASES CHANGES • Problem: How much heat is needed change 100 grams of ice at –10 0C to steam at 110 0C. (c for ice = 2.06 j / g 0C, c for water = 4.18 j / g 0C, c for steam = 2.02 j / g 0C, Heat of Fusion for ice = 6.02 Kj / mole, Heat of Vaporization = 40.7 Kj / mole) • Solution: qheating ice = m x c x T • qheating ice = 100 x 2.06 x (0 – (-10)) = 2060 j • qmelting = Heat of Melting x moles of ice melted • qmelting = (6020 j / mole)(100 / 18 ) mole = 3.34 x 104 j • qheating water = 100 x 4.18 x (100 – 0) = 4180 j • qvaporization = (40.7 x 103 j / mole)(100 / 18) mole = 2.26 x 105 j • qheating steam = 100 x 2.02 x (110 – 100) = 2020 j • qtotal = 2060 + 3.34 x104 + 4018 + 2.26 x 105 + 2020 = 2.67 x 105 j 13
  14. 14. CALCULATING HEAT EXCHANGES DURING PHASES CHANGES • IMPORTANT POINTS IN CALCULATING HEAT QUANTITIES: • (1) The specific heat of a substance is different for the same substance in a different physical state. For example, as we have seen, the specific heat of ice (2.06 j /g0C), that for water (4.18 j / g0C) and steam (2.02 j / g0C) are different. • (2) Heats of phase change are different for different changes involving the same material. For example Heat of Fusion for ice is 6.02 Kj / mole while Heat of Vaporization for water is 40.7 Jj / mole. Note that the Heat of Vaporization for a substance is always noticeably larger than its Heat of Fusion. • (3) Heat of phase change for a substance in the heating process are equal but opposite in sign as compared for the reverse phase change in the cooling process. For example, the Heat of Melting for ice is 6.02 Kj / mole while the Heat of Freezing for water is – 6.02 Kj / mole. 14
  15. 15. TEMPERATURE AND PRESSURE EFFECTS ON PHASE CHANGE • Phase changes occur as: • Melting or Fusion – solid to liquid (endothermic) • Freezing or Solidification – liquid to solid (exothermic) • Vaporization – liquid to gas (endothermic) • Condensation – gas to liquid (exothermic) • Sublimation – solid to gas (endothermic) • Deposition – gas to solid (exothermic) • All changes of physical state are affected by both temperatures and pressures. Gas state is favored by low pressures and high temperatures both allowing for free, random motion. Solids are favored by low temperatures and high pressures both contributing to an ordered low energy state. A phase diagram shows the relationship between temperature, pressure and physical state under a variety of conditions. 15
  16. 16. TEMPERATURE AND PRESSURE EFFECTS ON PHASE CHANGE • After a certain temperature (depending on the substance) no amount of pressure can cause a gas to form a liquid. This temperature is called the critical temperature of the substance. The pressure which the gas exhibits at this temperature is called the critical pressure.This temperature – pressure point is called the critical point. • At a specific temperature and pressure all three phases of a substance can coexist (solid, liquid and gas). This point is called the triple point. At this point a very slight change in temperature and / or pressure can cause a solid to melt, a liquid to vaporize, a solid to sublime, a liquid to freeze, etc. 16
  17. 17. P R E S S U R E TEMPERATURE GAS SOLID LIQUID PHASE DIAGRAM TRIPLE POINT a b c d e f g a = freezing b = melting c = vaporization d = condensation e= deposition f = sublimation g = critical point 17
  18. 18. P R E S S U R E GAS SOLID LIQUID PHASE DIAGRAM (for water) a b Notice that moving along the line a-b, as pressure rises, the solid melts without a temperature increase TEMPERATURE This behavior is typical for water but not most substances 18
  19. 19. P R E S S U R E GAS SOLID LIQUID PHASE DIAGRAM (for most substances) a b Notice that moving along the line a-b, as pressure rises, the solid Will not melt Increasing pressure without a temperature change will not melt the substance TEMPERATURE This behavior is typical for most substances19
  20. 20. VAPOR PRESSURE OF LIQUIDS • The measure of the tendency of a liquid to form a gas is indicated by its vapor pressure. Liquids which evaporate readily have high vapor pressures. The tendency of a liquid to enter the gaseous state varies based on several factors: • (1) the intermolecular forces which exist between the liquid molecules. Factors such as polar character, London Forces and hydrogen bonding (all of which have been discussed in other programs) affect vaporization.As the strength of intermolecular forces increases the vapor pressure of the liquid decreases. • (2) the Heat of Vaporization of a substance. Materials with large Hvap have low vapor pressures. • (3) the temperature of the liquid. As temperature rises the liquid molecules increase their kinetic energies and the tendency to form a gas increases. 20
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