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Part One Heat and Temperature What does it mean to have a temperature of 0 C? What is temperature? Is temperature the same thing as heat? Temperature is a measure of how “hot” or “cold” something is. Temperature is measured in arbitrary units, like Fahrenheit or Celsius. Temperature is proportional to the average kinetic energy of the molecules of the substance. Heat is the thermal energy transferred from a hot object to a cold object. Heat is measured in energy units -- Joules or calories. Heat has the symbol q and is calculated using … q = mcDT Quantity of heat specific heat capacity q = mcDT mass temperature change Quantity of heat specific heat capacity q = mcDT The specific heat capacity of water is 4.18 J/gC How much heat is needed to raise the temperature of 25.6 grams of water from 20.0 C to 50.0 C? q = m c DT q = (25.6g)(4.18J/gC)(30.0C) q = 3210 J What is the final temperature of 27.0 grams of liquid water, initially at 0C, after it absorbs 700.0 J of energy? q = m c DT Hint: start by solving for DT. q DT = m c Answer: 6.20 C Part Two Phase Changes We now know that heat is either absorbed or released during a phase change. A process that absorbs heat is called endothermic. A process that gives off heat is called exothermic. Melting (fusion) Endothermic: Vaporization Heat is absorbed. Sublimation Freezing Exothermic: Condensation Heat is released. Deposition Heat is absorbed by the ice. Ice And melts. Heat is absorbed by the ice. One gram of ice at 0C absorbs 334 J as it melts to form water at 0C. … making liquid water Heat is released by the water as it freezes. Ice water 334 joules is released when one gram of water freezes at 0C. Ice absorbs 334 J per gram as it melts at 0C Ice Water releases 334 J per gram as it freezes at 0C Steam releases 2260 J/g as it condenses at 100 C Water absorbs 2260 J/g as it boils at 100 C Hotplate The heat gained or lost in phase changes can be calculated using … q = mHf q = mHv Heat of fusion (melting) Heat of vaporization The values for water are … Hf = 334 J/g Hv=2260 J/g Heat of fusion (melting) Heat of vaporization How much heat is absorbed by 150.0 g of ice as it melts at 0C? q = m Hf q = (150.0 g)(334 J/g) q = 50,100 J or 50.1 kJ How much heat is released by 20.0 grams of steam as it condenses at 100C? q = m Hv q = (20.0 g)(2260 J/g) q = 45,200 J or 45.2 kJ Part Three Phase Diagrams Which phase is in each region? Pressure Hint: What happens to ice as temperature increases? 2 1 3 Temperature The phase diagram has three distinct regions. Pressure The point where all three phases exist in equilibrium is called the triple point. S L G Temperature At a pressure of 1 atm, most substances go through all three phases, as the temperature increases, S L 1 atm G Temp. Solids melt to form liquids, which vaporize to form gases. S L 1 atm G MP BP Temp. Notice the melting point and boiling point. But the phase diagram for CO2 is a little different. Notice that the triple point is L above 1 atm. 5 atm S 1 atm G Temperature At 1 atm CO2 goes directly from solid to vapor as the temperature increases. S L The sublimation point is –78.5 C 1 atm G Temperature Pressure What phase change is occurring? S L Melting (fusion) G Temperature Pressure What phase change is occurring? S L Vaporization G Temperature Pressure What phase change is occurring? S L Condensation G Temperature Pressure What phase change is occurring? S L Sublimation G Temperature Pressure What phase change is occurring? S L G Temperature Liquefying a gas by increasing the pressure. Part Four Heating and Cooling Curves Look at the different regions of the heating curve for water. Temp 100 Water Ice and water 0 Water and steam Steam Phase changes? Ice Time Part Five Calorimetry and Specific Heat Capacity Calorimetry is a collection of laboratory procedures used to investigate the transfer of heat. In calorimetry experiments, one might be looking for a final temperature or a specific heat capacity. What is the law of conservation of energy? Energy is neither created nor destroyed, only changed in form. The law of conservation of energy suggests that the heat lost by the hot object as it cools is equal to the heat gained by the cool water as it warms up. Investigate: To put it mathematically: qlost = -qgained Heat lost by the hot object Heat gained by the cold water = And since q = mcDT then mocDTo = -mwcDTw Investigate: The convention for DT is final temperature minus initial temperature or Tfinal – Tinitial mhcDTo = -mccDTw becomes moc(Tf -Ti) = -mwc(Tf -Ti) Use your algebra skills, to solve for Tf , the final temperature. Specific heat capacity … • …varies from one substance to another. • …a measure of how much heat something can “hold”. • …the amount of heat needed to raise one gram of a substance by one Celsius degree. Specific heat capacity lab suggestions: 1. Heat a metal to a known temp. 2. Transfer the metal to a known quantity of water at a known temperature. 3. Measure the equilibrium temperature. 4. Use qlost = -qgained to compute the specific heat of the metal. Get the initial temperature of the metal. metal The temperature of boiling water. hotplate Get initial temp of water in calorimeter cup. Transfer the metal to the calorimeter. Continue stirring. Data: Mass of metal Initial temp of metal Mass of water Initial temp of water Final temp of water and metal qlost = -qgained mmcmDTm = -mwcwDTw -mwcwDTw cm = mmDTm Mass of metal Initial T of metal Mass of water in calorimeter Initial T of water Final T of water and metal 40.0 g 98.0 C 60.0 g 20.0 C 22.9 C Calculate the specific heat capacity of the metal. Table of selected specific heats. What is the unknown metal? Substance c in J/g K Aluminum Bismuth Copper Brass Gold Lead Silver Tin Zinc Mercury Ethanol Water Ice 0.900 0.123 0.386 0.380 0.126 0.128 0.233 0.225 0.387 0.140 2.400 4.186 2.050