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## Last rev. 022007a

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**Heat, Temperature**And Phase Changes Pisgah High School Chemistry Mr. Jones Last rev. 022007a**Part One**Heat and Temperature**Yep.**It is snowing.**Why would snow cause the air temperature to be at**precisely 0C? What occurs at 0C? Waterfreezes and icemelts.**Ice, in the form of snow, falls through the slightly warmer**air. The snow melts and absorbs heat from the air, causing the air to cool. Ice melts at 0 C, so the air cools to that temperature.**So why is there snow on the ground if it is melting?**Yep. That’s what allows the snow to accumulate.**As the snow melts, it absorbs heat and cools the ground, the**car, and the grill.**This allows more snow to lay. It doesn’t melt because the**ground is now at 0C.**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. T µ KE KE =½ mv2**Velocity or speed**Temperature is therefore proportional to the speed of the molecules of a substance. T µ KE KE =½ mv2 T µv**Velocity or speed**The higher the temperature, the greater the average speed of the molecules. T µ KE KE =½ mv2 T µv**Heat is the thermal energy transferred from a hot object to**a cold object. Heat is measured in energy units -- Joules or calories.**The heat transferred is proportional to the mass of the**object, the specific heatcapacity of the object and the temperature change the object undergoes.**specific heat capacity**Quantity of heat mass temperature change q = mcDT**specific heat capacity**Quantity of heat 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**q**DT = m c 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. Answer: 6.20 C**Part Two**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.**Investigate:**Suppose two different masses of water at different temperatures are mixed. Can you predict the final temperature?**Investigate:**Will the final temperature be cooler than the cool water, or will it be warmer than the warm water? Or will the final temperature be somewhere in between?**Investigate:**Develop a procedure where you mix a known mass of cool water with a different mass of water at an elevated temperature and measure the final (equilibrium) temperature. What equipment will you need?**Investigate:**Develop a procedure where you mix a known mass of cool water with a different mass of water at an elevated temperature and measure the final (equilibrium) temperature. You could use a balance, a thermometer, a coffee cup calorimeter, and a hot plate.**Investigate:**What do we need to record in a data table? Mass of calorimeter cup _________ Mass of cool water and cup _________ Mass of cool water _________ Initial temperature of cool water _________ Initial temperature of hot water _________ Final temperature after mixing _________ Mass of mixed water and cup _________ Mass of hot water _________**Investigate:**Whenever we design an experiment we make some assumptions. Here are a couple, can you add any more? The calorimeter cup is a perfect insulator and no heat is exchanged with the surroundings. Warning: Hot plates and boiling water can cause severe burns.**Investigate:**You might need a hint about how to calculate the results. What is the law of conservation of energy? Energy is neither created nor destroyed, only changed in form.**Investigate:**You might need a hint about how to calculate the results. The law of conservation of energy suggests that the heat lost by the hot water 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 water Heat gained by the cold water = And sinceq = mcDT then mhcDTh = -mccDTc**Investigate:**The convention for DT is final temperature minus initial temperature or Tfinal – Tinitial mhcDTh = -mccDTc becomes mhc(Tf -Th) = -mcc(Tf -Tc) Use your algebra skills, to solve for Tf , the final temperature.**In the next investigation you will …**develop a method to find the specific heat capacity of a metal.**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:**• Heat a metal to a known temp. • Transfer the metal to a known quantity of water at a known temperature. • Measure the equilibrium temperature. • Use qlost = -qgained to compute the specific heat of the metal.**Get the initial temperature of the metal.**The temperature of boiling water. metal hotplate**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 40.0 g**Initial T of metal 98.0 C Mass of water in calorimeter 60.0 g Initial T of water 20.0 C Final T of water and metal 22.9 C Calculate the specific heat capacity of the metal.**Table of selected specific heats.**What is the unknown metal?**Part Three**Calorimetry and Phase Changes**Is heat is absorbed or released during a phase change?**How could you measure the heat absorbed or released as substances change phase?