Theory of Calorimetry - Whitney High School - Home
Thermochemistry over BREAK!! Terms to review: (know what they are and what units they are measured in)HeatTemperatureSpecific heatΔHrxnqEquations to remember: q=mcΔTC water= 4.18 J/ g°CC ice= 2.03 J/ g°CC steam= 1.97 J/ g°CΔHvap=m(2259 J/g)ΔHfus=m(334 J/g)Phase change curve:Place the equations listed above on the correct position on the graph below. Complete the following two worksheets over break. I have included some notes to help remind you of your first year chemistry course. There will be a quiz on the first day back from break. Enjoy PHASE CHANGE PRACTICEUse the following information to solve the phase change problems:Specific heat of water= 4.18 J/g·CHfusion = 334 J/gSpecific heat of ice= 2.03 J/ g·CHvaporization= 2259 J/gSpecific heat of steam= 1.97 J/g·CRecall that it requires additional energy to change phases:SOLID Hfusion LIQUID Hvaporization GAS -Hfusion -Hvaporization--------------------------------------------------------------------------------------------------------------------------------1. How much heat is required to melt 233.0 grams of ice into water, from -15.00C to room temperature (25.00C)? ANSWER: heating ice = 7095 J, heating water = 24349 J, heat of fusion = 77822 J, TOTAL HEAT = 110 kJ2. How much heat is required to change 32.5 grams water into steam, from room temperature (25.0C) to 115C? ANSWER: heating water = 10189 J, heating steam = 960 J, heat vaporization = 73418 J, TOTAL HEAT = 85 kJ3. How much heat is needed to melt 1.43 grams of ice into water from -5.34C to 84.3C? ANSWER: heating ice = 15.5 J, heating water = 504 J, heat fusion = 478 J, TOTAL HEAT = 998 J4. How much heat is needed to convert 0.232 grams water into steam, from 32.5C to 112C? ANSWER: heating water = 65.5 J, heating steam = 5.48 J, heat vaporization = 524 J, TOTAL HEAT = 595 J5. How much heat do you need to add to 3.22 grams H2O to raise the temperature from -23.0C to 152C? ANSWER: heating ice=150 J, heating water=1346 J, heating steam=330 J, heat fusion=1075 J, heat vaporization=7274 J, TOTAL=10. kJ 6. How much heat is needed to raise the temperature of 199 grams H2O from -10.3C to 154C? ANSWER: heating ice=4161 J, heating water=83182 J, heating steam=21170 J, heat fusion=66466 J, heat vaporization=449541 J, TOTAL = 625 kJCalorimetryCalorimeterA device used to measure enthalpy changes in chemical rxns. Usually a thick-walled vessel that keeps the reaction system well-insulated from the surroundings, equipped with a stirrer and a thermometer. In our case, we will use high-tech ultra-expensive styrofoam cups with cardboard covers. Using a calorimeter, you can measure the temperature change (T) of the rxn, and relate this to the enthalpy (H) of the rxn. Calories: Relating Energy to FoodThe food you consume contains energy that your body needs to survive. Food energy is measured in calories, which are really kilocalories in the United States. Potato chips that have 180 calories actually have 180 kcal, or 180,000 calories!1 calorie = 4.18 joulesHow much energy does a 235 calorie candy bar have? Remember that in the United States, one calorie is actually 1 kilocalorie.Theory of CalorimetryCalorimetry is based upon of conservation of energy, the First Law of Thermodynamics. Consider an exothermic rxn run in a calorimeter. Since the system is closed (completely isolated from the surroundings), all heat evolved by the reaction will be absorbed by the calorimeter. Thus:Hrxn = -HcalorimeterThe magnitude of the observed temperature change depends on the heat capacity of the calorimeter. So what’s heat capacity anyways?Heat Capacity and Specific HeatHeat Capacity (Cp) The heat capacity of any substance is the amount of energy (heat) required to raise the temperature of that substance 1C (or 1K). Heat capacity thus measures the substance’s ability to absorb energy as heat. Water, for example, has a relatively high specific heat.What do you think has a greater heat capacity, oven mitts or a piece of metal? How do you know this?The units for heat capacities are: kJ (heat absorbed)Cp =----- K (increase in temperature)In a calorimetry experiment, we can say that: Hrxn = Cp(calorimeter) TSpecific Heat (Csp)If we want to get more specific, we can speak of a substance’s specific heat. This is the amount of heat required to raise the temperature of 1 gram of that substance 1C. Units for specific heat are usually: kJCsp = --------- K gThe heat involved in any thermal process can be calculated using this formula:q = m Csp Tq = heatm = mass (moles for Molar Specific Heat)Csp = Specific Heat – pay attention to the unitsT = Change in temperatureSample ProblemsExample: When 1000 J of heat is added to 50.5 g of ammonia, the temperature increases by 9.8C. What is the specific heat of ammonia? Example: A 117 g sample of sodium chloride is dissolved in 1 L of water in a calorimeter whose total heat capacity (water + vessel) is 1000 J/K. The temperature decreases by 12.8 K. Calculate the heat of reaction per mole for this process.Example: A 36.7 g sample of aluminum is heated to 102C and placed in a calorimeter containing 150.0 g water at 23.4C. After the metal cools, the final temperature of metal and water is 25.0C. Calculate the specific heat capacity of aluminum, assuming that no heat escapes to the surroundings or is transferred to the calorimeter. The specific heat of water is 4.18 J/CgExample: In a calorimeter, 50.0 mL of 0.300 M AgNO3 and 50.0 mL of 0.300 M HCl are mixed to yield the following reaction:Ag+(aq) + Cl-(aq) AgCl(s)The two solutions were initially at 23.0C and the final temperature is 25C. Calculate the heat that accompanies this reaction in kJ/mol of AgCl formed. Assume that the combined solution has a mass of 100.0 g and has a specific heat capacity of 4.18 J/CgPhases of MatterThere are 3 common phases of matter: SOLID, LIQUID, GASDraw pictures of a solid, liquid, and gas, on a molecular level:The types of phase changes are listed below:SOLID melting LIQUID evaporationGAS solidification condensationBreaking bonds requires energy. Forming bonds gives off energy. Going from a solid liquid gas therefore requires energy. Going from a gas liquid solid therefore releases energy. Are the following phase changes exothermic or endothermic?ice cubes meltingfreezing water into iceboiling a liquidsteam from your shower condensing into water droplets on the mirrorA substance may never change temperature and phase simultaneously!This means that if a substance is heated, the heat may be used to do one of two things:-The temperature can increase-The phase can change (solidliquid, liquidgas) BUT NEVER BOTH AT THE SAME TIME!Draw a heat vs temperature graph of water being heated:The energy it takes to go from solid liquid is called the enthalpy of fusion (Hfusion) Going the reverse, a liquid solid, releases the same amount of energy (-Hfusion)The energy it takes to go from liquid gas is called the enthalpy of vaporization (Hvaporization)Going the reverse, a gas liquid, releases the same amount of energy (-Hvaporization)SOLID Hfusion LIQUID Hvaporization GAS -Hfusion -HvaporizationCalorimetry WorksheetHEAT GAINED BY WATER = HEAT LOST BY OBJECTSpecific Heat of Water = 4.18 J/G·°COneA 32.5 g sample of tin is heated to 132°C and placed in a calorimeter containing 125 g of water at 25.00°C. After the metal cools, the final temperature of metal and water is 27.00°C. Calculate the specific heat capacity of tin. a) Calculate the amount of heat gained by the water.Calculate the specific heat of the tin. Remember that the heat gained by the water is equal to the heat lost by the tin. TwoA 28.2 g sample of nickel is heated to 99.8°C and placed in a coffee-cup calorimeter containing 150.0 g water at 23.5°C. After the metal cools, the final temperature of the metal and water is 25.0°C. Calculate the specific heat capacity of nickel, assuming that no heat escapes to the surroundings or is transferred to the calorimeter. a) Calculate the amount of heat gained by the water.Calculate the specific heat of the nickel. Remember that the heat gained by the water is equal to the heat lost by the nickel. ThreeA 46.2 g sample of copper is heated to 95.4°C and then placed in a calorimeter containing 75.0 g water at 19.6°C. The final temperature of the metal and water is 21.8°C. Calculate the specific heat capacity of copper, assuming that no heat escapes to the surroundings or is transferred to the calorimeter. a) Calculate the amount of heat gained by the water.Calculate the specific heat of the nickel. Remember that the heat gained by the water is equal to the heat lost by the nickel. FourA 32.55 g sample of a greenish, unknown substance is heated to 54.54°C and the placed in a calorimeter containing 100.00 g water at 10.00°C. The final temperature of the metal and water is 12.32°C. What is the specific heat capacity of the unknown substance?FiveA 0.231 g piece of brass alloy is heated to 135.03°C and then placed in a calorimeter containing 50.0 g water at 25.00°C. The final temperature of the metal and water is 30.00°C. What is the specific heat capacity of the brass alloy?SixThe specific heat capacity of aluminum is 0.900 J/°C·g. Calculate the energy needed to raise the temperature of a 850. g block of aluminum from 22.8°C to 94.6°C. Calculate the molar heat capacity (heat/mol) of aluminum. SevenThe specific heat capacity of graphite (pure carbon) is 0.71 J/°C·g. Calculate the energy required toRaise the temperature of 1.0 mol graphite by 1.0°C. Raise the temperature of 850 g graphite by 150°C. ................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related searches
- gadsden high school home page
- list of high school codes
- examples of high school essays
- university of missouri high school online
- purpose of high school education
- copy of high school transcripts
- curtis high school home page
- scranton high school home page
- home high school help
- chaparral high school home page
- southeast high school home page
- letter of recommendation for high school student