Calorimetry
ChemistryName:_______________________CalorimetryDate:__________________q = m T CpWhere: q = total heat flowm = massT = change in temp. Cp = specific heatExample 1:Calculate the number of joules (and kilojoules) required to warm 1.00 x 102grams of water from 25.0 oC to 80.0 oC.Heat energy = mass x change in temperature x specific heatSpecific Heat:Many times Calorimetry problems involve solving for one of the other quantities such as specific heat of temperature change. This is done by simply using algebra to rearrange the formula q = mTCp.Example 2:Calculate the specific heat of gold if it required 48.0 joules of heat to warm 25.0 grams of gold from 40.0 oC to 55.0 oC. Problems: Show all work for full credit!How many joules are needed to warm 25.5 grams of water from 14 oC to 22.5 oC? Calculate the number of joules released when 75.0 grams of water are cooled from 100.0oC to 27.5 oC. How much heat is absorbed by 60.0 g of copper when it is heated from 20.0 oC to 80.0 oC? The specific heat of copper is 0.385 J/gCo.A 38kg block of lead is heated from 26.0oC to 180oC. How much heat does it absorb during the heating? The specific heat of lead is .138 J/g Co. How many joules are needed to warm 45.0 grams of water from 30.0oC to 75.0oC? What would be the final temperature if 3.31 x 103 joules were added to 18.5 grams of water at 22.0oCA sample of lead, specific heat 0.138J/gCo, released 1.20 x 103J when it cooled from 93.0oC to 29.5oC. What was the mass of this sample of lead? Energy TransferOne of the major uses of Calorimetry is to measure the specific heats of metals and various metallic alloys. Chemists use specific heats in their study of the structure and attractive forces in the materials. Specific heats are useful for engineers in planning structures or processes. This use of Calorimetry is based upon the law of conservation of energyenergy is neither created nor destroyed but can be transformed from one form to another. In this application, hot metal is added to cold water in an insulated container called a calorimeter. The cold water gains the heat lost by the hot metal as they come to the same final temperature. If you measure the temperature changes you can calculate the heat gained by the water and thus the heat lost by the metal, and finally can calculate the specific heat of the metal.Example 3:A 175 gram sample of a metal at 93.5oC was added to 105 grams of water at 23.5oC in a perfectly insulated calorimeter. The final temperature of the water and metal was 33.8oC. Calculate the specific heat of the metal in joules/gram oC.heat lost by the metal = heat gained by the watermmCpmtm = mwCpwtwProblems: Show all work for full credit.A 185 gram sample of copper at 98.0 oC was added to 102 grams of water at 20.0 oC in a perfectly insulated calorimeter. The final temperature of the copperwater mixture was 31.2 oC. Calculate the specific heat of copper using this data. A chemistry student added 225 grams of aluminum at 85.0 oC to 115 grams of water at 23.0 oC in a perfect calorimeter. The final temperature of the aluminum water mixture was 41.4 oC. Use this student's data to calculate the specific heat of aluminum in joules/gram Co. A student was given a sample of a silvery gray metal and told that it was bismuth, specific heat 0.122 J/goC, or cadmium, specific heat 0.232 J/g oC. The student measured out a 250 gram sample of the metal, heated it to 96.0oC and then added it to 98.5 grams of water at 21.0oC in a perfect calorimeter. The final temperature in the calorimeter was 30.3oC. Use the student's data to calculate the specific heat of the metal sample and then identify the metal. ................
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