1. A calorimeter is used to measure the amount of heat absorbed or released during a chemical reaction.A calorimeter is used to measure the amount of heat absorbed or released during a chemical reaction. In today’s experiment you will measure the heat of a reaction in a calorimeter that is made out of 2 Styrofoam coffee cups.In today’s experiment you will measure the heat of a reaction in a calorimeter that is made out of 2 Styrofoam coffee cups.  The inner cup holds an aqueous reaction mixture.  The outer cup gives additional thermal insulation from the surrounding environment.  The thermometer is used to measure the initial and final temperatures of the reaction mixture. Calorimetry

2. CALORIMETRY The first law of thermodynamics says that energy is conserved in all processes. Therefore, the heat lost (q lost ) by a system equals the heat gained (q gained ) by the surroundings.The first law of thermodynamics says that energy is conserved in all processes. Therefore, the heat lost (q lost ) by a system equals the heat gained (q gained ) by the surroundings. q lost + q gained = 0 In this experiment,In this experiment,  Heat lost by reaction = Heat gained by water q is negative ifq is negative if heat is lost. q is positive ifq is positive if heat is gained. q is measuredq is measured in joules (J).

3. CALORIMETRY 4.184 joules (J) are needed to heat 1 gram (g) of water 1 degree Celsius(°C). This is the specific heat of water. Therefore, the heat gained by the water is4.184 joules (J) are needed to heat 1 gram (g) of water 1 degree Celsius(°C). This is the specific heat of water. Therefore, the heat gained by the water is q = Cm∆Tq = Cm∆T Heat for the reaction is then -q waterHeat for the reaction is then -q water

4. Experimentally Determine Specific Heat A hot piece of Aluminum that was 0.65g and 105.0°C was placed into a calorimeter containing 200.0 mL of 23.6°C water. The temperature of the water rose to 25.1°C. If the specific heat of water is 4.184 J/g°C, determine the specific heat of Aluminum. –Step One: Determine the amount of heat gained by the water –Step Two: This is the amount of heat lost by the metal –Use q to determine the specific heat (C) of the metal

5. Heat gained by the water (1g = 1mL) –q = CmΔT –q = 4.184 J/g°C x 200.0 g x (25.1-23.6°C) –q = 4.184 J/g°C x 200.0 g x 1.5°C –q = 1300 J Heat lost by the metal –q = -1300J Specific Heat of metal –C = q / mΔT* be careful, mass and temp. and different than in step 1 –C = -1300 J/ (0.65g x (25.1 -105.0°C)) –C = -1300J/ (0.65g x -79.9°C) –C = 25 J/ g°C