Specific Heat lab

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Question: Find a method to find the specific heat capacity of an unknown object. Hypothesis: Using the equation q=mc∆T, the heat capacity of the unknown object can be determined. Assuming heat loss by water = heat gained by the object, we can use two simultaneous equations to solve for the c value of the object. Variables: Controlled – time for water to cool, mass of metal, volume of water Dependent – change in temperature Materials: • Hot plate • The unknown object • Thermometer • Mass balance • Calorimeter • Water • Beaker Procedure: 1. Carefully measure 200 mL of water for the beaker and the calorimeter. 2. Note down the water in both containers 3. weigh the unknown object using the mass balance 4. Place the unknown object in the beaker with the water. 5. Heat the water in the beaker until it boils. 6. Remove the object from the beaker and put it into the calorimeter filled with water. 7. Cover the lid and place the thermometer inside to measure the temperature change. Data Collection: Trial #1: Mass of the unknown object Initial temperature T1 (◦C) ± 0.5 Final temperature T2 (◦C) ± 0.5 Calorimeter 200mL of H2O 20 32 Beaker 200mL of H2O 20 86 Trial #2: Mass of the unknown object Initial temperature T1 (◦C) ± 0.5 Final temperature T2 (◦C) ± 0.5 Calorimeter 200mL of H2O 20 29 Beaker 200mL of H2O 20 87 Data Processing: We know that the heat lost by the water is equal to the heat gained by the unknown object. Therefore mwcw∆Tw = mobjectcobject∆Tobject is true. Trial #1: mw = 200g cw = 4.18 kJ/kgK, ∆Tw = 97 – 20 = 77◦C mobject = 21.7g cobject = ? ∆Tobject = 32 – 20 = 12◦C Therefore, (200)(4.18)(77) = (21.7)( cobject )(12) cobject = t1 = water t2 = water + metal Conclusion In this experiment, the

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