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AP Chem Lab 1.docx

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Department
Chemistry
Course
CHEM 1951
Professor
Dr.Acio
Semester
Fall

Description
Curt Kim, Jonathan Lee, Keshav Mantha & Bobbie Sheng, AP Chem Pd.1, 09/16/12 Title: Specific Heat of a Metal Goal: The purpose of this lab was to determine the specific heat of a metal sample by calculating the heat transfer from the metal sample to water inside a calorimeter. Procedure: First, a 250 mL beaker half-filled with water was set to boil. Meanwhile, the mass of the metal sample was measured on a scale which was zeroed with the sample’s cup. Then, the metal sample was placed into a test tube, and the tube was placed into the boiling water. The styrofoam cup calorimeter was then filled with 100mL of water, and the water’s temperature was taken. The test tube containing the metal was removed from the boiling water 15 minutes after the water started bubbling, at which point the water was very close to 100C. The metal sample was then poured into the calorimeter. The highest temperature of the water following the addition of the metal sample was recorded. The metal sample was then removed from the calorimeter and dried off. The volume of the metal sample was measured using water displacement and was used to calculate the density of the metal sample later on. Data: Trial 1 Trial 2 Mass of metal (g) 37.309 28.967 Mass of water (g) 100. 100. Temp. of boiling water (ºC) 100 (exact) 100 (exact) Temp. of water in cup before 23.3 22.6 metal added (ºC) Temp. of water in cup after 24.2 23.3 metal was added (ºC) Volume of water before metal 14.0 11.0 was added (mL) Volume of water after metal 17.0 13.7 was added (mL) Data Analysis: Heat gained by water = mass of water 4 184 g T Heat gained by water = Heat gained by metal eat gainedmetal Specific heat of metal = mass of metalT of metal Trial 1: Heat gained by water 100g 24 2-23 3 4 184 g 376 56 Heat gained by water = Heat lost by metal = 376.56J -376 56 Specific heat of metal 0 1331531273→ 0.133 /g* 37 309g 24 2 - 100 Trial 2: Heat gained by water 100g 23 3-22 6 4 184 g 292 88 Heat gained by water = Heat lost by metal = 292.88J -292 88 Specific heat of metal 28 967g 23 3 - 100 0 1318228912→ 0.132 /g* Average: Average of the two specific heats = (Trial 1 result + Trial 2 result) / 2 = (0.1331531273 + 0.1318228912)/2 = 0.1324880093 → 0.132 /g* Volume of metal = Volume of water and metal - Volume of water Density = mass/volume Trial 1: Volume of metal = 17.0mL - 14.0mL = 3.0mL = 3.0cm^3 Densit = 37.309g/3.0cm^3 = 12.43633333 → 12g/cm^3 Trial 2: Volume of metal = 13.7mL - 11.0mL = 2.7mL = 2.7cm^3 Densit = 28.967g/2.7cm^3 = 10.72851851 → 11g/cm^3 Average: Average of the two densities = (Trial 1’s densit + Trial 2’s densit )/2 = (12.43633333 + 10.72851851)/2 = 11.58242592 → 12g/cm^3 Discussion: This experiment was conducted in order to find the specific heat of a metal and identify the specific metal based on the specific heat of the metal. The basic theory that was proven through this experimental process was that the amount of heat lost is equivalent to the amount of heat gained. The calculations from the two trials show that because the amount of heat lost from the metal is equal to the amount of heat gained by the water at room temperature, and since the mass of the metal sample, change in temperature, specific heat of water (4.184 J/g*°C) are given, it is possible to calculate the specific heat of the metal sample. This specific heat of the metal sample, along with density, is then used to identify the metal sample. Heat Lost by Metalass of metal sample * change in Temperature( T)*cp metal sample
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