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CUNY College of Staten Island *

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BIO-225

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Chemistry

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Jan 9, 2024

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pdf

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2

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Report
Date:________________ Name: ________________________________ Experiment #4: Specific Heat of a Metal Data and results Data table: Complete and submit the highlighted sections to your instructor before the end of the class session. Aluminum (Al) Your Metal Choice: _______ Mass of metal (g) Volume of water (mL) Mass of water (g) Initial temperature of water (°C) Initial temperature of metal (°C) Final temperature of water + metal (°C) Results table Aluminum (Al) Your Metal Choice Change in temperature (ΔT) of water Heat absorbed by water ΔT of metal Specific heat of metal Show all calculations for the Specific heat of Aluminum below: Jachimma Onwuamaegbu 09/24/2022 Bronze 6.7870g 19.979 100.0 mL 100.0 mL 98.8g 98.9 g 25.00 C 25.00 C 200 C 200 C 27.38 C 26.75 C 2.38 C 1.75 C 67.52 J 146.29 J 172.62 C 172.62 0.06 J/g C 0.42 J/g C Heat absorbed (Al): Q= m*c*change in T Q= 6.7870g* 4.184J/g degrees C* 2.38 C= 67.52 Heat absorbed (Bronze): Q= 19.979g * 4.184J/ g C *1.75 C Q = 146.2862 J Q= 146.29 Specific heat (Al)= C= Q/ (change in T * mass) C= -67.52 J/ (-172.62 C * 6.7870g) = 0.05763 J/g C = 0.06 J/g C Specific heat (Bronze)= C= Q/ (change in T * mass) C= -146.29 J/ (-172.62 C * 19.979g) = 0.4241 J/g C = 0.42 J/g C
Postlab Questions 1. Differences in equipment used for obtaining measurements in the lab can introduce experimental errors into calculated values. To minimize the effects of these experimental errors on our calculated values we conduct multiple trials of an experiment and obtain an average of the values. Calculate the percent error in the specific heat value that you determined experimentally. The accepted specific heat value for Aluminum is 0.900 J/g°C. % 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = 𝑦𝑦𝑒𝑒𝑦𝑦𝑒𝑒 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑒𝑒𝑒𝑒 − 𝑎𝑎𝑎𝑎𝑎𝑎𝑒𝑒𝑎𝑎𝑎𝑎𝑒𝑒𝑎𝑎 𝑣𝑣𝑎𝑎𝑣𝑣𝑦𝑦𝑒𝑒 𝑎𝑎𝑎𝑎𝑎𝑎𝑒𝑒𝑎𝑎𝑎𝑎𝑒𝑒𝑎𝑎 𝑣𝑣𝑎𝑎𝑣𝑣𝑦𝑦𝑒𝑒 � 𝑥𝑥 100% 2. When conducting this experiment, you are usually instructed to transfer the metal quickly from the oven or hot water bath to the calorimeter. Why would you need to do this? _______________________________________________________________________________ _______________________________________________________________________________ ________________________________________________________________________ __ _____ 3. A metal object with a mass of 1 45 .25 g is heated to 97.5 °C and then transferred to a calorimeter containing 84.58 g of water at 2 1 .5 °C. If the water and metal reach a final (equilibrium) temperature of 3 1 . 8 °C, what is the specific heat of the metal object? Show all work for full credit. You are usually instructed to transfer the metal quickly from the oven or hot water bath to the calorimeter because if you take a long time, the heat can be going out in the air. Basically losing the heat. % error = (0.06 - 0.900/ 0.900) * 100= -93.33% Heat absorbed: Q= m*c*change in T 145.25g * 4.184J/ g C * 10.3 C = 6,259.58 J C= Q/ (change in T * mass) C= 6,259.58 J / (65.7 C * 145.25g) = 0.6559 J/g C Specific heat of the object is 0. 7 J/g C
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