# Expt 7 Physics Lab

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Expt 7 of Physics Lab
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Leader: Sollano, Frederiko Ma. Alessandro T. 2B - MT Member: Somo, John Carlo R. Date Performed: Oct 14, 2014 Sopoco, Marc Andrew M. Date Submitted: Oct 21, 2014 Sotong, Ma. Patricia Gabriel J. Tia, Queenie Rica Chen R. Venturina, Flian Lyra L. Expteriment # 7 Specific Heat of Aluminum Abstract: Specific heat is the amount of heat per unit of mass required to increase the temperature of an object by one degree Celsius. The specific heat differs depending on the object’s  composition. In this experiment, aluminum was computed for its specific heat using a calorimeter set-up. Utilizing a string tied to an aluminum block, it was placed in boiling water until it reached a certain temperature, which then was placed into cold water within a calorimeter set-up and was recorded of its temperature. Before this, the calorimeter, the aluminum cube, ice water was weighed, and measurement of temperature changes in the calorimeter system, and its components were noted. Upon receiving all the information previously mentioned, experimental value of the specific heat of the metal: aluminum was computed and was compared to its standard value. Question & Answer: 1. Define the ff: a. Heat  –  (In physics) heat is a form of energy defined by the movement of molecules and atoms of an object and is indicated by change of temperature. The higher the temperature, the faster the movements of the molecules are. b. Heat Capacity  –  The measurable physical quantity defined by the amount of heat required to raise an object’s temperature for a given amount. 2. Why is it desirable to have the water a few degrees colder than room temperature when the initial temperature is taken? It is desirable to have the water a few degrees colder than room temperature when initial temperature is taken because the temperature of the final calorimeter system, which also equals to final temperature of water, will be above the room temperature by the same amount; therefore having the initial temperature of water colder will allow an equal absorption of heat from surroundings, and releasing of heat after it exceeds room temperature. In effect, the experiment will yield a much more accurate calculation of specific heat.  3. Why is the mass of the outer shell of the calorimeter and the insulating ring is not included in the data for this experiment? The mass of the outer shell of the calorimeter and the insulating ring is not included in the data for this experiment because it is assumed that the inner shell is in a closed or isolated system. This means that no heat is assumed to exchange between the surroundings and the calorimeter, which in effect will help in the calculation of the specific heat of the material, as it needs specifically changes only happening between the calorimeter system and its components. 4. What does the experiment show about the specific heat of water? Based on the experiment, it was confirmed that water has a higher specific heat as it requires more energy or amount of heat absorbed to raise temperature compared to the aluminum block. 5. How does the conductivity of the metal used in this experiment affect the accuracy of the results? The conductivity of the metal used in this experiment determines the increase or decrease accuracy of results. This accuracy may be determined by several factors relating to the rate of conductivity of the metal. This includes the nature (element), the surface area (exposure to heat) and thickness. 6. Why should the hot metal be dry before it is introduced into the cold water? The hot metal should be dry before introducing into the cold water because the water clinging to the hot metal (right after removal from the boiling water) also contains heat and has higher amount of heat, which will yield a higher final temperature reading on the calorimeter system leading to huge errors in calculation for the specific heat of the metal. Problem: 837 calories of heat are required to heat 100.00 grams of copper from 0.0 °C to 100.0 °C. What is the specific heat of copper? Q  copper = (m copper )(c copper )(∆T)  837 cal = (100g)(C copper )(100°C - 0°C) 837 cal = (100 g)(100°C)(C copper ) 837 cal = 10,000 g*C°(C copper ) 837 cal / 10000 g * C° = C copper 0.0837 g * C° = C copper 0.08 g* C° ≈ C copper  0.08 g* C° is the specific heat of copper

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