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Heat capacity of solids

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ITU Termodynamics Lab Report

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Heat capacity of solids

  1. 1. PHYSICS ENGINEERING DEPARTMENT FIZ341E - Statistical Physics and Thermodynamics Laboratory Name of Exp. : Heat Capacity of Solids Date of Exp. : 10.12.2014 BARIŞ ÇAKIR 090100235
  2. 2. Hocam ben Sercan hocanın odasına bırakmıştım raporumu ama tedbir olsun diye buraya da atıyorum. Introduction Heat Capacity: is required heat energy to increase temperature of a certain amount of matter for 1 o C. It can be calculated with the following formula; = In this formula, c is specific heat and M is mass also heat capacity is different for every matter. The theory of experiment comes from the energy conservation law of closed systems, all formulas and calculations, extracted from this principle; = According this principle, aim of the experiment is adding some materials to the calorimeter vessel, for producing heat transfer, measuring temperature change and calculating heat capacity values for each matter; however, calorimeter is also have a heat capacity and it can not be neglected. So at the beginning of the experiment, heat capacity of vessel is calculated with adding cold and hot water in it (just heat transfer between hot and cold water). = − − ( − ) Heat capacity can be extracted from formula above; = − − ( − ) − We also know released heat equals, multiplication of specific heat, mass, and temperature change so, = ( − ) Released heat equals, absorbed heat so if we plug in this definition to equation above; = − ( − )
  3. 3. Specific heat can extracted. Finally to reach specific heat capacity expression we need to add heat capacity of vessel, and our equation will be like, = + − ( − ) Specific heat is different for each matter so, by final equation, type of material which is used in experiment can be founded. Experimental Procedure Tools and devices: Calorimeter vessel, kettle, balance, water, thermometer, graphite, copper and porcelain. First, experiment started with calculating heat capacity of the calorimeter vessel, cold water added to vessel mass of vessel and water is calculated. We add hot water on cold water and equilibrium temperature is measured and heat capacity of vessel calculated with the equation which is given at information part. Second, specific heat of some solids calculated. We already got hot and cold water mix at equilibrium temperature from first part, with putting solids into this water, we calculated their mass and equilibrium temperature of solid and water. Notice that all solids are heated to 200 o C. With measured mass of solid and equilibrium temperature of solid water mix specific heat of solid can be calculated. This step is repeated for copper, porcelain and graphite. Finally, absolute errors calculated with comparing experimental and real values. Data Analysis The weight of vessel without water is, = 865.45 when we add the cold water, = + = 1002.75 cold water’s mass can be extracted as m, = 137.30 . After 10 minutes equilibrium temperature of cold water measured as,
  4. 4. = 19 we heated another water with kettle, = 97 and we pour it in the vessel, equilibrium temperature of hot and cold water mix is measured as, = 44 and also hot water’s mass measured by, extracting total mass of vessel with and without hot water, = 1094.10 = − = 91.35 . Heat capacity of the vessel can be calculated with all of these measurements and heat capacity formula which is given at introduction part; = 1 ∗ 91.35 ∗ (97 − 44) − 137.30 ∗ 1 ∗ (44 − 19) 44 − 19 = 56.362 / After finding heat capacity absolute error (∆C) and relative error (∆C/C) can be calculated; (∆ ) = 39,64 ∆ = 39,64 56.362 = 0,70 For, specific heat of solids we prepared a table for measurements, Porcelain Copper Graphite Mwater(g) 1094.15 1093.25 1080.1 T2 44 44 42 T1 200 200 200 Tf 45 45 43.5 Msolid(g) 31.1 25.35 9.55 c (joule/gC) 0.23867 0.3304 1.0402 Table1. Measurements with solids
  5. 5. T2 - initial temperature of vessel T1 - initial temperature of solid Tf - equilibrium temperature Msolid - mass of solid Mwater - initial mass of water Real specific heat values of solids are, = 0.82 / = 0.385 / = 0.26 / Conclusion We calculated all of specific heat capacities of used metals in the experiment, by calculating the temperature difference and calculating heat difference. According to error percentages we can discuss our result’s reliability. % = %26 % = %14.2 % = %8.9 Except graphite we get reliable results, the issue at graphite can be explained with number of graphite used in experiment, we put many graphite and calculated it as one big graphite this can cause some extra error rate. Also we did not calculate heat loss between air and solid, when we take solid out of the oven it starts to translate heat with air, we know temperature difference between air and solid is too high when we take solid from the oven, so there is a huge loss of heat in this phase. For a perfect result relative error should equal to 1, our relative error is 0.70 so we can say, we calculate heat capacity of calorimeter vessel successfully. Resources  Thermodynamics LAB FÖY  https://www.google.com.tr/search?q=heat+capacity+of+solids+experimen t&source=lnms&tbm=isch&sa=X&ei=T2uIVNiPHYWBU- Xlg5gO&ved=0CAgQ_AUoAQ&biw=826&bih=883  http://www.engineeringtoolbox.com/specific-heat-solids-d_154.html
  6. 6. Answers 1- In this condition T1 is initial temperature and T2 will be the equilibrium temperature, we can guess which elements used by calculating specific heat, and we can calculate how much energy taken from T1 to T2 with following formulas; = ( 2 − 1) = We know T2,T1 and heat capacity values so we can calculate how much energy taken from T1 to T2 with first formula and second formula will give which material we used. 2- In this question we can make assumption on same formula; = − = ( − ) The subject is about cooling times, so we can say, relation between final temperatures are, > So same heat given to both, they got same mass and initial temperature so, > This condition explains, toffee’s long cooling period.

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