HEAT CAPACITY

SummaryThe experiment aimed at determining the value of specific heat capacity of tap water. The calorimeter was used to achieve the objective. It was mainly consist of a thermometer and a heating element. The vessel was used to hold water and avoid losing thermal energy. The formulaQ=mc ∆Twas used to obtain the value of specific heat capacity of water. Additionally, Q was found by the electric power that resistance wire transformed.W=P t=V2 RtTo make the value more accurate, experiment was conducted twice for the different volume of water. At last, the value was calculated by the equation∆Q=C∙(m b−m a)∙∆Tthat minus out the loss of heat of two experiment. The experimental value of specific heat capacity of tap was is 4933.3J kg∙k⁄Contents pageObjectiveDetermine the specific heat capacity of tap water by using a calorimeter.Introduction and TheoryThe whole experiment is based on the law of conservation of energy. Because it is impossible to observe the amount of heat transferred from resistance wire to the tap water, and the thermal energy the resistance transformed electrical energy into can be calculated. At last, the specific heat capacity of tap can be calculated by the equation which will be explained next. Additionally, Young and Geller (2007) suggest that the value of specific heat capacity of tap water should be close to 4.19×103J(kg∙k)⁄.Energy transfer that take place solely because of a temperature difference is called heat flow, or heat transfer, and energy transferred in this way is called heat (Young and Geller, 2007). The SI unit of quantity of heat is the joule(J). The symbol Q is used for quantity of heat. The quantity of heat Q increases the temperature T of a mass m of a certain material. The change in the temperature of a certain mass of substance from 10 ℃to 20 ℃(∆T=10 K) need twice as much as the amount needed to change it fro, 10 ℃to 15 ℃. When the temperature change and the material are the same, the heat that a substance of 20 g needed is twice as much as one of 10 g. So the definition of specific heat capacity can be obtained by putting all these things together. The amount of heat Q needed for a certain temperature change ∆T is proportional to the temperature change and to the mass m of substance being heated that is,Q=mc ∆T,where c is a quantity, different for different materials, called the specific heat capacity of the material (Young and Geller, 2007). When the increase in is 10 ℃, ∆T=10 K. The SI unit of specific heat capacity is the joule per kilogram per kelvin[J kg∙K⁄](Young and Geller, 2007).When current flows through a resistor, electrical energy is transformed into thermal energy. An electric toaster is an obvious example. The power dissipated through a pure resistance can be calculated. The electric power delivered by the circuit isW=P t=V2 Rtwhere the unit of V is voltage, R is ohm and W is JAs motioned before, the whole experiment is based on the law of conservation. The law of conservation of energy states that energy may neither be created nor destroyed. Therefore the sum of all the energies in the system is a constant. Therefore, the heat tap water obtained is equal to the thermal energy resistor transformed. So the thermal energy resistance transformed from electrical energy was calculated to obtain the heat that tap water received.So, when put the theories together, the value of specific heat capacity of can be calculated by the method shown below.C=Q m∙∆TQ=∆t∙V2 RThe actual heat flow to water can be written as Q=m∙C∙∆T+K∙∆T K∙∆T stands for the thermal energy loss by the vessel.Putting two data in the two experiments(consider K∙∆T are same)Q=C∙m∙∆TApparatusThermometerCalorimeterStirring RodLidVesselFigure 1●The thermometer was used to measure the increase in temperature oftap water●Stirring rod was used to make water heated evenly.●Lib fixed thermometer, stirring rod and resistance wire.●The vessel was used to hold water. There was supposed to be a outershell, so this caused the loss of heat.Graduated CylinderFigure 2●Graduated cylinder was used to amount the certain volume of waterStopwatchFigure 3●Stopwatch was used to time the heat heating process.MultimeterFigure 4●Multimeter was used to measure the resistance of resistance wireDC GeneratorFigure 5●DC generator provided the circuit with a constant voltage.Electronic ScaleFigure 6Electronic scale was used to find the exact mass of water.Procedure●Took a clean and dry vessel and weighted it.●Took 100mL of water by using a graduated cylinder and added the waterto the vessel.●Weighted the vessel and water again to find the exact mass(m) of thewater.●Placed the lid carefully on the vessel to make sure that the resistancewire is fully immersed in the water.●Placed the thermometer into the vessel and fixed it so that the tip of thebulb is about 1 cm from the bottom of the vessel.●When the temperature reading did not change, wrote the value down asthe temperature before heating.●Used the multimter to record the resistance across the heater terminals.●Adjusted the supply voltage and copied it down the value.●Connected the circuit elements.●Turned in the DC generator and started the stop watch at the same time.●Recorded the value of temperature every thirty seconds until reached1050 s●Voltage and resistance was recorded again by the end of experiment.●Measured 120 mL water and repeated the procedures above again.●Used the Excel to plot the graph whose y-axis stood for the Q and x-axisstood for T.Results and CalculationsTable 1 Element dataFigure 7 Temperature-heat(T-Q) graphFigure 8 Time-temperature graphTable 2 Experimental dataThe x-axis of figure 7 stands for the temperature of water, and y-axis stands for the thermal energy heat element produced.The y-axis of figure 8 shows the values of temperature of water, andx-axis is time.Data in table 1 shows the element parameters.Data in table 2 shows the exact changes in temperature.As motioned before in theory the specific heat capacity can be calculated by following steps.C=Q m∙∆TQ=∆t∙V2 RThe actual heat flow to water can be written as Q=m∙C∙∆T+K∙∆TK∙∆T stands for the t hermal energy loss by the vessel.Q a=m a∙C∙∆T+K∙∆TQ b=m b∙C∙∆T+K∙∆TPutting two data in experiment a and b (consider K∙∆T are same)∆Q=C∙(m b−m a)∙∆TIn the experiment a when temperature changed from 27℃to 29℃,∆T=3k. In the experiment b when temperature change from 25℃to 29℃, ∆T=2k. So, K∙∆T are same.Q a=m a∙C∙∆T+K∙∆T1065.6=0.099∙C∙2+2K 1.Q b=m b∙C∙∆T+K∙∆T1243.2=0.117∙C∙2+2K 2.Use equation 2. –equation 1., we can get177.6=0.036∙C C=4933.3J kg∙k⁄DiscussionThe experiment aimed at determining the value of specific heat capacity of tap water. It is can be seen from calculation section that the experimental value of specific heat capacity of water is 4933.3 J kg∙k⁄, which is 17.7% higher than the theoretical value. The result was obtained by a clever method which was to find the same temperature and minus out the loss of heat. But there are some uncertainly of component characters of the equipment used that caused the experimental errors.The first factor was the thermal energy loss from the vessel. The vessel was supposed to be placed in a shell, but there is no shell to use during the experiment. Additionally, the lid can not fix the thermometer very well. It was impossible to make the thermometer to be 1 cm from the bottom of the vessel. The stirring rod was not fixed very well, the rod may put thermometer very close the resistance wire making the temperature increase faster than it was expected. Besides, the plug of the lid was not totally closed when the thermometer was put in. The second factor was the evaporation from water surface. As temperature increased, more and more water was lost caused by evaporation, and it led the mass of water became smaller. The third factor was the increase in resistance made the value of Q was not accurate. The forth factor was the thermometer was not fixed very well and it touched the resistance wire sometime. The last factor was the vessel remained some thermal energy after the experiment a, this made the experiment b obtained extra heat.Seen from the result of the experiment, the experiment was well designed and operated. But there was a lot disadvantage of the element that was used during the experimentConclusionsThe experimental value of specific heat capacity of tap was is 4933.3J kg∙k⁄, which was 17.7% higher than the theoretical value 4190 J(kg∙k)⁄. Except for some obvious disadvantages of the element, in general, the experiment was well designed and operated. by a clever way which was to minus out the loss of thermal energy. Therefore, the experiment met the objectives and agreed with the expected theoretical valueReferenceYoung H.D. &Geller R.M. (2007) ‘ College Physics’, Young & Geller, 8th Eidtion , Person Education. Inc。