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### Lab19 Latent heat

Course: PHYSICS 1A+1B, Fall 2010
School: Laney College
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Heat Instructor: Dr. Lab3 Latent Mohebi Jiajie Huo(Jacky) Member: Ruihong Xiao TEJ Shiwakoti Anthony Hernandez Dana Saiez Rneiang Ye WAJIH RAHMAN 2. introduction. In this experiment, we are generally focusing on Measuring the latent heat of fusion for water. 1. Q=mc^T, where Q is the heat, m is the mass, C is the specific heat capacity, and ^T is the change in temperature. 2. In this experiment, the final...

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Heat Instructor: Dr. Lab3 Latent Mohebi Jiajie Huo(Jacky) Member: Ruihong Xiao TEJ Shiwakoti Anthony Hernandez Dana Saiez Rneiang Ye WAJIH RAHMAN 2. introduction. In this experiment, we are generally focusing on Measuring the latent heat of fusion for water. 1. Q=mc^T, where Q is the heat, m is the mass, C is the specific heat capacity, and ^T is the change in temperature. 2. In this experiment, the final equation that we would use is miCi(0-Ti)+miCw(Tf-0)+miL=mwCw(Tw-Tf)+mcCc(Tw-Tf), while mw refers to the Mass of the water, Cw refers to the specific heat of the water, mi refers to the mass of ice, Tf refers to the final temperature while the water reach the heat equilibrium, Tw refers to the temperature of water in normal atmosphere before the heating. Tf refers to the temperature to which the water had been heated. 3. Procedure.(additional: all temperature are in degree.) a. we found the mass of the calorimeter, so, mc=55.8g. b. We heated the water in a beaker to some degree above room temperature, , and the room temperature we measure is 22C. c. We transferred some of the heated water into the calorimeter and weigh the total mass of the container and the water, and the result is Tw=33C, mc+w=205.1g, so the mass of the water that had been transferred mw=149.3g. e. We recorded the temperature of the ice, so Ti=-1C. f. We added some ice to the calorimeter until the water reaches an equilibrium temperature and do not move any more, so Tf=11C. g. We weigh the calorimeter and mc+w+i=255.7g, so the mass of ice that had been added is mi=50.6g h. we calculated the latent heat of fusion of water and the above repeat procedures for three more times. 3. Date sheets, spread sheets. Measured Date Mw(g Tw(C ) Mi(g) ) Ti(C) Sample Trail Calculation Tf(C) Lf Water 1 149.3 50.6 33 -1 11 246 Water 2 137 41.9 37 -1 Water 3 181.9 50.8 38 -1 Water 4 133.8 43 35 -1 12 320 36 12 3 31 10 1 Lfo (Lf-Lfo)/Lfo 33 3 26% 33 3 4.00% 33 3 9% 33 3 6.70% Sample calculation: mw=149.3g, Cw=4186J/Kg*K, mc=55.8g,Cc=910J/Kg*K, mi=50.6g,Ti=-1C, Tw=33C, Tf=11C, and miCi(0-Ti)+miCw(Tf-0)+miLf=mwCw(Tw-Tf)+mcCc(Tw-Tf), So, Lf= [mwCw(Tw-Tf)+mcCc(Tw-Tf)- miCi(0-Ti)-miCw(Tf-0)]/mi =246J/Kg*K. As the standard heat of fusion for water is 333, So Error=(Lfo-Lf)/Lfo=(333-246)/333=26%. 6. Discussion and conclusion. In this experiment, we repeat four times and the final result are 246, 320,363,311, the error would be 26%,4%,9% and 6.7% respectively. During the whole experiment, there are many behaviors that would make the errors larger. First, the container can not completely rid the energy transfer from the outside, so there would be some energy lost during the processes. Second, at the first trail, we forget to place the container into the calorimeter after we place the ice into the water, so the energy transfer between the water and the outside atmosphere is much amplified. However, after that, we corrected the mistake and the error is much smaller. Third, after placing the ice into the water, we could not determine the time when its precisely in energy equilibrium, so the error is larger, above all, there are two trails that the final error are 4% and 6.7%, and it is within the acceptable range.
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